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QUESTION 1.
Let W be the set of all nonnegative integers, and G the class of allW
subsets of W ( 2 ). For A e G let P(A) = 1 if A has a finite number of
elements and P(A) = 0 otherwise. Does P define a probability measure on
(W,G) ? Provide a detailed answer with the defintion of a probability
measure.
Solution question 1
Let ( W, G ) be a measurable space. A set function P(W) ( P: G -----L R )
defined on G, is called a probability measure ( or simply
probability) if it satisfies the following conditions:
1) P(A) > 0 for all A e G.
2) P(W) = 1.
3) For every { A , j e N}, A e G, j=1,2,3,..., a sequence of disjoint setsj j
( A nA =o if j $ k) we havej k
8& 8 *
P7 u A 8 = S P(A ) (= P(A )+P(A )+...).k=1 k 1 2
k=1
In our case W = set of all nonegative integers, so it is infinite.
By definition of P P(W) = 0 which is contradictory to the condition 2)
above, so P given is not a probability measure.
QUESTION 2.
Let A ,A ,... be a infinite sequence of events such that1 2
A C A C A C A C ... ( all A e G from fixed (W,G,P)).1 2 3 4 i
Prove the following
8v
P( A ) = lim P(A )u k n
n[-----L 8k=1
Solution question 2.
Let us define
B = A , B = A \A ,...,B = A \A ,...1 1 2 2 1 n n n-1
The sequence of events {B } is a sequence of disjoint events such thatn
8 8 n nv v v v
B = A , B = A = A . n = 1,2,3,...u k u k u k u k n
k=1 k=1 k=1 k=1
By the 3) (countable additivity) from the definition of a probability
measure we can write:
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8 8 n8 n
v v vP( A ) = P( B ) = S P(B ) = lim S P(B ) = lim P( B ) =
u k u k k k u kk=1 n[-----L 8 k=1 n[-----L 8
k=1 k=1 k=1
nv
= lim P( A ) = lim P(A ) which ends the proof.u k n
n[-----L 8 n[-----L 8k=1
QUESTION 3.
Formulate and prove the Bayes Rule.
Solution question 3.
Let (W,G,P) be a probability space.
Theorem (Bayes Rule). Let { H , n = 1,2,...} be a disjoint sequencen
vof events such that P(H ) > 0, n = 1,2,... and H = W. Let B e G with
n u n
nP(B) > 0 Then
P(H )WP(B/H )k k
(1) P(H /B) = ---------------------------------------------------------------------------k
8s
P(H )WP(B/H )t k kk=1
Proof. From the definition of a conditional probability
P(BnH )k
P(BnH ) = P(H )W----------------------------------- = P(H )WP(B/H )k k k k
P(H )k
and it imply that
P(H )WP(B/H )k k
P(H /B) = -----------------------------------------------------------------k
P(B)
vNote now that B = (BnH ) and sets (BnH ) are disjoint, the by 3) from
u n n
n
the definition of a probability measure (countable additivity) we can write
8 8 8P(BnH )
v s s n sP(B) = P( (BnH )) = P(BnH )= ----------------------------------- P(H )= P(H )P(B/H ).
u n t n t n t n nP(H )
nn n=1 n=1 n=1Hence, we get
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P(H )WP(B/H )k k
P(H /B) = ---------------------------------------------------------------------------k
8s
P(H )WP(B/H )t k kk=1
which ends the proof.
QUESTION 4
Let W = [0,1], and G be the Borel s-field of subsets of W. Define X
on W as follows: X(w) =w if 0 < w < 1. Verify that X is a random
variable. Find the event { w; X(w) e (1/4,1/2)}.
Solution question 4.
The function X mapping W into R is a random variable if and only
if for each x e R-1
a) X ( (-8,x]) = { w 1 X(w) < x } = { X < x} e G.
In our case-1
X ( (-8,x]) = o if x < 1-1
X ( (-8,x]) = [ 0,x] if 0 < x < 1-1
X ( (-8,x]) = [0,1] if x > 1
G - is a borel s field which contain all subintervals of [0,1]
and o therefore a) holds and X is a random variable.
{ w; X(w) e (1/4,1/2)} = (1/4,1/2)
QUESTION 5.
Suppose that the density function of a random variable X is as follows:& 2(9 - x )/36 for -3 < x < 3
f(x) = {{7 0 otherwise
Find: a) P(X < 0)b) The distribution function F(x)
c) Var(X).
Solution question 5.
Question 5.
2f(x) = 0 (9-x )/36 0
[-----------------------------------------------------------------------------------------------k------------------------------------------------------------k----------------------------------------------------------------------k--------------------------------------------------------------------------------------------------------------L
-3 0 30
0 -3 0 2 & 3 1 *P(X < 0) = i f(x)dx = i 0dx + i (9-x )/36dx = 1/36 9x - x /3 = 0.5
7 1 8-8 -8 -3 -3
xF(x) = P( X < x) = i f(t)dt
-8
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x xif x < -3 then F(x) = P( X < x) = i f(t)dt = i 0dt = 0
-8 -8x -3 x 2
if -3 < x < 3 then F(x) = P( X < x) = i f(t)dt = i 0dt + i (9-t )/36dtx -8 -8 -3
& 3 1 * 31/36 9t - t /3 = (9x - x /3+18)/36
7 1 8-3 8x -3 3 2 =if x > 3 then (x) = P( X < x) = i f t)dt = i 0dt + i ( -t )/36dt+ i0dt 1
-8 -8 -3 3&
0 if x < -32
3so F(x) = { (9x - x /3+18)/36 if -3 < x < 3
7 1 if x > 3
2 2 2Var(X) = E(X-EX) = E(X ) - (EX) = 324/180
8 8-3 3 2
E(X) = ixf(x)dx = i 0dx + i x(9-x )/36dx + i 0dx =-8 -8 -3 3
3 3 2 4 3= 1/36i 9x - x dx = 1/36( 9x - x /4 1 ) = 0
/2 -3-3
8 82 2 -3 3 2 2
E(X ) = ix f(x)dx = i 0dx + i x (9-x )/36dx + i 0dx =-8 -8 -3 3
3 2 4 3 5 3 324 324= 1/36i 9x - x dx = 1/36( 3x - x /5 1 ) = -------------------- = ---------------
-3-3 5W36 180
QUESTION 6.
The distribution of a random vector (X,Y) is given by
X \ Y 2 4 5[----------i---------------------------------------------------------------------------------------------------------]1 p 0.1 0.1 0.3
p2 p 0.1 0.09 0.01
p3 p 0.15 0.05 0.1
Find: a) P( X = 2) b) P(Y > 1) c) P(X < 2, Y < 4.5)‘ ‘
d) P( 2 2 < 2) e) F(x).X-Y% %
Solution question 6.
P(X = 2) = 0.1 + 0.09 + 0.01 = 0.2
P(Y > 1) = 1
P(X < 2, Y < 4.5) = 0.1 + 0.1 = 0.2‘ ‘
P(2 2 < 2) = 0.1 + 0.1 + 0.15 + 0.05 = 0.4X-Y% %
F(x) = P( X < x) = 0 if x < 1
= 0.1 + 0.1 + 0.3 = 0.5 if 1 < x < 2
= 0.1 + 0.1 + 0.3 + 0.1 + 0.09 + 0.01 = 0.7 if 2 < x < 3
= 1 if x > 3
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so& 0 if x < 1
F(x) = { 0.5 if 1 < x < 22 0.7 if 2 < x < 37 1 if x > 3
QUESTION 7.
Suppose that the joint density function for a random vector (X,Y) isgiven by:
2& c(x + y) for 0 < x <1, 0 < y < 1-x
f(x,y) = {7 0 otherwise
Find: a) The constant c.b) P( Y < X +1)
c) f(y)
Solution question 7.
2& c(x + y) for 0 < x <1, 0 < y < 1-x
f(x,y) = {7 0 otherwise
2D = { (x,y);0 < x < 1, 0 < y < 1-x }
A = { (x,y); y < x + 1} D C A
Since f(x,y) is a density function then
8 8i i f(x,y)dxdy = 1
-8 -8
8 8i i f(x,y)dxdy = i if(x,y)dxdy + i if(x,y)dxdy =
-8 -8 D cD
2 21 1-x 1 2 1-x
y 1=c i ix+y dydx = c i(xy + ---------- )dx
10 0 0 2 01
3 2 4= ci x-x +1/2-x +x /2 dx
012 4 3 5 1 ) = c31/60
= c( x /2 - x /4 +x/2 - x /3 +x /1010
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c31/60 = 1 therefore c = 60/31.
P(Y < X + 1 ) = P((X,Y) e A) > P((X,Y) e D) = 1
so P(Y < X + 1 ) = 1.
8f(y) = yi f(x,y)dx = 0 if y < 0 or y > 1
-8--------------- ---------------
8 r1-y r1-y2 1
f(y) = yi f(x,y)dx = 60/31i x+ydx = 60/30 (x /2 +xy )1
-8 00
( )2 260
= ----------21-y ---------------2 if 0 < y < 1--------------- + yr1-y2 231
9 2 0( )2 260
f(x) = ----------21-y ---------------2 for 0 < y < 1--------------- + yr1-y2 231
9 2 0and f(x) = 0 otherwise
QUESTION 8
Suppose that the joint density function for a random vector (X,Y) isgiven by:
-y& 2xe for 0 < x < 1 and 0 < y < 8
f(x,y) = {0 otherwise
7
Using marginal densities of X and Y verify the independence of X and Y ?
Solution question 8
1 1-y1 2xe 1p p
---------------------------------------------------------------------------i-----------------------------------------------------------------i---------------------------------------------p p0 1
8 8f(x) = i f(x,y)dy = i 0dy if x < 0 or x > 0
-8 -88 8 b
-y -y -y bf(x) = i f(x,y)dy = i 2xe dy = 2x lim ie dy = 2xlim ( -e 1 ) =
0-8 0 b[L8 0 b[L8
-b= 2xlim (-e +1) = 2x if 0 < x < 1
b[L8
& 2x for 0 < x < 1so f(x) = {
0 otherwise7
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8 8f(y) = i f(x,y)dx = i 0dx if y < 0
-8 -88 0 1 8
-y -y 2 1 -yf(y) = i f(x,y)dx = i 0dx + i2xe dx + i0dx = e (x 1 ) = e for y > 0
0-8 -8 0 1
-y& e for 0 < y < 8
so f(y) = {0 otherwise
7
-yTherefore f(x)Wf(y) = 2xe for 0 < x < 1, 0 < y < 8
= 0 otherwise
and f(x)Wf(y) = f(x,y) so X and Y are independent.
QUESTION 9.
Let (W,G) be a measurable space and P the probability measure
defined on (W,G). Prove the following statements:
a) A if A c B then P(B\A) = P(B) - P(A).A,BeG
b) A P(AuB) < P(A) + P(B).A,BeG
Solution question 9.
a) A c B +++++6 B = (B\A) u A and the sets on the right side are
disjoint, so P(B) = P(B\A) + P(A). Solving we get
P(B\A) = P(B) - P(A).
b) AuB = (A\(AnB) u (B\(AnB) u (AnB) and the sets on the right side
are disjoint and A = (A\(AnB) u (AnB), B = (B\(AnB) u (AnB)
hence
P(AuB) = P(A\(AnB) + P(B\(AnB) + P(AnB) =
= [P(A\(AnB) + P(AnB)] + [P(B\(AnB) + P(AnB)] - P(AnB) =
= P(A) + P(B) - P(AnB) < P(A) + P(B) since P(AnB) > 0.
QUESTION 10.
Prove the following theorem:
If the moments of order t exists for a random variable X of continuous
type with density function f(x), t > 0 then the moments of order s, 0 < s < t
exist.
Solution question 10
8s s
We have to show that E1X1 = i 1x1 f(x)dx < 8-8
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8t t
We know that E1X1 = i 1x1 f(x)dx < 8-8
Therefore
8 1 1 8s s s s s
E1X1 = i 1x1 f(x)dx = i1x1 f(x)dx + i1x1 f(x)dx + i1x1 f(x)dx <-8 -8 -1 1
1 1 8t t
i1x1 f(x)dx + i 1Wf(x)dx + i1x1 f(x)dx <-8 -1 1
1 1 8 1t t t t
i1x1 f(x)dx + i1x1 f(x)dx + i1x1 f(x)dx + i 1Wf(x)dx = E1X1 + P{1X1 < 1} < 8.-8 -1 1 -1
t sWe used that for 1x1 > 1 the following 1x1 > 1x1 and
sfor 1x1 < 1 the following 1x1 < 1.
QUESTION 11.
Let W be the set of all natural numbers, and G the class of allW
subsets of W ( 2 ).For A e G, let P(A) = 1 if A is a finite set
and P(A) = 0 otherwise. Does P define a probability measure on (W,G) ?
Solution question 11.
Definition: A set function P(W) ( P: G L R ) defined on G,
is called a probability measure ( or simply
probability) if it satisfies the following conditions:
1) P(A) > 0 for all A e G.
2) P(W) = 1.
3) For every {A , j e N }, A e G, j = 1,2,3,... , a sequence of disjointj j
sets ( A n A = o if j $ k) we havej k
88& *P u A = S P(A ).7 k8 kk=1 k=1
In our case P(A) = 0 or 1 so P(A) > 0 and 1) holds.
Since our W = N - infinite set so by definition of P P(W) = 0
and it contradicts 2).
Since second condition for the probability meaure is not satisfied
therefore P is not a probability measure.
QUESTION 12.
Suppose that the random variable X has the following probability density
function
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&c2x2 for -2 < x < 2
2f(x)={{
2 0 otherwise7
Find a) the constant c
b) P{ -0.5 < X < 1.5 }
c) The distribution function
Solution question 12
Density function have to satisfy:
a) f(x) > 0 for all x
8b) i f(x)dx = 1
-88 -2 0 2 8
from b) i f(x)dx = i 0dx + ic(-x)dx + icxdx + i0dx =-8 -8 -2 0 0
2 2x 2 0 x 22
= -c ---------- + c ---------- = 4c = 1 so c = 1/42 2
2 -2 2 0
If c = 1/4 a) holds too.
1.5 0 1.5 2 0 2 1.5x 2 x 2
P{ -0.5 < X < 1.5 } = i f(x)dx = i-x/4dx + ix/4dx = ----- ---------- + ---------- =2 2
-0.5 -0.5 0 8 -0.5 8 0
= 10/24
f(x) 0 -x/4 x/4 0------------------------------------------------------------------------------------------k------------------------------k------------------------------k-----------------------------------------------------------------
-2 0 2
xF(x) = if(t)dt
-8 x xif x < -2 then F(x) = if(t)dt = i0dt = 0
-8 -8x -2 x
2 2xi i i t 1 - xif -2 < x < 0 then F(x) = 2 f(t)dt = 2 0dt + 2 -t/4dt = - -------------------- 1 = -----------------------------------
j j j 8 8-2
-8 -8 -2x -2 0 xi i i i
if 0 < x < 2 then F(x) = 2 f(t)dt = 2 0dt + 2 -t/4dt + 2t/4dt =j j j j-8 -8 -2 0
2 0 2 x 2t 2 t 2 1 x
= ----- ---------- + ---------- = ----- + ----------2 2
8 -2 8 0 2 8
if x > 2 then F(x) = 1
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&0 if x < -2
2F(x) = { 2
1 x2 --------------- ----- --------------- if -2 < x < 0
2 822 2
1 x2 --------------- + --------------- if 0 < x < 2
2 822
1 if x > 27
QUESTION 13.
Let (X,Y) be a random vector with the joint density function given
(2 2 22 cx y for x < y < 1
by: f(x,y) = {2 0 otherwise29
Find: a) The constant c, b) P( X < 0.5,Y < 0.2 ), c) f(y)
Solution question 13
y2
y = x
y = 1
D
-1 1 x
2D = {(x,y); x < y < 1}
Since f(x,y) is a density function then
8 8i i2 2 f(x,y)dxdy = 1j j-8 -8
8 8i i i i i i2 2 f(x,y)dxdy = 2 2f(x,y)dxdy + 2 2f(x,y)dxdy =j j j j j j
c-8 -8 D D1 1 1i i 2 i 2 2 1
= 2 2 cx ydydx = c 2 (x y /21 )dx =j j j 2
x-1 2 -1
x1
1i 2 6 3 7 4= c/2 2 x - x dx = c/2 (x /3 - x /7)1 = c --------------------
j 21-1-1
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Therefore c = 21/4
y2
y = x
y = 1
q==========6-e0.2 xA
12
D = {(x,y); x < y < 1}q==========6 q==========6 2
A = {(x,y);-e0.2 < x < e0.2 ,x < y < 0.2}
i i i i 2P(X < 0.5,Y < 0.2) = 2 20dxdy + 2 2x ydxdy 21/4 =
j j j jA A1
q==========6 q==========6e0.2 0.2 e0.2
1i i 2 i 2 2= 2 2 x ydydx = 2 (x y /21 )dx =
j j j 2q==========6 q==========6 x2
-e0.2 x -e0.2-----q==========6 5 6
e0.2r0.2
i 2 4 3 5= 21/4 2 (0.02x - x /2)dx = 21/4 ( 0.02x /3 - x /101 )
j 5 6q==========6 _____----- r0.2-e0.2
= 0.043826
8i
f(y) = 2 f(x,y)dx = 0 if y < 0 or y > 1j-8
q6===== q6=====8 -ey ey 8i i i i
f(y) = 2 f(x,y)dx = 2 f(x,y)dx + 2 f(x,y)dx + 2 f(x,y)dx =j j j j
q6===== q6=====-8 -8-ey ey
q6=====eyi 2 5/2
= 21/4 2 x ydx = 7y /2jq6=====
-ey5/2
f(y) = 7y /2 if 0 < y <1
f(y) = 0 otherwise
QUESTION 14.
Let X be a random variable with density function given by
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&x/c for 0 < x < 4
f(x) = {7 0 otherwise
and Y be a random variable independent of X with density function
given by
& y if y e (0,1)f(y) = { 2-y if y e [1,2)
7 0 otherwise
Find a) c
b) f(x,y)
Solution question 14.
f(x) is a density if and only if
a) A f(x) > 0xeR8i
b) 2 f(x)dx = 1j-88 0 4 8i i 1 i i2 f(x)dx = 2 0dx + --------------- 2 xdx + 2 0dx =j j c j j-8 -8 0 4
2 41 x 8--------------- (--------------------1 ) = --------------- = 1c 2 c
0
Hence c = 8 and a) holds too.
&x/8 for 0 < x < 4
f(x) = {7 0 otherwise
Since X and Y are independent then
f(x,y) = f(x)f(y)
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y
0 0 00
A
0 2-y x(2-y)/8 0
B0y0 xy/8
f(x) = 0 x/8 4 0
0
0 0 0@
f(y)
(2 xy/8 if 0 < x < 4, 0 < y < 1222
f(x,y) = { x(2-y)/8 if 0 < x < 4, 1 < y < 22222 0 otherwise9
QUESTION 15.
The distribution of a random vector (X,Y) is given byX \ Y 2 4 5[----------i---------------------------------------------------------------------------------------------------------]1 p 0.1 0.1 0.2
p2 p 0.2 0.3 0.1
pFind: a) P( X = 2), b) P(Y > 1), c) P(X < 2/ Y < 4.5)Solution question 15.
a) P( X = 2) = P( (X,Y) e{(2,2),(2,4),(2,5)} = 0.2 + 0.3 + 0.1 = 0.6
b) P(Y > 1) = P((X,Y)e { (1,2),(1,4),(1,5),(2,2),(2,4),(2,5)}) = 1
c) P(X < 2/Y < 4.5) = P(X < 2, Y < 4.5)/ P( Y< 4.5) =
= P((X,Y) e{(1,2),(1,4)})/P( (X,Y) e {(1,2),(1,4),(2,2),(2,4)})
= 0.2/0.7 = 2/7
QUESTION 16.
Let (W,G,P) be a probability space.
a) Prove the following statement:
A, B e G, P(AuB) = P(A) + P(B) - P(AnB).
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1 1b) Consider two events C and D such that P(C) = ---------------, P(D) = ---------------.
4 2c
Determine the value P(CnD ) for each of the following conditions
1i) CnD = o ii) C C D iii) P(CnD) = ---------------
8
Solution question 16.
a) AuB = (A\(AnB) u (B\(AnB) u (AnB) and the sets on the right side
are disjoint and A = (A\(AnB) u (AnB), B = (B\(AnB) u (AnB)
hence
P(AuB) = P(A\(AnB) + P(B\(AnB) + P(AnB) =
= [P(A\(AnB) + P(AnB)] + [P(B\(AnB) + P(AnB)] - P(AnB) =
= P(A) + P(B) - P(AnB)
c c 1b) i) CnD = o 6 CnD = C so P(CnD ) = P(C) = ---------------
4c c
ii) CCD 6 CnD = o so P(CnD ) = P(o) = 0
c 1 1 1iii) P((CnD ) = P(C) - P(CnD) = --------------- - --------------- = ---------------
4 8 8
QUESTION 17.
Let W = [0,1], and G be the Borel s-field of subsets of W. Define
X on W as follows X(w) = w for w e W. Verify that X is a random
variable. Find the event {w; X(w) e (1/4,1/2)}
Solution question 17.
The function X mapping W into R is a random variable if
and only if for each x e R-1
1) X ((-8, x]) ={w; X(w) < x} = {X < x} e G.
In our case-1
X ((-8, x]) = o if x < 0-1
X ((-8, x]) = [0,x] if 0 < x < 1-1
X ((-8, x]) = [0,1] if x > 0
G - is a Borel s-field which contain all subintervals
of [0,1] and o therefore 1) holds and X is a random variable
QUESTION 18.
Suppose that the random variable X has the following probability density
function
&c2x2 for -2 < x < 2
2f(x)={{
2 0 otherwise7
Page 15
Find a) the constant c
b) P{ -0.5 < X < 1.5 }
c) The distribution function
Solution question 18.
The function X mapping W into R is a random variable if
and only if for each x e R-1
1) X ((-8, x]) ={w; X(w) < x} = {X < x} e G.
In our case-1
X ((-8, x]) = o if x < 0-1
X ((-8, x]) = [0,x] if 0 < x < 1-1
X ((-8, x]) = [0,1] if x > 0
G - is a Borel s-field which contain all subintervals
of [0,1] and o therefore 1) holds and X is a random variable
QUESTION 19.
Suppose that the random variable X has the following probability density
function
&c2x2 for -2 < x < 2
2f(x)={{
2 0 otherwise7
Find a) the constant c
b) P{ -0.5 < X < 1.5 }
c) The distribution function
Solution question 19.
Density function have to satisfy:
a) f(x) > 0 for all x
8b) i f(x)dx = 1
-88 -2 0 2 8
from b) i f(x)dx = i 0dx + ic(-x)dx + icxdx + i0dx =-8 -8 -2 0 0
2 2x 2 0 x 22 V
= -c ---------- + c ---------- = 4c = 1 so c = 1/42 2
2 -2 2 0
If c = 1/4 a) holds too.
1.5 0 1.5 2 0 2 1.5x 2 x 2
P{ -0.5 < X < 1.5 } = i f(x)dx = i-x/4dx + ix/4dx = ----- ---------- + ---------- =2 2
-0.5 -0.5 0 8 -0.5 8 0
= 10/24
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f(x) 0 -x/4 x/4 0------------------------------------------------------------------------------------------k------------------------------k------------------------------k-----------------------------------------------------------------
-2 0 2
F(x) = if(t)dt-8 x x
if x < -2 then F(x) = if(t)dt = i0dt = 0-8 -8
x -2 x2 2xi i i t 1 - x
if -2 < x < 0 then F(x) = 2 f(t)dt = 2 0dt + 2 -t/4dt = - -------------------- 1 = -----------------------------------j j j 8 8
-2-8 -8 -2x -2 0 xi i i i
if 0 < x < 2 then F(x) = 2 f(t)dt = 2 0dt + 2 -t/4dt + 2t/4dt =j j j j-8 -8 -2 0
2 0 2 x 2t 2 t 2 1 x
= ----- ---------- + ---------- = ----- + ----------2 2
8 -2 8 0 2 8
if x > 2 then F(x) = 1&
0 if x < -22
F(x) = { 21 x
2 --------------- ----- --------------- if -2 < x < 02 8
22 2
1 x2 --------------- + --------------- if 0 < x < 2
2 822
1 if x > 27
QUESTION 20.
Suppose that a random variable X is of a discrete type with the following
distribution given by:
P(X = k) = ck for k = 1,2,3,4,5 and P(X e {1,2,3,4,5} ) = 1.
Find : a) the value of c.
b) The distribution function F(x).
c) P( X < 3,5 / X > 0.5)
Solution question 20.
a) Since P(X e {1,2,3,4,5} ) = 1 then V = {1,2,3,4,5}is a set of all
values. Let us denote p = P(X = k), k = 1,2,3,4,5.k
{P } constitute an assignment of a probability if and only ifk
1) each p > 0k
2) S p = 1k
By 2) we have S p = c + 2c + 3c + 4c + 5c = 15c = 1 so c = 1/15.k
Page 17
If c = 1/15 then a) also holds so indeed c = 1/15 gives a probability
distribution.
b) F(x) = P(X < x)
if x < 1 then F(x) = P(X < x) = P(o) = 0
if 1 < x < 2 then F(x) = P(X < x) = P(X = 1) = 1/15
if 2 < x < 3 then F(x) = P(X < x) = P(X e {1,2}) = 3/15
if 3 < x < 4 then F(x) = P(X < x) = P(X e {1,2,3}) = 6/15
if 4 < x < 5 then F(x) = P(X < x) = P(X e {1,2,3,4}) =10/15
if 5 < x then F(x) = P(X < x) = P(X e {1,2,3,4,5}) = 1
Then & 0 if x < 12
1/15 if 1 < x < 22
F(x) = { 3/15 if 2 < x < 32
6/15 if 3 < x < 422 10/15 if 4 < x < 52
1 if 5 < x then7
P(X < 3.5, X > 0.5) P(X e {1,2,3})c) P(X < 3.5 / X > 0.5) = --------------------------------------------------------------------------------------------------------- = ----------------------------------------------------------------------------------------------- =
P(X > 0.5) P(X e {1,2,3,4,5})
= (6/15)/(1) = 6/15QUESTION 21.
The distribution of a random vector (X,Y) is given byX \ Y 2 4 5[----------i---------------------------------------------------------------------------------------------------------]1 p 0.1 0.1 0.2
p2 p 0.2 0.3 0.1
pFind: a) P( X = 2), b) P(Y > 1), c) P(X < 2/ Y < 4.5)
Solution question 21.
a) P( X = 2) = P( (X,Y) e{(2,2),(2,4),(2,5)} = 0.2 + 0.3 + 0.1 = 0.6
b) P(Y > 1) = P((X,Y)e { (1,2),(1,4),(1,5),(2,2),(2,4),(2,5)}) = 1
c) P(X < 2/Y < 4.5) = P(X < 2, Y < 4.5)/ P( Y< 4.5) =
= P((X,Y) e{(1,2),(1,4)})/P( (X,Y) e {(1,2),(1,4),(2,2),(2,4)})
= 0.2/0.7 = 2/7
sWe use the following P((X,Y)eA) = p
t i,ji,j;(x ,y )eA
i j
QUESTION 22.
Let W be the set of all natural numbers, and G the class of allW
subsets of W ( 2 ).For A e G, let P(A) = 1 if A is a finite set
and P(A) = 0 otherwise. Does P define a probability measure on (W,G) ?
Page 18
Solution question 22.
Definition: A set function P(W) ( P: G L R ) defined on G,
is called a probability measure ( or simply
probability) if it satisfies the following conditions:
1) P(A) > 0 for all A e G.
2) P(W) = 1.
3) For every {A , j e N }, A e G, j = 1,2,3,... , a sequence of disjointj j
sets ( A n A = o if j $ k) we havej k
88& *P u A = S P(A ).7 k8 kk=1 k=1
In our case P(A) = 0 or 1 so P(A) > 0 and 1) holds.
Since our W = N - infinite set so by definition of P P(W) = 0
and it contradicts 2).
Since second condition for the probability meaure is not satisfied
therefore P is not a probability measure.
QUESTION 23.
Let W be the set of all nonnegative integers and G the class of allW
subsets of W ( G = 2 ). For A e G let
s kP(A) = p(1-p) 0 < p < 1.
tkeA
Does P define a probability measure on (W,G) ?
Give detail answer with all definition required.
nn+1
s k 1 - tHint: t = -------------------------------------------------- ( t =/ 1)
t 1 - tk=0
Solution question 23.
Let ( W, G ) be a measurable space. A real set function P(W)
( P: G L R ) defined on G, is called a probability measure ( or simply
probability) if it satisfies the following conditions:
1) P(A) > 0 for all A e G.
2) P(W) = 1.
3) For every {A , j e N }, A e G, j = 1,2,3,... , a sequence of disjointj j
sets ( A n A = o if j $ k) we havej k
88& *P u A = S P(A ).7 k8 kk=1 k=1
Page 19
s kIn our case P(A) = p(1-p) 0 < p < 1
tkeA
and each term in the sum is > 0 so P(A) > 0 and 1) holds.
8 ns k s k
P(W) = P({0,1,2,3,...}) = p(1-P) = p lim (1-p) =t t
n-----L8k=0 k=0
n+11 - (1-p) 1
= p lim ---------------------------------------------------------------------- = p ------------------------- = 11 - (1-p) 1-1+p
n-----L8
so 2) holds.
Let {A , j e N } be as in 3).j
( )2 8 22 2
v s kP2 A 2 = p(1-p) = { series converges absolutely can be rearranged} =
u j t2 22j=1 2 k eUuA
j9 0
8s k s k s k
= p(1-p) + p(1-p) + ... + p(1-p) + ... = S P(A )t t t j
j=1keA keA keA
1 2 2
so 3) holds. Since 1), 2) and 3) are satisfied therefore P is a probability
measure.
QUESTION 24.
Let (W,G) be a measurable space and P the probability measure
defined on (W,G). Prove the following statements:
a) A if A c B then P(B\A) = P(B) - P(A).A,BeG
b) A P(AuB) < P(A) + P(B).A,BeG
Solution question 24.
a) A c B +++++6 B = (B\A) u A and the sets on the right side are
disjoint, so P(B) = P(B\A) + P(A). Solving we get
P(B\A) = P(B) - P(A).
b) AuB = (A\(AnB) u (B\(AnB) u (AnB) and the sets on the right side
are disjoint and A = (A\(AnB) u (AnB), B = (B\(AnB) u (AnB)
hence
P(AuB) = P(A\(AnB) + P(B\(AnB) + P(AnB) =
= [P(A\(AnB) + P(AnB)] + [P(B\(AnB) + P(AnB)] - P(AnB) =
= P(A) + P(B) - P(AnB) < P(A) + P(B) since P(AnB) > 0.
QUESTION 25.
Page 20
Let (W,G,P) be a probability space.
Prove the following statements:c
a) If A and B are independent events then A and B
are independent events.
b) If A and B are independent events and P(B) > 0 then
P(A/B) = P(A)
Solution question 25.
a) A and B are independent if P(AnB) = P(A)P(B)c c
We have to show that P(BnA ) = P(B)P(A ).c
B = (BnA) u (BnA ) and the sets on the right side are disjoint
hencec
P(B) = P(BnA) + P(BnA ), solving we getc
P(BnA ) = P(B) - P(BnA) = P(B) - P(B)P(A) =c
= P(B)[1 - P(A)] = P(B)P(A )
P(AnB) P(A)P(B)b) P(A/B) = ---------------------------------------- = -------------------------------------------------- = P(A).
P(B) P(B)
QUESTION 26.
Does the following function define distribution function(2 0 if x < 0
F(x) = {-x22 1 - e if x > 0.
9Solution question 26.
A real-valued function F defined on R ( (-8,8) ) that is
a) nondecreasing, b) right continuous and
c) lim F(x) = 0 d) lim F(x) = 1x-----L -8 x-----L8
is called a distribution function.
y = F(x)y
x
Our function is continuous so b) holds.
Page 21
if x < y < 0 then F(x) = 0 = F(y)-y
if x < 0 < y then 0 = F(x) and F(y) = 1 - e > 0
so F(x) < F(y)-x -y -x -y
if 0 < x < y then e > e so 1 - e < 1 - e
so F(x) < F(y)
Therefore if x < y +++++6 F(x) < F(y) so F(x) is nondecreasing
function and a) holds
lim F(x) = lim 0 = 0 so c) holdsxL -8 xL -8
-xlim F(x) = lim (1 - e ) = 1 so d) holds.
xL 8 xL 8
Since all conditions are satisfied therefore F(x) is
a distribution function.
QUESTION 27.
Suppose that a joint density function for a random vector (X,Y) is given by(2 c(x+y) for 0 < x < y < 1
f(x,y) = {2 0 otherwise9
Find: a) The constant c.
b) F(y)
c) P( Y < X + 0.5)
Solution question 27.
8 8i i
f(x,y) - joint density function 46 a) 2 2 f(x,y)dxdy = 1j j-8 -8
b) A f(x,y) > 0.x,y
Page 22
D = { (x,y); 0 < y < 1, 0 < x < y}
8 8i i i i i i2 2 f(x,y)dxdy = 2 2f(x,y)dxdy + 2 2f(x,y)dxdy =j j j j j j
c-8 -8 D D1 y 1
2( ) ( y )i i i x= 2 {2 2 c(x,y)dx }2dy = c 2 {2 -------------------- + xy 1 }2 dy =
j j j 29 0 9 000 0 0
1 12 2 3 1i y 2 i 3y y c
= c 2 -------------------- + y dy = c 2 ------------------------- dy = c --------------------1 = --------------- = 1j 2 j 2 2 200 0
so c = 2.yi
F(y) = P( Y < y) = 2 f (t)dtj 2-8
where8i
f (y) = 2 f(x,y)dx2 j
-8
Page 23
8 0 y 8i i i i
y e (0,1) +++++6 f (y) = 2 f(x,y)dx = 2 0 dx + 22(x+y)dx + 20 dx2 j j j j
-8 -8 0 y
2 yx 2= 2 ( -------------------- + xy 1 ) = 3y
20
8i
y e/ (0,1) +++++6 f (y) = 2 0 dx = 02 j
-8(2 2
3y if 0 < y < 1f (y) = {2
22 0 otherwise9yi
F(y) = 2 f (t)dtj 2-8
yi
y < 0 +++++6 F(y) = 2 0dt = 0j-8
0 yi i 2 3 y 3
0 < y < 1 +++++6 F(y) = 2 0dt + 23t dt = t 1 = yj j 0-8 0
0 1 yi i 2 i 3 1
y > 1 +++++6 F(y) = 2 0dt + + 23t dt + 20dt = t 1 = 1j j j 0-8 0 1
Page 24
(2 0 if y < 022 3
F(y) = { y if 0 < y < 12222 1 if y > 19
P(Y < X + 0.5) = P((X,Y) e C) + P((X,Y) e C ) =1
i i i i= 2 2f(x,y)dxdy + 2 2f(x,y)dxdy =
j j j jC C
1
C C1
C = {(x,y); 0 < y < 0.5,0 < x < y } u {(x,y); 0.5 < y < 1,y - 0.5 < x < y}
i i i i2 22(x+y)dxdy + 2 20dxdy =j j j jC C
10.5 y 1 yi i i i
= 2 22(x+y)dxdy + 2 2 2(x+y)dxdy =j j j j0 0 0.5 y-0.50.5 1
y yi 2 i 2= 2 (x +xy1 )dy + 2 (x +xy1 )dy =
j j0 y-0.5
0 0.50.5 1i 2 i 2 2
= 2 3y dy + 2 3y - (y-0.5) - 2(y-0.5)ydy =j j0 0.5
0.5 10.5 1i 2 i 3 2
= 2 3y dy + 2 2y -0.25dy = y 1 + (y - 0.25y)1 = 0.75.j j 0 0.50 0.5
Page 25
P(Y < X +0.25) = 0.75.
QUESTION 28.
Let X,Y be independent identically distributed random variables
with common density function(2 -z2 e if z > 0
f(z) = {222 0 otherwise9
Find the distribution function of M = max{X,Y}.
Solution question 28.
X and Y are independent so for all x,y P(X < x,Y< y) = P(X < x)P(Y < y)
and also F(x,y) = F(x)F(y)
Distribution function for X and Y is the same and is equal to
z zi i
F(z) = P(X < z) = P( Y < z) = 2 f(t)dt = 2 0dt = 0 if z < 0j j
-8 -80 zi i -t -t z -z
= 2 0dt + 2e dt = - e 1 = 1 - e if z > 0j j
0-8 0
(2 0 if x < 02
F (x) = P( max{X,Y} < x) = P(X < x,Y < x) = F(x)F(x) ={M 2 -x 222 (1 - e ) if x > 1
9QUESTION 29.
Let (X,Y) be a random vector with joint density given by(2 22 3x
2xy + ------------------------------ if 0 < x < 1, 0 < y < 1f(x,y) = { 2
222 0 otherwise9
Find the line of regression of Y on X
Solution question 29.
cov(X,Y)y - EY = ----------------------------------------(x - E)
Var(X)
I1j---------------------------------------------o1 11 2 1
3x12xy+--------------- 1 D = {(x,y);0<x<1,0<y<1}
21 1---------------------------------------------k---------------------------------------------k---------------------------------------------L
1 1
Page 26
8i
f(x) = 2 f(x,y)dy = 0 if x e/ (0,1)j
-88 0 1 8
2 2i i i 3x i 2 3yx 1
f(x) = 2 f(x,y)dy = 2 0dy + 2 2xy+---------------dy + 20dy = xy + --------------------1 =j j j 2 j 2
0-8 -8 0 1
23x
= x + --------------- if x e (0,1)2
8 0 1 82
i i i 3x iEX = 2 xf(x)dx = 2 0dx + 2x(x + --------------- )dx + 20dx =
j j j 2 j-8 -8 0 1
3 4 1 17= x /3 + 3x /8 1 = -------------------- = 0.7083
2408 0 1 8
22 i 2 i i 2 3x i
EX = 2 x f(x)dx = 2 0dx + 2x (x + --------------- )dx + 20dx =j j j 2 j
-8 -8 0 1
4 5 1 11= x /4 + 3x /10 1 = --------------------
200
2 2 11 17 2Var(X) = EX - (EX) = -------------------- - (--------------------) = 0.048
20 24
8i
f(y) = 2 f(x,y)dx = 0 if y e/ (0,1)j
-88 0 1 8
2i i i 3x i
f(y) = 2 f(x,y)dx = 2 0dx + 2 2xy+---------------dx + 20dx =j j j 2 j
-8 -8 0 12 3 1
= x y + x /2 1 = y + 1/2 if y e (0,1)0
8 0 1 8i i i 2 y i 3 2 1 7
EY = 2 yf(y)dy = 2 0dy + 2y + ----- dy + 20dy = y /3 + y /4 1 = -------------------- = 0.5833j j j 2 j 120
-8 -8 0 1
cov(X,Y) = E(X - EX)(Y - EY) = EXY - EXEY
8 8 11 12 3 4
i i ii 3x i 2 x 3yx 1EXY = 2 2 xyf(x,y)dxdy = 22xy(2xy+---------------)dxdy = 2(2y -------------------- + -------------------- 1 )dy =
j j jj 2 j 3 8 0-8 -8 00 01i 2 3 2 1
= 2 2y /3 + 3y/8 dy = 2y /9 - 3y /16 1 = 0.4097j 00
cov(X,Y) = E(X - EX)(Y - EY) = EXY - EXEY = 0.4097 - 0.7083Q0.5833 = -0.0034
cov(X,Y)y - EY = ----------------------------------------(x - E)
Var(X)
Page 27
y - 0.5833 = -0.0034/0.048(x - 0.7083)
y = -0.0708x +0.6334
QUESTION 30.
Let (W,G) be a measurable space and P the probability measure
defined on (W,G). Prove the following statements:
1) A, B e G, P(AuB) = P(A) + P(B) - P(AnB)c
2) A, B e G, P(AnB ) = P(A) - P(AnB)
Solution question 30
1) AuB = (A\(AnB) u (B\(AnB) u (AnB) and the sets on the right side
are disjoint and A = (A\(AnB) u (AnB), B = (B\(AnB) u (AnB)
hence
P(AuB) = P(A\(AnB) + P(B\(AnB) + P(AnB) =
= [P(A\(AnB) + P(AnB)] + [P(B\(AnB) + P(AnB)] - P(AnB) =
= P(A) + P(B) - P(AnB) < P(A) + P(B)c
2) A = (AnB)u(AnB ) and the sets on the right side are disjoint thereforec
P(A) = P(AnB) + P(AnB ) solving we getc
P(AnB ) = P(A) - P(AnB)
QUESTION 31.
Let X be an r.v. Is Z =1X1 a random variable ?
Solution question 31.
Let (W,G,P) be a probability space.
Definition: A real function X : W -------------------------L R is called a random variable (r.v.)
if the inverse images under X of all Borel sets in R are events, that is,-1
1) X (B) = { w; X(w) e B } e G for all B e B(R)
This condition is equivalent to the following:
X is an r.v. if and only if for each x e R-1
2) X ((-8,x]) = { w; X(w) < x} = { X < x } e G
We have to show that-1
Z ((-8.x]) e G .
We have-1
Z ((-8.x]) = { w;Z(w) < x} = { w; 1X(w)1 < x } = {w; -x < X(w) < x } =-1
X ( [-x,x]) J-------------------- this set is in G since interval [-x,x]
is a borel set and 1) holds (because X is a random variable).
Therefore Z = 1X1 is a random variable.
QUESTION 32.
Suppose that the density function of a random variable X is
Page 28
as follows:(2 22 (9 - x )/36 for -3 < x < 3
f(x) = {22 0 otherwise9
Find: a) P(X < 0)
b) The distribution function F(x)
c) P( X > -0.5/ X < 2)
Solution question 32.
2f(x) = 0 (9-x )/36 0
[-----------------------------------------------------------------------------------------------k------------------------------------------------------------k----------------------------------------------------------------------k--------------------------------------------------------------------------------------------------------------L
-3 0 3
00 -3 0 2 & 3 1 * 1
P(X < 0) = i f(x)dx = i 0dx + i (9-x )/36dx = 1/36 9x - x /3 = --------------------7 1 8 2
-8 -8 -3 -3
xF(x) = P( X < x) = i f(t)dt
-8x x
if x < -3 then F(x) = P( X < x) = i f(t)dt = i 0dt = 0-8 -8
x -3 x 2if -3 < x < 3 then F(x) = P( X < x) = i f(t)dt = i 0dt + i (9-t )/36dt
x -8 -8 -3& 3 1 * 3
1/36 9t - t /3 = (9x - x /3+18)/367 1 8
-3 8x -3 3 2
if x > 3 then F(x) = P( X < x) = i f(t)dt = i 0dt + i (9-t )/36dt+ i0dt=1-8 -8 -3 3
&0 if x < -3
23
so F(x) = { (9x - x /3+18)/36 if -3 < x < 37 1 if x > 3
P(-0.5 < X < 2 ) 0.549P( X > -0.5 / X < 2) = -------------------------------------------------------------------------------- = ------------------------- = 0.593
P( X < 2 ) 0.9252 2
2 3P(-0.5 < X < 2) = i (9 - x )/36 dx = 1/36 ( 9x - x /3)1 = 0.549
-0.5-0.5
P( X < 2) = P( X < 2) = F(2) = 0.925
QUESTION 33.
Suppose that a random variable X is of a discrete type with the following
distribution given by:
P(X = k) = ck for k = 1,2,3,4,5 and P(X e {1,2,3,4,5} ) = 1.
Find : a) the value of c.
Page 29
b) The distribution function F(x).
c) P( X < 3,5 / X > 0.5)
Solution question 33.
a) Since P(X e {1,2,3,4,5} ) = 1 then V = {1,2,3,4,5}is a set of all
values. Let us denote p = P(X = k), k = 1,2,3,4,5.k
{P } constitute an assignment of a probability if and only ifk
1) each p > 0k
2) S p = 1k
By 2) we have S p = c + 2c + 3c + 4c + 5c = 15c = 1 so c = 1/15.k
If c = 1/15 then a) also holds so indeed c = 1/15 gives a probability
distribution.
b) F(x) = P(X < x)
if x < 1 then F(x) = P(X < x) = P(o) = 0
if 1 < x < 2 then F(x) = P(X < x) = P(X = 1) = 1/15
if 2 < x < 3 then F(x) = P(X < x) = P(X e {1,2}) = 3/15
if 3 < x < 4 then F(x) = P(X < x) = P(X e {1,2,3}) = 6/15
if 4 < x < 5 then F(x) = P(X < x) = P(X e {1,2,3,4}) =10/15
if 5 < x then F(x) = P(X < x) = P(X e {1,2,3,4,5}) = 1
Then & 0 if x < 12
1/15 if 1 < x < 22
F(x) = { 3/15 if 2 < x < 32
6/15 if 3 < x < 422 10/15 if 4 < x < 52
1 if 5 < x then7
P(X < 3.5, X > 0.5) P(X e {1,2,3})c) P(X < 3.5 / X > 0.5) = --------------------------------------------------------------------------------------------------------- = ----------------------------------------------------------------------------------------------- =
P(X > 0.5) P(X e {1,2,3,4,5})
= (6/15)/(1) = 6/15QUESTION 34.
The distribution of a random vector (X,Y) is given by
X \ Y 2 4 5[----------i---------------------------------------------------------------------------------------------------------]1 p 0.1 0.1 0.2
p2 p 0.2 0.3 0.1
p
Find: a) P( X = 2)
b) P(Y > 1)
c) P(X < 2/ Y < 4.5)
Solution question 34.
a) P( X = 2) = P( (X,Y) e{(2,2),(2,4),(2,5)} = 0.2 + 0.3 + 0.1 = 0.6
Page 30
b) P(Y > 1) = P((X,Y)e { (1,2),(1,4),(1,5),(2,2),(2,4),(2,5)}) = 1
c) P(X < 2/Y < 4.5) = P(X < 2, Y < 4.5)/ P( Y< 4.5) =
= P((X,Y) e{(1,2),(1,4)})/P( (X,Y) e {(1,2),(1,4),(2,2),(2,4)})
= 0.2/0.7 = 2/7
QUESTION 35.
Let (W,G,P) be a probability space.
a) Prove the following statement:
A, B e G, P(AuB) = P(A) + P(B) - P(AnB).
1 1b) Consider two events C and D such that P(C) = ---------------, P(D) = ---------------.
4 2c
Determine the value P(CnD ) for each of the following conditions
1i) CnD = o ii) C C D iii) P(CnD) = ---------------
8
Solution question 35.
a) AuB = (A\(AnB) u (B\(AnB) u (AnB) and the sets on the right side
are disjoint and A = (A\(AnB) u (AnB), B = (B\(AnB) u (AnB)
hence
P(AuB) = P(A\(AnB) + P(B\(AnB) + P(AnB) =
= [P(A\(AnB) + P(AnB)] + [P(B\(AnB) + P(AnB)] - P(AnB) =
= P(A) + P(B) - P(AnB)
c c 1b) i) CnD = o 6 CnD = C so P(CnD ) = P(C) = ---------------
4c c
ii) CCD 6 CnD = o so P(CnD ) = P(o) = 0
c 1 1 1iii) P((CnD ) = P(C) - P(CnD) = --------------- - --------------- = ---------------
4 8 8
QUESTION 36.
Let (W,G,P) be a probability space. A , B e G such that P(A) > 0,
P(B) > 0 and AnB = o. Define X(w) = -I (w) + 2I (w)A B
& &1 if w e A 1 if w e B
where I (w) = { and I (w) = { .A B
7 0 if w m A 7 0 if w m B
Verify that X is a random variable. Find the event { w; X(w) e (-2,1)}.
Solution question 36.
The function X mapping W into R is a random variable if and only
if for each x e R-1
a) X ( (-8,x]) = { w 1 X(w) < x } = { X < x} e G.
X(w) = -I (w) + 2I (w)A B
I (w) = 0 ! 0 ! 0 ! 1 ! 0B
Page 31
I (w) = 0 ! 1 ! 0 ! 0 ! 0A
-----------------------------------------------------------------k--------------------------------------------------k-------------------------k-----------------------------------k-------------------------------------------------------A B (A and B disjoint)
Therefore(2 -1 if w e A2 c
X(w) = { 0 if w e (AuB)22 2 if w e B9
In our case-1
X ( (-8,x]) = o if x < -1-1
X ( (-8,x]) = A if -1 < x < 0-1 c
X ( (-8,x]) = A u (AuB) if 0 < x < 2-1
X ( (-8,x]) = W if x > 2
Since G is a s-field so o e G and by our assumption A,B eG soc
AuB e G and (AuB) , therefore a) holds and X is a random variable.c
{ w; X(w) e (-2,1)} = {w: X(w) = -1 or 0} = A u (AuB)
QUESTION 37.
Prove the following theorem:
Let X be a random variable on a probability space (W,G,P) of thek
continuous type with the density function f(x). Let E1X1 < 8 for
some k > 0. Thenk
n P(1X1 >n) ----------L 0 as n ----------L 8.
Solution question 37.
We have
8 nk k
8 > i 1x1 f(x)dx = lim i 1x1 f(x)dx .-8 n[-----L 8 -n
It follows that-n 8
& k k *lim i 1x1 f(x)dx + i 1x1 f(x)dx = 0
7 8n[-----L 8 -8 n
But-n 8 -n 8
k k k ki1x1 f(x)dx + i1x1 f(x)dx > i n f(x)dx + i n f(x)dx
-8 n -8 n-n 8 -n 8
k k k k> i n f(x)dx + i n f(x)dx = n ( i f(x)dx + i f(x)dx) = n P{1X1 > n}
-8 n -8 ncompleting the proof.
QUESTION 38.
Does the following function define distribution function
Page 32
(2 0 if x < 0
F(x) = {-x22 1 - e if x > 0.
9If yes find: EX, P(-1 < X < 4)
Solution question 38.
A real-valued function F defined on R ( (-8,8) ) that is
a) nondecreasing, b) right continuous and
c) lim F(x) = 0 d) lim F(x) = 1x-----L -8 x-----L8
is called a distribution function.
y y = F(x)
x
Our function is continuous so b) holds.
if x < y < 0 then F(x) = 0 = F(y)-y
if x < 0 < y then 0 = F(x) and F(y) = 1 - e > 0
so F(x) < F(y)-x -y -x -y
if 0 < x < y then e > e so 1 - e < 1 - e
so F(x) < F(y)
Therefore if x < y +++++6 F(x) < F(y) so F(x) is nondecreasing
function and a) holds-x
Or since F’(x) = e or 0 so is > 0 therefore F(x) is nondecreasing
and a) holds.
lim F(x) = lim 0 = 0 so c) holdsxL -8 xL -8
-xlim F(x) = lim (1 - e ) = 1 so d) holds.
xL 8 xL 8
Since all conditions are satisfied therefore F(x) is
a distribution function.
Page 33
8i
EX = 2 xf(x)dxj-8
(2 0 if x < 0
f(x) = F’(x) = {-x22 e if x > 0.
98 0 8i i i -x
EX = 2 xf(x)dx = 2 0dx + 2xe dx =j j j-8 -8 0
q===============================================================================================eb b2 -x 2
2v’ = e u = x 2 bi -x -x i -x= lim 2xe dx = 2 2 = lim [-xe 1 + 2e dx]
j jbL8 2 -x 2 bL8 02v = -e 20 2 u’= 0 2 0z===============================================================================================c
b-b -x -b -b= lim [-be + (-e 1 )] = lim [-be -e + 1] = 1
bL8 bL80
-4P(-1 < X < 4) = F(4) - F(-1) = 1 - e
QUESTION 39.
Suppose that the density function of a random variable X is
as follows:(2 x--------------- if 0 < x < 4
cf(x) = {22 0 otherwise9
Find: a) c2
b) distribution of Y = X
Solution question 39.
a) f(x) is a density function if and only if f(x) > 0 for all x
8i
and 2 f(x)dx = 1j-8
8 0 4 82 4i i i x i 1 x
2 f(x)dx = 2 0dx + 2--------------- dx + 2 0dx = --------------- [ -------------------- 1 ] =j j j c j c 2
0-8 -8 0 4
8= --------------- = 1 so c = 8.
c(2 x--------------- if 0 < x < 4
8f(x) = {22 0 otherwise9
b) f(x) > 0 if x e (0,4) , so (0,4) - set of all values taken by X.
Page 34
2Y = g(X) = X .g’(x) = 2x > 0 for x e (0,4).
Therefore Y takes values from (g(0), g(4)) = (0,16)q6===== q6=====2 -1
y = x so x = ey for x e (0,4) so g (y) = ey.
Y is a continuous random variable with density function given by:
-1( -1 dg (y)2 f(g (y))1----------------------------------1 if y e (0,16)
dyh(y) = {2 0 otherwise9
-1dg (y) 1
------------------------------------------------ = ---------------q6=====dy2ey
(2 62 ry 12 --------------- -------------------- if y e (0,16)6h(y) = { 8 2ry222 0 otherwise9
( 12 --------------- if 0 < y < 162 16
Then h(y) = {22 0 otherwise9
QUESTION 40.
Formulate and prove the Chebychev’s inequality.
Solution question 40.2
Chebychev’s inequality If EX = m, var(X) = s < 8, then for
any e > 0
2s
P(1X - m1 > e ) < ---------2e
Proof.2 2 2s = var(X) = E(X - m) = E((X - m) I ) =
W2 2
E((X - m) (I + I )) > E((X-m) I ))[w;1X-m1<e] [w;1X-m1>e] [w;1X-m1>e]
2 2> Ee I = e P(1X-m1 > e).
[w;1X-e1>e]
Hence
2s
P(1X - m1 > e) < -------------------2e
QUESTION 41.
Suppose that a joint density function for a random vector (X,Y)
is given by:
Page 35
(2 c(x+y) for 0 < x < y < 1
f(x,y) = {2 0 otherwise9
Find: a) The constant c.
b) F(y)
c) P( Y < X + 0.5)
Solution question 41.
8 8i i
f(x,y) - joint density function 46 a) 2 2 f(x,y)dxdy = 1j j-8 -8
b) A f(x,y) > 0.x,y
D = { (x,y); 0 < y < 1, 0 < x < y}
8 8i i i i i i2 2 f(x,y)dxdy = 2 2f(x,y)dxdy + 2 2f(x,y)dxdy =j j j j j j
c-8 -8 D D1 y 1
2( ) ( y )i i i x= 2 {2 2 c(x,y)dx }2dy = c 2 {2 -------------------- + xy 1 }2 dy =
j j j 29 0 9 000 0 0
1 12 2 3 1i y 2 i 3y y c
= c 2 -------------------- + y dy = c 2 ------------------------- dy = c --------------------1 = --------------- = 1j 2 j 2 2 200 0
so c = 2.
yi
F(y) = P( Y < y) = 2 f (t)dtj 2-8
Page 36
where8i
f (y) = 2 f(x,y)dx2 j
-8
8 0 y 8i i i i
y e (0,1) +++++6 f (y) = 2 f(x,y)dx = 2 0 dx + 22(x+y)dx + 20 dx2 j j j j
-8 -8 0 y
2 yx 2= 2 ( -------------------- + xy 1 ) = 3y
20
8i
y e/ (0,1) +++++6 f (y) = 2 0 dx = 02 j
-8
(2 2
3y if 0 < y < 1f (y) = {2
22 0 otherwise9
yi
F(y) = 2 f (t)dtj 2-8
yi
y < 0 +++++6 F(y) = 2 0dt = 0j-8
Page 37
0 yi i 2 3 y 3
0 < y < 1 +++++6 F(y) = 2 0dt + 23t dt = t 1 = yj j 0-8 0
0 1 yi i 2 i 3 1
y > 1 +++++6 F(y) = 2 0dt + + 23t dt + 20dt = t 1 = 1j j j 0-8 0 1
(2 0 if y < 022 3
F(y) = { y if 0 < y < 12222 1 if y > 19
P(Y < X + 0.5) = P((X,Y) e C) + P((X,Y) e C ) =1
i i i i= 2 2f(x,y)dxdy + 2 2f(x,y)dxdy =
j j j jC C
1
C C1
C = {(x,y); 0 < y < 0.5,0 < x < y } u {(x,y); 0.5 < y < 1,y - 0.5 < x < y}
i i i i2 22(x+y)dxdy + 2 20dxdy =j j j jC C
1
0.5 y 1 yi i i i
= 2 22(x+y)dxdy + 2 2 2(x+y)dxdy =j j j j0 0 0.5 y-0.5
Page 38
0.5 1y yi 2 i 2
= 2 (x +xy1 )dy + 2 (x +xy1 )dy =j j
0 y-0.50 0.5
0.5 1i 2 i 2 2
= 2 3y dy + 2 3y - (y-0.5) - 2(y-0.5)ydy =j j0 0.5
0.5 10.5 1i 2 i 3 2
= 2 3y dy + 2 2y -0.25dy = y 1 + (y - 0.25y)1 = 0.75.j j 0 0.50 0.5
P(Y < X +0.25) = 0.75.
QUESTION 42.
The distribution of a random vector (X,Y) is given by
X \ Y 2 4 5[----------i---------------------------------------------------------------------------------------------------------]1 p 0.1 0.1 0.2
p2 p 0.2 0.3 0.1
p
Find: a) P( X = 2), b) P(Y > 1), c) P(X < 2/ Y < 4.5)
Solution question 42.
a) P( X = 2) = P( (X,Y) e{(2,2),(2,4),(2,5)} = 0.2 + 0.3 + 0.1 = 0.6
b) P(Y > 1) = P((X,Y)e { (1,2),(1,4),(1,5),(2,2),(2,4),(2,5)}) = 1
c) P(X < 2/Y < 4.5) = P(X < 2, Y < 4.5)/ P( Y< 4.5) =
= P((X,Y) e{(1,2),(1,4)})/P( (X,Y) e {(1,2),(1,4),(2,2),(2,4)})
= 0.2/0.7 = 2/7
QUESTION 43.
Let X, Y be two independent identically distributed random variables with
common density function(2 -t
e if t > 0f(t) = {
22 0 otherwise9
Find the distribution function of Z = max {X,Y}.
Solution question 43.
X and Y are identically distributed so thay have the same distribution function F(
x 2 0 if2 x < 0
i xF(x) = P(X < x) = 2 f(t)dt = { -t -t x -xj i e dt = -e 1 = 1 - e if x > 0
2-8 2 0 0
9X and Y are independent so for all x,y
P( X < x,Y < y) = P(X < x)P(Y < y) = F (x)F (y) =F(x)F(y)X X
Therefore
Page 39
F (z) = P(Z < z) = P( max{X,Y} < z) = P(X < z, Y < z) = P(X < z)P(Y < z) =Z
(2 0 if z < o
2= F (z)F (z) = [F(z)] = {
X Y -z 22 (1 - e ) if z > 09
QUESTION 44.
A point (x,y) is to be selected from the square S containing all points
(x,y) such that 0 < x < 1 and 0 < y < 1. Suppose that each point has the
same chance to be selected. Find the probability that the selected point
will be from the area given by 1/2 < x + y < 3/2.
Solution question 44.
1
0.5 DD = {(x,y);0.5 < x + y < 1.5}
0.5 1 x
Applying geometric probability we have
area of D 1 - 2W0.5W0.5W0.5 3P(D) = --------------------------------------------- = ------------------------------------------------------------------------------------------ = ---------------
area of W 1 4
QUESTION 45.
Let A and B be two independent events such that P(A) = 1/3, P(B) = 1/2.
Find P(AuB).
Solution question 45.
P(AuB) = P(A) + P(B) - P(AnB)
Since A and B are independent
P(AnB) = P(A)P(B)
Therefore
P(AuB) = P(A) + P(B) - P(A)P(B) =
1 1 1 1 2= --------------- + --------------- - --------------- --------------- = ---------------
3 2 3 2 3
QUESTION 46.
Suppose that the distribution function of a random variable X is given by:
Page 40
& 0 for x < -12 0.25 for -1 < x < 0
F(x) = { 0.45 for 0 < x < 22 0.75 for 2 < x < 42 0.95 for 4 < x < 67 1 for x > 6
Find: a) The distribution of X
b) P( -1 < X < 3.5)
c) P( X > 1.5)
d) EX and Var(X)
Solution question 46.
Since F(x) has a jumps at certain points and is constant in between
therefore X is a random variable of discrete type with values equal
to the points where F has jumps and corresponding probabilities equal
to the size of jumps.
Hence distribution is given by
x 1-1 1 0 1 2 1 4 1 6i-------------------------k--------------------k--------------------k--------------------k--------------------k-------------------------
p 10.2510.2010.3010.2010.05i
P(a < X < b) = F(a) - F(b)
P(-1 < X < 3.5) = P(Xe{-1,0,2}) = 0.75
P(X > 1.5) = 1 - P(X < 1.5) = 1 - F(1.5) = 1 - 0.45 = 0.55
EX = S x p = 1.45i i
2 2 2 2Var(X) = E(X - EX) = E(X ) - (EX) = 6.45 - (1.45) = 4.39
2 2E(X ) = S x p = 6.45
i iQUESTION 47.
Let (X,Y) be a random vector with a common density function given by:
1 + xy& ------------------------------ if -1 < x < 1, -1 < y < 1f(x,y) = { 4
7 0 otherwise
Find a) P( X < Y/ Y > 0.1 )
b) Var(X)
Solution question 47.
P( X < Y, Y > 0.1 )P( X < Y/ Y > 0.1 ) = ---------------------------------------------------------------------------------------------------------
P(Y > 0.1)
Page 41
yC
y = xy = 1
BA
y = 0.1
1 +xy 1 +xyP(X < Y, Y > 0.1) = Ii i -----------------------------------dxdy + i i -----------------------------------dxdy + i i0dxdy =
4 4A B C
0.1 1 1 1i i 1 +xy i i 1 +xy
= 2 2 -----------------------------------dydx + 2 2 -----------------------------------dydx =j j 4 j j 4
-1 0.1 0.1 x0.1 1
11 i 2 i 2 1= ---------------[ 2 {y + xy /21 }dx + 2 {y + xy /21 }dx] =
4 j j x0.1-1 0.10.1 1
1 i i 3= --------------- [ 2 0.9 + 0.99 x/2 dx + 2 1 - x/2 - x /2dx] =
4 j j-1 0.1
0.1 11 2 2 4= ---------------[(0.9 x + 0.99 x /41 ) + ( x - x /4 - x /81 )]=
4 -10.1
= 0.3181
Page 42
C
D
1 11 + xy i i 1 + xy
P(Y > 0.1 ) = Ii i----------------------------------------dxdy + i i0dxdy = 2 2 ---------------------------------------- dydx =4 j j 4
D C-1 0.1
1 11 i 2 1 1 i
= --------------- 2 { y + xy /21 }dx = --------------- 2 {0.9 + 0.99 x/2 }dx =4 j 4 j
0.1-1 -1
11 2= --------------- (0.9 x + 0.99 x /41 ) = 0.45
4 -1
0.3181P( X < Y/ Y > 0.1 ) = ----------------------------------- = 0.7068
0.452 2 2
Var(x) = E(X - EX) = EX - (EX)
8i
EX = 2 xf(x)dxj-8
8i
f(x) = 2 f(x,y)dyj-8
Page 43
y
x
8i
if -1 < x < 1 then f(x) = 2 f(x,y)dy =j-8
-1 1 81i i 1 + xy i 1 2
= 2 0dy + 2 ----------------------------------------dy + 20dy = ---------------( y + xy /21 ) =j j 4 j 4 -1-8 -1 1
1= --------------- ( 1 + x/2 +1 -x/2) = 0.5
4
8 8i i
if x < -1 or x > 1 then f(x) = 2 f(x,y)dy = 2 0dy = 0j j-8 -8
Hence(2 0.5 if -1 < x < 1
f(x) = {2 0 otherwise98 1i i 2 1
EX = 2 xf(x)dx = 2 0.5xdx = 0.5 (x /21 ) = 0j j -1-8 -18 1
2 i 2 i 2 3 1EX = 2 x f(x)dx = 2 0.5x dx = 0.5 ( x /31 )= 0.33
j j -1-8 -1
Hence Var(X) = 0.33 - 0 = 0.33
Page 44
QUESTION 48.
Suppose that the density function of a random variable X is
as follows:(2 22 (9 - x )/36 for -3 < x < 3
f(x) = {22 0 otherwise9
Find: a) P(X < 0)
b) The distribution function F(x)
c) P( X > -0.5/ X < 2)
Solution question 48.
2f(x) = 0 (9-x )/36 0
[-----------------------------------------------------------------------------------------------k------------------------------------------------------------k----------------------------------------------------------------------k--------------------------------------------------------------------------------------------------------------L
-3 0 3
00 -3 0 2 & 3 1 * 1
P(X < 0) = i f(x)dx = i 0dx + i (9-x )/36dx = 1/36 9x - x /3 = --------------------7 1 8 2
-8 -8 -3 -3
xF(x) = P( X < x) = i f(t)dt
-8x x
if x < -3 then F(x) = P( X < x) = i f(t)dt = i 0dt = 0-8 -8
x -3 x 2if -3 < x < 3 then F(x) = P( X < x) = i f(t)dt = i 0dt + i (9-t )/36dt
x -8 -8 -3& 3 1 * 3
1/36 9t - t /3 = (9x - x /3+18)/367 1 8
-3 8x -3 3 2
if x > 3 then F(x) = P( X < x) = i f(t)dt = i 0dt + i (9-t )/36dt+ i0dt=1-8 -8 -3 3
&0 if x < -3
23
so F(x) = { (9x - x /3+18)/36 if -3 < x < 37 1 if x > 3
P(-0.5 < X < 2 ) 0.549P( X > -0.5 / X < 2) = -------------------------------------------------------------------------------- = ------------------------- = 0.593
P( X < 2 ) 0.9252 2
2 3P(-0.5 < X < 2) = i (9 - x )/36 dx = 1/36 ( 9x - x /3)1 = 0.549
-0.5-0.5
P( X < 2) = P( X < 2) = F(2) = 0.925
QUESTION 49.
Let X be a random variable. Is Z =1X1 a random variable ?
Solution question 49.
Page 45
Let (W,G,P) be a probability space.
Definition: A real function X : W -------------------------L R is called a random variable (r.v.)
if the inverse images under X of all Borel sets in R are events, that is,-1
1) X (B) = { w; X(w) e B } e G for all B e B(R)
This condition is equivalent to the following:
X is an r.v. if and only if for each x e R-1
2) X ((-8,x]) = { w; X(w) < x} = { X < x } e G
We have to show that-1
Z ((-8.x]) e G .
We have-1
Z ((-8.x]) = { w;Z(w) < x} = { w; 1X(w)1 < x } = o if x < 0
For x > 0 we have-1
Z ((-8.x]) = { w;Z(w) < x} = { w; 1X(w)1 < x } = {w; -x < X(w) < x } =-1
X ( [-x,x]) J-------------------- this set is in G since interval [-x,x]
is a borel set and 1) holds (because X is a random variable).
Therefore Z = 1X1 is a random variable.
QUESTION 50.
Let X, Y be two independent identically distributed random variables with
common density function(2 -t
e if t > 0f(t) = {
22 0 otherwise9
Find the distribution function of Z = max {X,Y}.
Solution question 50.
X and Y are identically distributed so they have the same distribution function F(
x 2 0 if2 x < 0
i xF(x) = P(X < x) = 2 f(t)dt = { -t -t x -xj i e dt = -e 1 = 1 - e if x > 0
2-8 2 0 0
9X and Y are independent so for all x,y
P( X < x,Y < y) = P(X < x)P(Y < y) = F (x)F (y) =F(x)F(y)X X
Therefore
F (z) = P(Z < z) = P( max{X,Y} < z) = P(X < z, Y < z) = P(X < z)P(Y < z) =Z
(2 0 if z < o
2= F (z)F (z) = [F(z)] = {
X Y -z 22 (1 - e ) if z > 09
Page 46
QUESTION 60.
Prove the following theorem:
Let X be a random variable on a probability space (W,G,P) of thek
continuous type with the density function f(x). Let E1X1 < 8 for
some k > 0. Thenk
n P(1X1 >n) ----------L 0 as n ----------L 8.
Solution question 60.
We have
8 nk k
8 > i 1x1 f(x)dx = lim i 1x1 f(x)dx .-8 n[-----L 8 -n
It follows that-n 8
& k k *lim i 1x1 f(x)dx + i 1x1 f(x)dx = 0
7 8n[-----L 8 -8 n
Let us notice that for x e (-8,-n) or x e (n,8)k k
1x1 > nUsing it we get
-n 8 -n 8k k k k
i1x1 f(x)dx + i1x1 f(x)dx > i n f(x)dx + i n f(x)dx-8 n -8 n
-n 8 -n 8k k k k
> i n f(x)dx + i n f(x)dx = n ( i f(x)dx + i f(x)dx) = n P{1X1 > n}-8 n -8 n
completing the proof.
QUESTION 61.
Let (X,Y) be the random vector with the joint density function given by:
&1 for 0 < x < 1, 0 < y < 1
f(x,y)= {7 0 otherwise
Find the distribution of U(X,Y) = ( X + Y, X - Y).
Solution question 61.
Let U(X,Y) = ( X + Y, X - Y) ( u (x,y) = x + y, u (x,y) = x - y)1 2
& u = x + y u + u u - u1 1 2 1 2
Solving { we have x(u ,u ) = ---------------------------------------------, y(u ,u )= --------------------------------------------- .1 2 2 1 2 2u = x - y7 2
Then the Jacobian of the inverse function is given by ( h = x(u ,u ),1 1 2
h = y(u ,u ))2 1 2
Page 47
q e2 2dh dh q e2 21 12 2 2 2--------------- --------------- 1 12 2 2 2du du --------------- ---------------2 2 2 21 2 2 2 1
J = det2 2 = det 2 2 = - ---------------2dh dh 1 12 2 2 22 2 --------------- ----------------2 2 2 2--------------- --------------- 2 22 2 2 2du du2 21 2 z c2 2
z cand the joint density of U = (U ,U ) is given by
1 2f (u ,u ) = 1J1Wf (h (u ,u ),h (u ,u )) =U 1 2 X 1 1 2 2 1 2
u + u u - u1 2 1 2& 1/2 if 0 < ---------------------------------------- < 1, 0 < ---------------------------------------- < 12 2= {
7 0 otherwise
QUESTION 62.
Let X be a random variable with the moment generating function-n/2
given by M (t) = (1 - 2t)X
3Find variance of X.
Solution question 62.2 2 2
Var(X) = E(X - EX) = EX - (EX)
-n/2 n -n/2 -1EX = (M (t))’ = ((1 2t) )’ = (------(1-2t) (-2)) = n1X 10 10 2 0
2 -n/2 -1EX = (M (t))" = (n(1-2t) )’ =
X 10 10n -n/2 - 2 2
= n(------ - 1)(1-2t) (-2) = n + 2n2 102 2
Var(X) = n + 2n - n = 2n
QUESTION 63.
Let (X,Y) be a random vector with joint density given by(2 22 3x
2xy + ------------------------------ if 0 < x < 1, 0 < y < 1f(x,y) = { 2
222 0 otherwise9
Find the line of regression of Y on X
Solution question 63.
cov(X,Y)y - EY = ----------------------------------------(x - EX)
Var(X)
I1j---------------------------------------------o1 11 2 1
3x12xy+--------------- 1 D = {(x,y);0<x<1,0<y<1}
21 1---------------------------------------------k---------------------------------------------k---------------------------------------------L
Page 48
1 18i
f(x) = 2 f(x,y)dy = 0 if x e/ (0,1)j
-88 0 1 8
2 2i i i 3x i 2 3yx 1
f(x) = 2 f(x,y)dy = 2 0dy + 2 2xy+---------------dy + 20dy = xy + --------------------1 =j j j 2 j 2
0-8 -8 0 1
23x
= x + --------------- if x e (0,1)2
8 0 1 82
i i i 3x iEX = 2 xf(x)dx = 2 0dx + 2x(x + --------------- )dx + 20dx =
j j j 2 j-8 -8 0 1
3 4 1 17= x /3 + 3x /8 1 = -------------------- = 0.7083
2408 0 1 8
22 i 2 i i 2 3x i
EX = 2 x f(x)dx = 2 0dx + 2x (x + --------------- )dx + 20dx =j j j 2 j
-8 -8 0 1
4 5 1 11= x /4 + 3x /10 1 = --------------------
200
2 2 11 17 2Var(X) = EX - (EX) = -------------------- - (--------------------) = 0.048
20 24
8i
f(y) = 2 f(x,y)dx = 0 if y e/ (0,1)j
-88 0 1 8
2i i i 3x i
f(y) = 2 f(x,y)dx = 2 0dx + 2 2xy+---------------dx + 20dx =j j j 2 j
-8 -8 0 12 3 1
= x y + x /2 1 = y + 1/2 if y e (0,1)0
8 0 1 8i i i 2 y i 3 2 1 7
EY = 2 yf(y)dy = 2 0dy + 2y + ----- dy + 20dy = y /3 + y /4 1 = -------------------- = 0.5833j j j 2 j 120
-8 -8 0 1
cov(X,Y) = E(X - EX)(Y - EY) = EXY - EXEY
8 8 11 12 3 4
i i ii 3x i 2 x 3yx 1EXY = 2 2 xyf(x,y)dxdy = 22xy(2xy+---------------)dxdy = 2(2y -------------------- + -------------------- 1 )dy =
j j jj 2 j 3 8 0-8 -8 00 01i 2 3 2 1
= 2 2y /3 + 3y/8 dy = 2y /9 - 3y /16 1 = 0.4097j 00
cov(X,Y) = E(X - EX)(Y - EY) = EXY - EXEY = 0.4097 - 0.7083Q0.5833 = -0.0034
Page 49
cov(X,Y)y - EY = ----------------------------------------(x - E)
Var(X)
y - 0.5833 = -0.0034/0.048(x - 0.7083)
y = -0.0708x +0.6334
QUESTION 64.
Let X ,X ,...,X ,... be i.i.d. random variable with joint1 2 n
density function given by(2 12--------------- if 0 < x < 42
4f(x) = {
22 0 otherwise29
PLet M = max {X ,X ,...,X ). Prove that M ------------------------------L 4 as n --------------------L 8.
n 1 2 n nSolution question 64.
We have to show that
A lim P(1M - 41 > e) = 0ne>0 nL8
Let e > 0 ( e < 4) be fixed
P(1M - 41 > e) = 1 - P(1M - 41 < e) = 1 - P(4-e < M < 4 + e) =n n n
1 - P( 4 - e < M ) { since values taken by each X are in (0,4)n i
so values taken by M ) = P( M < 4 - e) =n n
P(max{X ,X ,...,X } < 4 - e) = P(X < 4-e,X < 4-e,...,X < 4 - e)1 2 n 1 2 n
= P(X < 4-e)P(X < 4-e)...,P(X < 4 - e) ={since identical dist.)1 2 n
4-e 0 4-en i n i i 1 n
= P(X < 4-e) = ( 2 f(x)dx) = ( 2 0dx + 2 ---------------dx) =1 j j j 4
-8 -8 0n4 - e 4 - e
= (-------------------------) ------------------------------L 0 as n --------------------L 8 since 1-------------------------1 < 1.4 4
PHence M ------------------------------L 4 as n --------------------L 8.
n
QUESTION 65.
Let X be a random variable with the density function f(x) given by(2 1/2 for -1 < x < 1
f(x) = {2 0 otherwise9
2Find the distribution of Z = X
Solution question 65.Our random variables take values from (-1,1). Z has the values [0,1)Our function can be restricted to two 1:1 functions
Page 50
2 2Let Y = X ( g(x) = x ). D = (-1,0), D = [0,1)
1 2-1 -1 6 -1 -1 6
g (y) = g (y) = - ry, g (y) = g (y) = ry , y e [0,1) (C = [0,1)).1 1D 2 1D
1 2& -1 d -1 -1 d -1
f(g (y))W1--------------- g (y)1 + f(g (y))W1--------------- g (y)1 if y e [0,1)2 1 dy 1 2 dy 2h(y) = {
7 0 otherwise&
1 1 1 12 --------------- 1--------------------1 + --------------- 1---------------1 if y e (0,1]2 q6===== q6=====2 2-2ey 2ey= {
27 0 otherwise
1&------------------------- if y e (0,1]q==========2 e2 y= {
7 0 otherwise
QUESTION 66.
Let X be a random variable with the moment generating function( )-12 2t
given by M (t) = 21 - ---------------2 t < l.X l2 2
9 0Find variance of X.
Solution question 66.2 2 2
Var(X) = E(X - EX) = EX - (EX)
( ) ( )( ) ’ ( )2 2 2 2-1 -22 2 2 2 2 2 2 2t t 1 1
EX = (M (t))’ = 2 21 - ---------------2 2 = 2 21 - ---------------2 --------------- 2 = ---------------X 10 l l l l2 2 2 2 2 2 2 2
2 2 2 29 0 9 09 010 9 010
( ) ( )( ) ’ ( )2 2 2 2-2 -32 2 2 2 2 2 2 22 t 1 t 1 2
EX = (M (t))" = 2 21 - ---------------2 --------------- 2 = 2 221 - ---------------2 ------------------------- 2 = -------------------------X 10 l l l 2 22 2 2 2 2 2 2 2l l2 2 2 29 0 9 0
9 010 9 010( )
22 22 1 1Var(X) = ------------------------- - 2 --------------- 2 = -------------------------
2 l 22 2l l9 0
QUESTION 67.
Let X ,X ,...,X ,... be i.i.d. random variable with joint1 2 n
density function given by
Page 51
(2 12--------------- if 0 < x < 42
4f(x) = {
22 0 otherwise29
PLet M = max {X ,X ,...,X ). Prove that M ------------------------------L 4 as n --------------------L 8.
n 1 2 n nSolution question 67.
We have to show that
A lim P(1M - 41 > e) = 0ne>0 nL8
Let e > 0 ( e < 4) be fixed
P(1M - 41 > e) = 1 - P(1M - 41 < e) = 1 - P(4-e < M < 4 + e) =n n n
1 - P( 4 - e < M ) { since values taken by each X are in (0,4)n i
so values taken by M ) = P( M < 4 - e) =n n
P(max{X ,X ,...,X } < 4 - e) = P(X < 4-e,X < 4-e,...,X < 4 - e)1 2 n 1 2 n
= P(X < 4-e)P(X < 4-e)...,P(X < 4 - e) ={since identical dist.)1 2 n
4-e 0 4-en i n i i 1 n
= P(X < 4-e) = ( 2 f(x)dx) = ( 2 0dx + 2 ---------------dx) =1 j j j 4
-8 -8 0n4 - e 4 - e
= (-------------------------) ------------------------------L 0 as n --------------------L 8 since 1-------------------------1 < 1.4 4
QUESTION 68.
Prove the following theorem:
Let { X } be a sequence of random variables. Thenn
P -1 Pmax 1X 1---------------L 0 ++++++++++6 1n S 1---------------L 0
k n1<k<n
where S = X + ...+ Xn 1 n
Solution question 68.P
Let Z = max 1X 1. Since Z ---------------L 0 then we haven k n1<k<n
A P(1Z 1 > e) --------------------L 0 as n ---------------L 8ne>0
We have to show that-1
A P(1n S 1 > e) ---------------L 0 as n ----------L 8ne>0
-1Let us notice that if each 1X 1 < d then 1n S 1 < d
i nand als that P(AnB) < P(A)
Let e,d such that 0 < d < e < 8 be fixed.-1 -1
P(1n S 1 > e) = 1 - P(1n S 1 < e) =n n
-1 -1= 1 - P(1n S 1 < e, max 1X 1 < d ) - P(1n S 1 < e, max 1X 1 < d )
n k n k1<k<n 1<k<n
Page 52
Let us notice-1
P(1n S 1 < e, max 1X 1 < d ) = P( max 1X 1 < d ) ----------L 1 as n ----------L 8n k k1<k<n 1<k<n
and-1
0 < P(1n S 1 < e, max 1X 1 > d ) < P( max 1X 1 > d ) ----------L 0 as n ----------L 8.n k k1<k<n 1<k<n
Therefore we get-1
P(1n S 1 > e) ---------------L 0 as n ----------L 8 which end the proof.n
QUESTION 69.
Let X , X , X , ... be a sequence of random variables defined on the1 2 3
probability space (W,G,P) and F (x), F (x), F (x), ... be a sequence1 2 3
of corresponding distribution functions.D P
Prove that X -----------------------------------L a ++++++++++6 X -----------------------------------L a, as n -----------------------------------L 8, wheren n
a is a constant.
Solution question 69.
The limit random variable X is a constant with the distribution function
F(x) = 0 if x < a and F(x) = 1 if x > a. F(x) is continuous for all
x $ a. By the definition of convergence in distribution we have
lim F (x) = F(x) for all x $ a.n[L8 n
PIn order to prove the X -----------------------------------L a we have to show that
nA lim P(1X - X1 > e) = 0
n[L8 ne > 0
Let e > 0 be fixed. Let d > 0 be fixed such that 0 < d < e
P(1X - X1 > e) = 1 - P(1X - a1 < e) = 1 - P( -e < X - a < e)n n n
= 1 - P( a - e < X < a + e) < 1 - P( a - e < X < a + e - d) =n
= 1 - F (a + e - d) - F (a - e) [------------------------------L 1 - 1 + 0 as n [--------------------L 8 whichn n
ends the proof.
QUESTION 70.
Let X be a r.v. with density
& 0 if x < 0f(x) = {
1/2 if 0 < x < 12
27 1/(2x ) if 1 < x
Find the density of Y = U(X) = 1/X
Solution question 70.
Page 53
1
1
U(x) is 1:1 for x > 0 and all values taken by X are greater than 0
(since f(x) > 0 for x > 0).
(0,1] is transferred into [1,8) by U and (1,8) is transferred into (0,1)
by U.
-1 -1 1u = 1/x so x = 1/u therefore U (u) = 1/u and (U (u))’ = ----- ----------
2u
In this case
& -1 -1f(U (u))1(U (u))’1 if 0 < u < 8
2h(u) = {
2 0 otherwise7
& 1/2 if u > 1-1
f(U (u) = f(1/u) = {{ 2 21/(2(1/u) ) = u /2 for 0 < u < 1
7
Finally we get
& 0 if u < 0f(u) = {
1/2 if 0 < u < 12
27 1/(2u ) if 1 < u
QUESTION 71.
Suppose that the joint density function for a random vector (X,Y) isgiven by:
& 2(x+y) for 0 < x < y < 1f(x,y) = {
0 otherwise7
Find the distribution of U(X,Y) = ( X + Y, X - Y).
Solution question 71.
& 2(x+y) for 0 < x < y < 1f(x,y) = {
0 otherwise7
Page 54
U(X,Y) = ( X + Y, X - Y) ( u (x,y) = x + y, u (x,y) = x - y)1 2
(2 u = x + y2 1 u + u u - u
1 2 1 2solving { we have x = ---------------------------------------------, y = ---------------------------------------------
2 22 u = x - y22 29
Then the Jacobian of the inverse function is given by ( h = x(u ,u ), h = y(u ,u ))1 1 2 2 1 2
q e2 2dh dh q e2 21 12 2 2 2--------------- --------------- 1 12 2 2 2du du --------------- ---------------2 2 2 21 2 2 2 1
J = det2 2 = det 2 2 = - ---------------2dh dh 1 12 2 2 22 2 --------------- ----------------2 2 2 2--------------- --------------- 2 22 2 2 2du du2 21 2 z c2 2
z cand the joint density of U = (U ,U ) is given by
1 2f (u ,u ) = 1J1Wf (h (u ,u ),h (u ,u )) =U 1 2 X 1 1 2 2 1 2
& u for (u ,u ) e C1 1 2
= {7 0 otherwise
u +u u -u1 2 1 2
0 < ------------------------- < ------------------------- < 12 2
C = { (u ,u ): 0 < u + u < u - u < 2}1 2 1 2 1 2
C = { (u ,u ): 0 < u < 1, -u < u < 0}u{(u ,u ): 1< u < 2,u -2 < u < 0}1 2 1 1 2 1 2 1 1 2
u2
u = 02
2 uC 1
u u = u - 2= -u2 2 11
QUESTION 72.
Let X be a random variable with the moment generating function-n/2
given by M (t) = (1 - 2t) .Find variance of X.X
Solution question 72.2 2 2
Var(X) = E(X - EX) = EX - (EX)
Page 55
-n/2 n -n/2 -1EX = (M (t))’ = ((1 2t) )’ = (------(1-2t) (-2)) = n1X 10 10 2 0
2 -n/2 -1EX = (M (t))" = (n(1-2t) )’ =
X 10 10n n -n/2 - 2 2
= ------(------ - 1)(1-2t) (-2) = n + 2n2 2 10
2 2Var(X) = n + 2n - n = 2n
QUESTION 73.
Let X be a random variable with the density function f(x) given by(2 1/2 for -1 < x < 1
f(x) = {2 0 otherwise9
2Find the distribution of Y = X
Solution question 73.2
Our random variables take values from (-1,1). Y = X has the values2 2
in [0,1), Y = X ( g(x) = x )
Our function g(x) can be restricted to two 1:1 functions
by using the partition of R into D = (-1,0), D = [0,1)1 2
-1 -1 6 -1 -1 6g (y) = g (y) = - ry, g (y) = g (y) = ry , y e [0,1) (C = [0,1)).1 1D 2 1D
1 2& -1 d -1 -1 d -1
f(g (y))W1--------------- g (y)1 + f(g (y))W1--------------- g (y)1 if y e [0,1)2 1 dy 1 2 dy 2h(y) = {
7 0 otherwise&
1 1 1 12 --------------- 1--------------------1 + --------------- 1---------------1 if y e (0,1]2 q6===== q6=====2 2-2ey 2ey= {
27 0 otherwise
1&------------------------- if y e (0,1]q==========2 e2 y= {
7 0 otherwise
QUESTION 74.
Let X be a random variable with the moment generating function-n/2
given by M (t) = (1 - 2t)X
Find variance of X.
Solution question 74.2 2 2
Var(X) = E(X - EX) = EX - (EX)
-n/2 n -n/2 -1EX = (M (t))’ = ((1 2t) )’ = (------(1-2t) (-2)) = n1X 10 10 2 0
2 -n/2 -1EX = (M (t))" = (n(1-2t) )’ =
X 10 10
Page 56
n -n/2 - 2 2= n(------ - 1)(1-2t) (-2) = n + 2n
2 102 2
Var(X) = n + 2n - n = 2n
QUESTION 75.
Let (X,Y) be the random vector with the joint density function given by:
&1 for 0 < x < 1, 0 < y < 1
f(x,y)= {7 0 otherwise
Find the distribution of U(X,Y) = ( X + Y, X - Y).
Solution question 75.
Let U(X,Y) = ( X + Y, X - Y) ( u (x,y) = x + y, u (x,y) = x - y)1 2
& u = x + y u + u u - u1 1 2 1 2
Solving { we have x(u ,u ) = ---------------------------------------------, y(u ,u )= --------------------------------------------- .1 2 2 1 2 2u = x - y7 2
Then the Jacobian of the inverse function is given by ( h = x(u ,u ),1 1 2
h = y(u ,u ))2 1 2
q e2 2dh dh q e2 21 12 2 2 2--------------- --------------- 1 12 2 2 2du du --------------- ---------------2 2 2 21 2 2 2 1
J = det2 2 = det 2 2 = - ---------------2dh dh 1 12 2 2 22 2 --------------- ----------------2 2 2 2--------------- --------------- 2 22 2 2 2du du2 21 2 z c2 2
z cand the joint density of U = (U ,U ) is given by
1 2f (u ,u ) = 1J1Wf (h (u ,u ),h (u ,u )) =U 1 2 X 1 1 2 2 1 2
u + u u - u1 2 1 2& 1/2 if 0 < ---------------------------------------- < 1, 0 < ---------------------------------------- < 12 2= {
7 0 otherwise
QUESTION 76.
Let {X } be a sequence of independent identically distributed randomn
n2 s
variables with EX = 0 and EX = 1. Let S = X , n = 1,2,3,...1 n t k
k=1
Sn P
Prove that -------------------- ----------------------------------------L 0 as n ----------------------------------------L 8.n
Solution question 76.
We have to show that
A limP(1S /n - 01 > e) = 0ne>0 n-----L8
Let e > 0 be fixed but arbitrary. ES = E(X +X +...+X ) = 0n 1 2 n
Page 57
Var(X ) = 1 so Var(S ) = Var(X +X +...+X ) = Var(X )+Var(X )+...+Var(X ) = ni n 1 2 n 1 2 n
We have to evaluate P(1S - 01 > ne) = P(1S - ES 1 > ne)n n n
If in the Chebychev’s inequality
2s
P(1Z - m1 > e) < --------------------2e
we substitute S in place of Z we will getn
Var(S )n n 1
P(1S - 01 > ne) = P(1S - ES 1 > ne) < ----------------------------------- = -------------------- = --------------- ----------L 0n n n 2 2 2 2ne(ne) n e
as n ---------------L 8 which completes the proof.
QUESTION 77.
Let (X,Y) be a random vector with distribution given by
Y \ X1 0 1 11 1----------------------------------------------------------------------------------------------------1 11 0.25 0.251 1
2 1 0 1 0.5
Find the line of regression of Y on X.
Solution question 77.
cov(X,Y)y - EY = --------------------------------------------------W(x - EX)
var(X)
Y \ X1 0 1 1 pqj1 1
--------------------------------------------------------------------------------------------------------------------------------------------1 11 0.25 0.25 0.51 12 1 0 1 0.5 0.5
p 0.25 0.75iq
Cov(X,Y) = E(X - EX)(Y - EY) = EXY - EXEY
sEXY = x y p = 0*1*0.25 + 0*2*0 + 1*1*0.25 + 1*2*0.5 =
t i j iji,j
= 0.25 + 1 = 1.25
sEX = x p = 0*0.25 + 1*0.75 = 0.75
t i iqi
sEY = y p = 1*0.5 + 2*0.5 = 1.5
t j jqj
Cov(X,y) = 1.25 - 0.75*1.5 = 0.1252 2 2
Var(X) = E(X - EX) = EX - (EX)
2 s 2 2 2EX = x p = 0 *0.25 + 1 *0.75 = 0.75
t i iqi
Page 58
2Var(X) = 0.75 - 0.75 = 0.75 - 0.5625 = 0.1875
0.125y - 1.5 = ------------------------------(x - 0.75)
0.1875
y - 1.5 = 0.66(x - 0.75)