Section 8.3 Suppose X 1 , X 2 , ..., X n are a random sample from a distribution defined by the p.d.f. f(x) for a < x < b and corresponding distribution function F(x), The random variables which order the sample from smallest to largest Y 1 < Y 2 < ...< Y n are called the order statistics . Suppose n = 2. The space of (X 1 , X 2 ) is The space of (Y 1 , Y 2 ) is {(x 1 , x 2 ) | a < x 1 < b , a < x 2 < b} {(y 1 , y 2 ) | a < y 1 < y 2 < b } For a subset A of the space of (Y 1 , Y 2 ), we have that P[(Y 1 , Y 2 ) A] =
34
Embed
Section 8.3 Suppose X 1, X 2,..., X n are a random sample from a distribution defined by the p.d.f. f(x)for a < x < b and corresponding distribution function.
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Section 8.3
Suppose X1 , X2 , ..., Xn are a random sample from a distribution defined by the p.d.f.
f(x) for a < x < b and corresponding distribution function F(x),
The random variables which order the sample from smallest to largest Y1 < Y2 < ...< Yn are called the order statistics.
Suppose n = 2.
The space of (X1 , X2) is
The space of (Y1 , Y2) is
{(x1 , x2) | a < x1 < b , a < x2 < b}
{(y1 , y2) | a < y1 < y2 < b }
For a subset A of the space of (Y1 , Y2), we have that P[(Y1 , Y2) A] =
For a subset A of the space of (Y1 , Y2), we have that P[(Y1 , Y2) A] =
n [F(y)]n–1 f(y) =Observe that when k = r this second term is the negative of the preceding term when k = r + 1. This pattern continues until k = n – 1 when this second term is the negative of the isolated term.
Since this looks hard to integrate, we shall use an alternative approach:
P(Y3 1/2) = P[at least three of X1 , X2 , X3 , X4 , X5 are 1/2] =
5
3[ ]3 [1 – ]2 + [ ]4 [1 – ]1 + [ ]5 =
5
41/4 1/4 1/4 1/4 1/4
1 3 1 3 110 — — + 5 — — + — = 4 4 4 4 4
3 2 4 1 5 106 53—— = ——1024 512
Note that this probability can be read as 0.1035 from Table II in the appendix of the textbook.
Suppose the random sample X1 , X2 , … , Xn is from a U(0,1) distribution. Let Y1 , Y2 , … , Yn be the order statistics of the sample. (Note: Parts of this Exercise are the same as Text Exercise 8.3-6.)
Find the distribution function corresponding to the U(0, 1) distribution.
3.
(a)
(b)
F(x) =
if x 0
0
if 0 < x 1
x
if 1 < x1
Find the joint p.d.f. of the order statistics (Y1 , Y2 , … , Yn).
The joint p.d.f. of Y1 , Y2 , …, Yn is
g(y1 , y2 , …, yn) = n! if 0 < y1 < y2 < … < yn < 1
(c) Find the p.d.f. of Yr where r is any integer from 1 to n.
The p.d.f. of Yr is gr(y) = n!—————— yr–1 (1 – y)n–r if 0 < y < 1(r – 1)! (n – r)!
Realizing that (n + 1) = n! , (r) = (r – 1)! , and (n – r + 1) = (n – r)!, we find that Yr has a distributionbeta rwith = and = n – r + 1 .
This is essentially what Text Exercise 8.3-6(c) says to show.
3.-continued
(d) Find the mean and variance of Yr where r is any integer from 1 to n.
E(Yr) = —— = +
r——n + 1
Var(Yr) =
———————— =( + + 1)( + )2
r(n – r + 1)——————(n + 2)(n + 1)2
E(Yr+1 – Yr) =r + 1—— –n + 1
r—— =n + 1
1——n + 1
(e) Find E(Yr+1 – Yr) where r is any integer from 1 to n – 1.
4.
(a)
Let Q have a U(0, 1) distribution. For constants b > a, define the random variable X = (b – a)Q + a .
Find the distribution function for X, find the p.d.f. for X, and state what type of distribution X has.
The distribution function for Q is F(q) = P(Q q) =
if q 00
if 0 < q 1if 1 < q1
The space for X is The distribution function for X is
G(x) = P(X x) =
{x : a < x < b}.
P([b – a]Q + a x) =P(Q [x – a] / [b – a]) =
We see then that X has a distribution.U(a, b)
The p.d.f. for X is g(x) = for a < x < b
q
x – a——b – a
1——b – a
(b) Let Q1 , Q2 , Q3 be a random sample selected from the U(0, 1) distribution, and let V1 , V2 , V3 be the order statistics. Also, let
X1 = (b – a)Q1 + a , X2 = (b – a)Q2 + a , X3 = (b – a)Q3 + a ,
and let Y1 , Y2 , Y3 be the order statistics, which implies
Y1 = (b – a)V1 + a , Y2 = (b – a)V2 + a , Y3 = (b – a)V3 + a .
State why X1 , X2 , X3 is a random sample, use part (a) to find the type of distribution this random sample is from, and use Class Exercise #3 to find E(Y1) , Var(Y1) , E(Y2) , Var(Y2) , E(Y3) , Var(Y3) , and E(Y1Y3) .Since Q1 , Q2 , Q3 are independent, then X1 , X2 , X3 are independent
and this together with part (a) implies X1 , X2 , X3 is a random sample from a U(a, b) distribution.
Recall that the (100p)th percentile of the distribution defined by p.d.f. f(x) is a number p such that
–
p
f(x) dx = F(p) = pwhich motivates the following definition:
The (100p)th percentile of the sample X1 , X2 , …, Xn is defined to be
Yr where r = (n+1)p
a weighted average of Yr and Yr+1 where r = (n+1)p
Note: This definition is extended to an observed sample of values x1 , x2 , …, xn where the ordered values in the sample are represented by y1 , y2 , …, yn .
if (n+1)p is not an integer
if (n+1)p is an integer
The detailed definition of sample order statistics was given in Section 3.2.
1013 1019 1021 1024 1026 1028
1033 1035 1039 1040 1043 1047
The location of the 40th percentile is (n + 1)p = (13)(0.40) = 5.2 .
Find the 40th percentile and the 80th percentile for data of Text Example 8.3-5.
5.
The detailed definition of sample order statistics was given in Section 3.2, and an Excel spreadsheet was constructed to find sample order statistics. Recall that the Excel formulas were slightly different.