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Chapter 2. Survival models.

Manual for SOA Exam MLC.Chapter 2. Survival models.

Section 2.1. Survival models.

c2009. Miguel A. Arcones. All rights reserved.

Extract from:

Arcones Manual for SOA Exam MLC. Fall 2009 Edition,available athttp://www.actexmadriver.com/

c2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

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Chapter 2. Survival models. Section 2.1. Survival models.

Review of Probability theory

Definition 1Given a set, a probability P on is a function defined in thecollection of all (subsets) events of such that(i) P() = 0.

(ii) P() = 1.(iii) If{An}

n=1 are disjoint events, then

P{n=1An} =

n=1 P{An}.

is called the sample space.


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Review of Probability theory

Definition 1Given a set, a probability P on is a function defined in thecollection of all (subsets) events of such that(i) P() = 0.

(ii) P() = 1.(iii) If{An}

n=1 are disjoint events, then

P{n=1An} =

n=1 P{An}.

is called the sample space.

Definition 2A random variableX is function from the sample space intoR.

We will abbreviate random variable into r.v.


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Ageatdeath

Many insurance concepts depend on accurate estimation of the lifespan of a person. It is of interest to study the distribution of liveslifespan. The life span of a person (or any alive entity) can bemodeled as a positive (r.v.) random variable.To model the lifespan of a live, we use ageatdeath randomvariableX.For inanimate objects, ageatfailure is the age of an object at

the end of termination.


Ch 2 S i l d l S i 2 1 S i l d l

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Cumulative distribution function

Definition 3Thecumulative distribution function of a r.v. X isFX(x) =P{X x}, x R.

Theorem 1A function FX : R is the (c.d.f.) cumulative distributionfunction of a r.v. X if and only if:(i) FX is nondecreasing, i.e. for each x1 x2, FX(x1) FX(x2).(ii) FX is right continuous, i.e. for each x ,

limh0+ FX(x+h) =FX(x).(iii) lim

xFX(x) = 0.

(iv) limx

FX(x) = 1.


Ch t 2 S i l d l S ti 2 1 S i l d l

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The previous theorem gives the following for positive r.v.s.Theorem 2A function FX : R is the c.d.f. of a positive r.v. X if and onlyif:(i) F

X is nondecreasing, i.e. for each x

1x2

, FX

(x1

)

FX

(x2

).(ii) FX is right continuous, i.e. for each x ,

limh0+

FX(x+h) =FX(x).

(iii) For each x 0, FX(x) = 0.(iv) lim

x

FX(x) = 1.


Chapter 2 Survival models Section 2 1 Survival models

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Example 1

Determine which of the following function is a legitime cumulativedistribution function of an ageatdeath r.v.:(i) FX(x) =

x+1x+3 , for x 0.

(ii) FX(x) = x2x+1 , for x 0.

(iii) FX(x) = x

x+1

, for x 0.



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Example 1


x+1x+3 , for x 0.

(ii) FX(x) = x2x+1 , for x 0.

(iii) FX(x) = xx+1

, for x 0.

Solution: (i) FX(x) = x+1x+3 is not a legitime c.d.f. of an

ageatdeath because FX(0) = 13 = 0.



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Example 1


x+1x+3 , for x 0.

(ii) FX(x) = x2x+1 , for x 0.

(iii) FX(x) = xx+1

, for x 0.



(ii) FX(x) = x+1x+3 is not a legitime c.d.f. of an ageatdeath

because limxFX(x) =

1

2 = 1.



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Example 1


x+1x+3 , for x 0.

(ii) FX(x) = x2x+1 , for x 0.

(iii) FX(x) = xx+1

, for x 0.



(ii) FX(x) = x+1x+3 is not a legitime c.d.f. of an ageatdeath

because limxFX(x) =

1

2 = 1.(iii) FX(x) =

xx+1 is a legitime c.d.f. because it satisfies all

properties which a c.d.f. should satisfy.



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Chapter 2. Sur i al models. Section 2. . Sur i al models.

Discrete r.v.

Definition 4A r.v. X is calleddiscrete if there is a countable set C R suchthatP{X C} = 1.

IfP{X C} = 1, where C = {xj}j=1, then for any set A R,

P{X A} = P{X A C} = P{X A {xj}j=1}

=P{X j:j1,xjA{xj}} = j:j1,xjA

P{X =xj}.



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p

Definition 5The probability mass function (or frequency function) of thediscrete r.v. X is the function p: R R defined by

p(x) = P{X =x}, x R.

IfX is a discrete r.v. with p.m.f. pand A R, then

P

{X A} =x:xA

P

{X =x} =x:xA p(x).

Theorem 3Let p be the (p.m.f.) probability mass function of the randomvariable X. Then,(i) For each x 0, p(x) 0.(ii)

xRp(x) = 1.

If a function p: R R satisfies conditions (i)(ii) above, thenthere are a sample space S, a probability measureP on S and a

r.v. X :SR

such that X has p.m.f. p.c2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.


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Continuous r.v.

Definition 6A r.v. X is calledcontinuous continuous random variable if thereexists a nonnegative function f called a (p.d.f.) probability densityfunction of X such that for each A R,

P{X A} =A

f(x) dx=R

f(x)I(x A) dx.



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Continuous r.v.

Definition 6A r.v. X is calledcontinuous continuous random variable if thereexists a nonnegative function f called a (p.d.f.) probability densityfunction of X such that for each A R,

P{X A} =A

f(x) dx=R

f(x)I(x A) dx.

Theorem 4

A function f : is the probability density function of a r.v. Xif and only if the following two conditions hold:(i) For each x R, f(x) 0.(ii)

R

f(x) dx= 1.



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If a r.v. is positive and continuous, then fX(x) = 0, for each x

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Example 2

Determine which of the following function is a probability densityfunction of a ageatdeath:

(i) fX(x) = 1(x+1)2 , for x 0.

(ii) fX(x) = 1(x+1)3

, for x 0.

(iii) fX(x) = (2x 1)ex, for x 0.



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Example 2


(i) fX(x) = 1(x+1)2 , for x 0.

(ii) fX(x) = 1(x+1)3

, for x 0.


Solution: (i) fX is a density because for each x 0, 1(x+1)2

0,

and 0

1

(x+ 1)2 =

1

x+ 1

0

= 1.



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Example 2


(i) fX(x) = 1(x+1)2 , for x 0.

(ii) fX(x) = 1(x+1)3

, for x 0.



0,

and 0

1

(x+ 1)2 =

1

x+ 1

0

= 1.

(ii) fXis not a density function because

0

1

(x+ 1)3 =

1

2(x+ 1)2

0

=1

2= 1.



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Example 2


(i) fX(x) = 1(x+1)2 , for x 0.

(ii) fX(x) = 1(x+1)3

, for x 0.



0,

and 0

1

(x+ 1)2 =

1

x+ 1

0

= 1.

(ii) fXis not a density function because

0

1

(x+ 1)3 =

1

2(x+ 1)2

0

=1

2= 1.

(iii) fX is not a density function because (2x 1)ex

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Knowing the density fof a r.v. X, the cumulative distributionfunction ofX is given by

FX(x) = x

f(t) dt, x R.

Knowing the c.d.f. of a r.v. X, we can find its density using:

Theorem 5Suppose that the c.d.f. F of a r.v. X satisfies the followingconditions:(i) F is continuous in R.(ii) There are a1, . . . , an R such that F is continuously

differentiable on each of the intervals(, a1), (a1, a2), . . . , (an1, an), (an,).Then, X has a continuous distribution and the p.d.f. of X is givenby f(x) =F(x), except at a1, . . . , an.



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Example 3

The cumulative distribution function of the random variable X isgiven by

F(x) =

0 if x < 1,x+14 if 1 x

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Example 3

The cumulative distribution function of the random variable X isgiven by

F(x) =

0 if x < 1,x+14 if 1 x

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Mixed r.v.

Definition 7

A r.v. X has a mixed distribution if there is a function f andnumbers xj, pj, j 1, with pj >0, such that for each A R,

P{X A} =

Af(x) dx+

j:xjApj.

A mixed distribution Xhas two parts: a continuous part and adiscrete part. The function fin the previous definition is the p.d.f.of the continuous part ofX. The function p(x) = P[X =x],

x R, is the p.m.f. of the discrete part ofX.In order to have a r.v., we must have that f is nonnegative and

R

f(x) dx+

j=1pj= 1.



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Survival function

Definition 8Thesurvival function of a r.v. X is the functionSX(x) = P{X >x}, x .



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Survival function


Sometimes we will denote the survival function of a r.v. X bys.

Notice that for each x 0, SX(x) = 1 FX(x).



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Survival function


Sometimes we will denote the survival function of a r.v. X bys.

Notice that for each x 0, SX(x) = 1 FX(x).Theorem 6A function SX : [0,) is the survival function of a positiver.v. X if and only if the following conditions are satisfied:(i) SX is nonincreasing.

(ii) SX is right continuous.(iii) SX(0) = 1.(iv) lim

xSX(x) = 0.



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Theorem 7If the survival function SX of a r.v. X is continuous everywhereand continuously differentiable except at finitely points, then X has

a continuous distribution and the density of X is fX(x) = SX(x),whenever the derivative exists.



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Example 4

Find the density function for the following survival functions:

(i) s(x) = (1 +x)ex, for x 0.(ii)

s(x) =

1 x

2

10,000 for0 x 100,

0 for100

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Example 4


(i) s(x) = (1 +x)ex, for x 0.(ii)

s(x) =

1 x

2

10,000 for0 x 100,

0 for100

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Example 4


(i) s(x) = (1 +x)ex, for x 0.(ii)

s(x) =

1 x

2

10,000 for0 x 100,

0 for100

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Example 4


(i) s(x) = (1 +x)ex, for x 0.(ii)

s(x) =

1 x

2

10,000 for0 x 100,

0 for100

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Terminal age

Often, we will assume that the individuals do not live more than a

certain age. This age is called the terminal age or limiting ageof the population. So, S(t) = 0, for each t .



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Example 5

Suppose that the survival function of a person is given bySX(x) = 90x

90 , for0 x 90.(i) Find the probability that a person dies before reaching 20 yearsold.(ii) Find the probability that a person lives more than 60 years.



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Example 5



Solution: (i)

P{X 20} = 1 SX(20) = 190 20

90 =

2

9.



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Example 5



Solution: (i)

P{X 20} = 1 SX(20) = 190 20

90 =

2

9.

(ii)

P{X >60} =SX(60) =90 60

90 =1

3.



Indicator function

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Indicator function

Given a set A R, the indicator function ofA is the function

I(A) =I(x A) =

1 if x A0 if x A



Th

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Theorem 8(using the survival function to find an expectation) Let X be anonnegative r.v. with survival function s. Let h: [0,) [0,)

be a function. Let H(x) =x0 h(t) dt. Then,

E[H(X)] =

0

s(t)h(t) dt.



Th 8

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Theorem 8(using the survival function to find an expectation) Let X be anonnegative r.v. with survival function s. Let h: [0,) [0,)

be a function. Let H(x) =x0 h(t) dt. Then,

E[H(X)] =

0

s(t)h(t) dt.

Proof.Since H(x) =

0 I(x >t)h(t) dt,

E[H(X)] =E

0

I(X >t)h(t) dt =

0

E[I(X >t)]h(t) dt

=

0

s(t)h(t) dt.



R ll th t if H( )x

h(t) dt th

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Recall that ifH(x) = x0 h(t) dt, then

E[H(X)] =

0

s(t)h(t) dt.



R ll th t if H( )x

h(t) dt th

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Recall that ifH(x) =0 h(t) dt, then

E[H(X)] =

0

s(t)h(t) dt.

Corollary 1

Let X be a nonnegative r.v. with survival function s. Then,

E[X] = 0

s(t) dt.



Recall that if H(x)x

h(t) dt then

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Recall that ifH(x) =0 h(t) dt, then

E[H(X)] =

0

s(t)h(t) dt.

Corollary 1


E[X] = 0

s(t) dt.

Solution: Let h(t) = 1, for each t 0. Then,H(x) = x

0 h(t) dt=x, for each x 0. By Theorem8,

E[X] =E[H(X)] =

0

s(t)h(t) dt=

0

s(t) dt.



Example 6

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Example 6

Suppose that the survival function of X is s(x) =ex(x+ 1),x 0.

(i) Find E[X] using that E[X] =0 s(t) dt.

(ii) Find the density of X.(iii) Find E[X] using that E[X] =

0 xf(x) dx.



Example 6

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Example 6




0 xf(x) dx.

Solution: (i)

E[X] = 0

s(t) dt= 0

ex(x+ 1) dx= 2.



Example 6

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a p e 6




0 xf(x) dx.

Solution: (i)

E[X] = 0

s(t) dt= 0

ex(x+ 1) dx= 2.

(ii) The density ofX is

f(x) = s

(x) = e

x

(1)(x+ 1) e

x

(1) =e

x

x.



Example 6

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p




0 xf(x) dx.

Solution: (i)

E[X] = 0

s(t) dt= 0

ex(x+ 1) dx= 2.

(ii) The density ofX is

f(x) = s

(x) = e

x

(1)(x+ 1) e

x

(1) =e

x

x.

(iii)

E[X] =

0

xf(x) dx=

0

x2ex dx= 2.



Recall that ifH(x) =x0 h(t) dt, then

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( )0 ( ) ,

E[H(X)] =

0

s(t)h(t) dt.




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( )0 ( ) ,

E[H(X)] =

0

s(t)h(t) dt.

Corollary 2


E[X2] = 0

s(t)2t dt.


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E[H(X)] = 0

s(t)h(t) dt.



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E[H(X)] = 0

s(t)h(t) dt.

Corollary 3

Let X be a nonnegative r.v. with survival function s. Let p>0.

Then,

E[Xp] =

0

s(t)ptp1 dt.



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E[H(X)] = 0

s(t)h(t) dt.

Corollary 3

Let X be a nonnegative r.v. with survival function s. Let p>0.

Then,

E[Xp] =

0

s(t)ptp1 dt.

Solution: We take h(t) =ptp1, for each t 0. Hence,

H(x) =x0 h(t) dt=x

p, for each x 0. By Theorem8,E[Xp] =

0 s(t)pt

p1 dt.




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E[H(X)] =

0

s(t)h(t) dt.

Corollary 4

Let X be a nonnegative r.v. with survival function s. Let a 0.Then,

E[min(X, a)] = a

0

s(t) dt.




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E[H(X)] =

0

s(t)h(t) dt.

Corollary 4

Let X be a nonnegative r.v. with survival function s. Let a 0.Then,

E[min(X, a)] = a

0

s(t) dt.

Solution: Let h(t) =I(t [0, a]), for each t 0. For x 0,

H(x) = x

0

h(t) dt= x

0

I(t [0, a]) dt= min(x,a)

0

dt= min(x, a).

By Theorem8,

E[min(X, a)] =E[H(X)] =

0

s(t)h(t) dt=

a0

s(t) dt.



Example 7

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Example 7


(i) Find E[min(X, 10)] using thatE[min(X, 10)] =

0 min(x, 10)f(x) dx.

(ii) Find E[min(X, 10)] using that E[min(X, 10)] =100 s(t) dt.


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Example 7

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p


(i) Find E[min(X, 10)] using thatE[min(X, 10)] =

0 min(x, 10)f(x) dx.

(ii) Find E[min(X, 10)] using that E[min(X, 10)] =100 s(t) dt.

Solution: (ii)

100

s(t) dt=

100

et(t+ 1) dt=

100

ettdt+

100

etdt

= et(t+ 1) 10

0

et 10

0

= 1 11e10 + 1 e10 = 2 12e10.



Theorem 9Let X be a discrete r v whose possible values are nonnegative

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Let X be a discrete r.v. whose possible values are nonnegativeintegers. Let h: [0,) [0,) be a function. LetH(x) = x

0

h(t) dt. Then,

E[H(X)] =

k=1

P{X k}(H(k) H(k 1)).

Proof: We have that s(t) = P{X k}, fork 1 t

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E[H(X)] =

k=1

P{X k}(H(k) H(k 1)).

This implies that

E[X] =

k=1

P{X k},

E[X2] =k=1

P{X k}(k2 (k 1)2) =k=1

P{X k}(2k 1)

and

E[min(X, a)] =a

k=1

P{X k},

where a is positive integer.



Example 8

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Let X be a discrete r.v. with probability mass function given bythe following table,

k 0 1 2

P{X =k} 0.2 0.3 0.5

(i) Find E[X] and E[X2], using thatE[H(X)] =

k=0H(k)P{X =k}.

(ii) Find E[X] and E[X2], using that E[X] =

k=1 P{X k} andE[X2] =

k=1 P{X k}(2k 1).



Example 8

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k 0 1 2

P{X =k} 0.2 0.3 0.5


k=0H(k)P{X =k}.



k=1 P{X k}(2k 1).

Solution: (i) We have that

E[X] = (0)(0.2) + (1)(0.3) + (2)(0.5) = 1.3E[X2] = (0)2(0.2) + (1)2(0.3) + (2)2(0.5) = 2.3.



Example 8

L X b di i h b bili f i i b

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k 0 1 2

P{X =k} 0.2 0.3 0.5


k=0H(k)P{X =k}.



k=1 P{X k}(2k 1).

Solution: (ii) We have that P{X 1} = 0.8, P{X 2} = 0.5,and P{X k} = 0, for each k 3. Hence,

E[X] = P{X 1}+ P{X 2} = 0.8 + 0.5 = 1.3

E[X2] = P{X 1}((2)(1) 1) + P{X 2}((2)(2) 1)

=0.8 + 0.5(3) = 2.3.



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Definition 9Given 0

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Theorem 10If X has a uniform distribution on the interval(a, b), then the pthquantilepof X is a+ (b a)p.



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Theorem 10If X has a uniform distribution on the interval(a, b), then the pthquantilepof X is a+ (b a)p.

Proof: We have that

p= P{X p} =

p

a

1b a

+ dt= p a

b a.

So, p=a+ (b a)p.



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Definition 10A median m of a r.v. X is a value such that

P{X

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pp g y

fX(x) =5x4

k

5 if 0

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pp g y

fX(x) =5x4

k

5 if 0

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Theorem 11Let X be a continuous r.v. with density function fX. Let0

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Theorem 12Let X be a r.v. with range(a, b) and density fX. Let0

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fX(x) = 5x4

(84)5 if 0

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fX(x) = 5x4

(84)5 if 0

Mlc Guia

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