Probability spaces and random variables Exercises

Probability spaces and random variables Exercises

Exercise 1.1. This exercise refers to the random variables described in the slides from the course.

=n

1; 2; 3; 4; 5; 6o

. We already established that it is not possible to build a prob- ability measure on the sample space such that both random variables X and W have a uniform distribution. Is it possible to build a probability measure on the sample space such that both random variables X and W have the same distribution ?

Exercice 1.2. Let Card( ) < 1. Show that if X : ! R and Y : ! R are F measurable then minfX; Yg, maxfX; Yg and XY are also F measurable. In addi- tion, show that if 8!2 ,Y (!)6= 0 then X=Y is F measurable.

Exercice 1.3. The indicator function of an event C is de…ned as I^C : ! f0;1g

! 7!I^C(!) = 1 if !2C 0 if ! =2C Show that

(I1) If A belongs to the -algebraG then I^A is G measurable.

(I2) I^A\B =I^AI^B:

(I3) I^A[B =I^A+I^B I^A\B:

Now, letA₁; :::; A_n be a …nite partition of the sample space . Show that (I4) Pn

i=1I^Ai(!) = 1 for every ! 2 :

Exercise 1.4. Consider the following sample space and functionX : !R: X

!₁ 0

!₂ 0

!₃ 1

!₄ 1

!₅ 1

a) What is the smallest -algebraF such thatX is a random variable ? b) What is the smallest -algebraG such thatjXj is a random variable ? c) Is jXj aF random variable ? Justify your answer.

d) Is X aG random variable ? Justify your answer.

Exercise 1.5. Let =f1; :::;6g; A=ff1;3;5g;f1;2;3gg and B =ff2;4;6g;f4;5;6gg: a) DescribeF = (A), the -algebra generated byA.

b) List the atoms ofF.

c) DescribeG = (B), the -algebra generated byB. d) List the atoms ofG:

Exercise 1.6. Let X be a random variable de…ned on a probability space ( ;F;P) and Y(!) = exp [X(!)]; !2

a) Show thatY is also a random variable, and express its cumulative distribution function F_Y as a function of F_X.

b) AssumeX has a continuous distribution. Then,Y is continuous too. Expressf_Y, the density function ofY, as a function of f_X, the density function ofX.

c) Assume X N(0;1). Calculatef_Y and P(Y 1).

d) What is the name of the distribution of Y ?

Solutions

1 Exercise 1.1

Yes, it is possible. Indeed, let pi =P n

i o

. We are looking for values of p1; :::; p6 such that :

…rstly, P is a probability measure, meaning

(i) 0 p_i 1

(ii) p₁+p₂+p₃+p₄+p₅+p₆ = 1 ;

secondly, the distribution of the random variable X is arbitrary, meaning that (X1) PfX = 0g=Pn

1; 2o

=p₁+p₂ =q₀ (X2) PfX = 5g=Pn

3; 4o

=p₃+p₄ =q₅ (X3) PfX = 10g=Pn

5; 6o

=p₅+p₆ =q₁₀

and …nally, the distribution of the random variable W is the same asX, meaning (W1) PfW = 0g=Pn

5o

=p₅ =q₀ (W2) PfW = 5g=Pn

1; 2; 3; 4o

=p₁+p₂+p₃ +p₄ =q₅ (W3) PfW = 10g=P

n 6

o

=p6 =q10

The constraints(X3)and(W3)imply thatp₅ = 0andp₆ =q₁₀. Then, from(W1)we obtain that q₀ =p₅ = 0. This impliesq₅ = 1 q₀ q₁₀ = 1 q₁₀. Since p₁ and p₂ are non negative and because their sum is 0 (see X1), we conclude that p₁ = p₂ = 0. The constraint (X2) yieldsp₃ =q₅ p₄ = 1 q₁₀ p₄. Hence the solution is

x PfX =xg=PfW =xg

0 0

5 1 q₁₀

10 q₁₀ 2[0; 1]

:

! P(!) ! P(!)

1 0 4 p₄ 2[0; 1 q₁₀]

2 0 5 0

2 Exercise 1.2

SinceCard( )<1, the random variablesX and Y can only take a …nite number of values, sayx₁ < ::: < x_m and y₁ < ::: < y_n respectively.

Let us show that minfX; Yg is F measurable :8z 2R, f! 2 jminfX(!); Y (!)g zg

= f! 2 jX(!) z or Y (!) zg

= f|! 2 jX{z(!) zg}

2F

[ f|!2 jY{z(!) zg}

2F

2 F.

Let us show that maxfX; Yg is F measurable :8z 2R, f! 2 jmaxfX(!); Y (!)g zg

= f! 2 jX(!) z and Y (!) zg

= f|! 2 jX{z(!) zg}

2F

\ f|!2 jY{z(!) zg}

2F

2 F.

Let us show that XY isF measurable :8z 2R, f!2 jX(!)Y (!) zg

= [

xiyj z

f!2 jX(!) = x_i and Y (!) = y_jg

= [

xiyj z

f!2 jX(!) = x_ig

| {z }

2F

\ f!2 jY (!) = y_jg

| {z }

| {z 2F }

2F

2 F.

Let us show that if Y 6= 0; X=Y is F measurable :8z 2R,

!2 X(!) Y (!) z

= [

xi yj z

f!2 jX(!) =x_i and Y (!) =y_jg

= [

xi yj z

f!2 jX(!) =x_ig

| {z }

2F

\ f|! 2 jY{z(!) =y_jg}

| {z 2F }

2F

2 F.

3 Exercise 1.3

Proof of

(I1): To show : If A belongs to the -algebra G then I^A is G measurable.

We must show that 8x2R, f! 2 :I^A(!) =xg 2 G. However, f!2 :I^A(!) = 1g=A2 G, f!2 :I^A(!) = 0g=A^c 2 G and 8x =2 f0;1g, f!2 :I^A(!) = xg=?2 G.

Proof of

(I2): To show : I^A\B =I^AI^B:

I^A\B(!) = 1 ! 2A\B 0 ! =2A\B On the other hand,

I^A(!)I^B(!) = 1 , I^A(!) = 1 and I^B(!) = 1 , ! 2A and !2B

, ! 2A\B.

Proof of

(I3): To show : I^A[B =I^A+I^B I^A\B: Let us consider four cases :

if ! 2A\B then I^A[B(!) = 1 and I^A(!) +I^B(!) I^A\B(!)

= 1 + 1 1 = 1 if ! 2A\B^c then I^A[B(!) = 1 and I^A(!) +I^B(!) I^A\B(!)

= 1 + 0 0 = 1 if ! 2A^c\B then I^A[B(!) = 1 and I^A(!) +I^B(!) I^A\B(!)

= 0 + 1 0 = 1 if ! 2A^c\B^c then I^A[B(!) = 0 and I^A(!) +I^B(!) I^A\B(!)

= 0 + 0 0 = 0

Proof of

(I4). To show : Let A₁; :::; A_n be a …nite partition of the sample space . Show that Pn

i=1I^Ai(!) = 1 for every ! 2 :

Let! 2A_j be an arbitrary element from the sample space. Since A₁; :::; A_n is a partition we have

! 2A_j and ! =2A_i for every i6=j:

Then,

I^Aj(!) = 1 and I^Ai(!) = 0 for every i6=j:

Therefore,

Xn

4 Exercise 1.4

a) What is the smallest -algebraF such thatX is a random variable ? 8>

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:

?; ;f!₁; !₂g;f!₃g;f!₄; !₅g;f!₆g;f!₇g f!₁; !₂; !₃g;f!₁; !₂; !₄; !₅g;f!₁; !₂; !₆g;f!₁; !₂; !₇g f!₃; !₄; !₅g;f!₃; !₆g;f!₃; !₇g;f!₄; !₅; !₆g;f!₄; !₅; !₇g;f!₆; !₇g

f!₁; !₂; !₃; !₄; !₅g;f!₁; !₂; !₃; !₆g;f!₁; !₂; !₃; !₇g f!₁; !₂; !₄; !₅; !₆g;f!₁; !₂; !₄; !₅; !₇g;f!₁; !₂; !₆; !₇g f!₃; !₄; !₅; !₆g;f!₃; !₄; !₅; !₇g;f!₃; !₆; !₇g;f!₄; !₅; !₆; !₇g f!₁; !₂; !₃; !₄; !₅; !₆g;f!₁; !₂; !₃; !₄; !₅; !₇g;f!₁; !₂; !₃; !₆; !₇g

f!₁; !₂; !₄; !₅; !₆; !₇g;f!₃; !₄; !₅; !₆; !₇g

9>

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b) What is the smallest -algebra G such that jXj is a random variable ?

?; ;f!₁; !₂g;f!₃; !₄; !₅g;f!₆; !₇g

f!₁; !₂; !₃; !₄; !₅g;f!₁; !₂; !₆; !₇g;f!₃; !₄; !₅; !₆; !₇g c) Is jXj a F random variable ? Justify your answer.

Yes since

f!2 :jX(!)j xg= 8>

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:

?2 F if x <0 f!₁; !₂g 2 F if 0 x <1 f!₁; !₂; !₃; !₄; !₅g 2 F if 1 x <2

2 F if x 2

d) Is X a G random variable ? Justify your answer.

No since

f!2 :X(!) 2g=f!₇g2 G= :