Probability spaces and random variables Exercises Exercise 1.1. This exercise refers to the random variables described in the slides from the course. n o = 1 ; 2 ; 3 ; 4 ; 5 ; 6. We already established that it is not possible to build a probability 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 min fx; Y g, max fx; Y g and XY are also F measurable. In addition, show that if 8! 2, Y (!) 6= 0 then X=Y is F measurable. Exercice 1.3. The indicator functiof an event C is de ned as I C :! f0; 1g 1 if! 2 C! 7! I C (!) = 0 if! =2 C Show that (I1) If A belongs to the -algebra G then I A is G measurable. (I2) I A\B = I A I B : (I3) I A[B = I A + I B I A\B : Now, let A 1 ; :::; A n be a nite partitiof the sample space. Show that (I4) P n i=1 I A i (!) = 1 for every! 2 : Exercise 1.4. Consider the following sample space and function X :! R: X! 1 0! 2 0! 3 1! 4 1! 5 1! 6 2! 7 2 1
a) What is the smallest -algebra F such that X is a random variable? b) What is the smallest -algebra G such that jxj is a random variable? c) Is jxj a F random variable? Justify your answer. d) Is X a G 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) Describe F = (A), the -algebra generated by A. b) List the atoms of F. c) Describe G = (B), the -algebra generated by B. d) List the atoms of G: Exercise 1.6. Let X be a random variable de ned on a probability space (; F; P) and Y (!) = exp [X (!)] ;! 2 a) Show that Y is also a random variable, and express its cumulative distribution function F Y as a functiof F X. b) Assume X has a continuous distribution. Then, Y is continuous too. Express f Y, the density functiof Y, as a functiof f X, the density functiof X. c) Assume X N(0; 1). Calculate f Y and P(Y 1). d) What is the name of the distributiof Y? 2
1 Exercise 1.1 Solutions Yes, it is possible. Indeed, let p i = P i. We are looking for values of p 1 ; :::; p 6 such that : rstly, P is a probability measure, meaning (i) 0 p i 1 (ii) p 1 + p 2 + p 3 + p 4 + p 5 + p 6 = 1 ; secondly, the distributiof the random variable X is arbitrary, meaning that (X1) P fx = 0g = P 1 ; 2 = p 1 + p 2 = q 0 (X2) P fx = 5g = P 3 ; 4 = p 3 + p 4 = q 5 (X3) P fx = 10g = P 5 ; 6 = p 5 + p 6 = q 10 and nally, the distributiof the random variable W is the same as X, meaning (W 1) P fw = 0g = P 5 = p 5 = q 0 n o (W 2) P fw = 5g = P 1 ; 2 ; 3 ; 4 = p 1 + p 2 + p 3 + p 4 = q 5 (W 3) P fw = 10g = P 6 = p 6 = q 10 The constraints (X3) and (W 3) imply that p 5 = 0 and p 6 = q 10. Then, from (W 1) we obtain that q 0 = p 5 = 0. This implies q 5 = 1 q 0 q 10 = 1 q 10. Since p 1 and p 2 are non negative and because their sum is 0 (see X1), we conclude that p 1 = p 2 = 0. The constraint (X2) yields p 3 = q 5 p 4 = 1 q 10 p 4. Hence the solution is x P fx = xg = P fw = xg 0 0 5 1 q 10 10 q 10 2 [0; 1] :! P (!)! P (!) 1 0 4 p 4 2 [0; 1 q 10 ] 2 0 5 0 3 1 q 10 p 4 6 q 10 3
2 Exercise 1.2 Since Card() < 1, the random variables X and Y canly take a nite number of values, say x 1 < ::: < x m and y 1 < ::: < y n respectively. Let us show that min fx; Y g is F measurable : 8z 2 R, f! 2 jmin fx (!) ; Y (!)g z g = f! 2 jx (!) z or Y (!) z g = f! 2 jx (!) z g [ f! 2 jy (!) z g 2 F. Let us show that max fx; Y g is F measurable : 8z 2 R, f! 2 jmax fx (!) ; Y (!)g z g = f! 2 jx (!) z and Y (!) z g = f! 2 jx (!) z g Let us show that XY is F measurable : 8z 2 R, \ f! 2 jy (!) z g 2 F. f! 2 jx (!) Y (!) z g f! 2 jx (!) = x i and Y (!) = y j g x i y j z x i y j z f! 2 jx (!) = x i g \ f! 2 jy (!) = y j g 2 F. {z } Let us show that if Y 6= 0; X=Y is F measurable : 8z 2 R,! 2 X (!) Y (!) z f! 2 jx (!) = x i and Y (!) = y j g x i y j z x i z y j f! 2 jx (!) = x i g \ f! 2 jy (!) = y j g 2 F. 4
3 Exercise 1.3 Proof of (I1): To show : If A belongs to the -algebra G then I A is G We must show that 8x 2 R, f! 2 : I A (!) = xg 2 G. However, and 8x =2 f0; 1g, Proof of (I2): To show : I A\B = I A I B : On the other hand, f! 2 : I A (!) = 1g = A 2 G, f! 2 : I A (!) = 0g = A c 2 G f! 2 : I A (!) = xg =? 2 G. 1! 2 A \ B I A\B (!) = 0! =2 A \ B I A (!) I B (!) = 1, I A (!) = 1 and I B (!) = 1,! 2 A and! 2 B,! 2 A \ B. Proof of (I3): To show : I A[B = I A + I B I A\B : Let us consider four cases : if! 2 A \ B then if! 2 A \ B c then if! 2 A c \ B then if! 2 A c \ B c then I A[B (!) = 1 and I A[B (!) = 1 and I A[B (!) = 1 and I A[B (!) = 0 and measurable. = 1 + 1 1 = 1 = 1 + 0 0 = 1 = 0 + 1 0 = 1 = 0 + 0 0 = 0 Proof of (I4). To show : Let A 1 ; :::; A n be a nite partitiof the sample space. Show that P n i=1 I A i (!) = 1 for every! 2 : Let! 2 A j be an arbitrary element from the sample space. Since A 1 ; :::; A n is a partition we have! 2 A j and! =2 A i for every i 6= j: Then, Therefore, I Aj (!) = 1 and I Ai (!) = 0 for every i 6= j: nx I Ai (!) = 1: i=1 5
4 Exercise 1.4 a) What is the smallest -algebra F such that X is a random variable? 8 9?; ; f! 1 ;! 2 g ; f! 3 g ; f! 4 ;! 5 g ; f! 6 g ; f! 7 g f! 1 ;! 2 ;! 3 g ; f! 1 ;! 2 ;! 4 ;! 5 g ; f! 1 ;! 2 ;! 6 g ; f! 1 ;! 2 ;! 7 g f! 3 ;! 4 ;! 5 g ; f! 3 ;! 6 g ; f! 3 ;! 7 g ; f! 4 ;! 5 ;! 6 g ; f! 4 ;! 5 ;! 7 g ; f! 6 ;! 7 g >< >= f! 1 ;! 2 ;! 3 ;! 4 ;! 5 g ; f! 1 ;! 2 ;! 3 ;! 6 g ; f! 1 ;! 2 ;! 3 ;! 7 g f! 1 ;! 2 ;! 4 ;! 5 ;! 6 g ; f! 1 ;! 2 ;! 4 ;! 5 ;! 7 g ; f! 1 ;! 2 ;! 6 ;! 7 g f! 3 ;! 4 ;! 5 ;! 6 g ; f! 3 ;! 4 ;! 5 ;! 7 g ; f! 3 ;! 6 ;! 7 g ; f! 4 ;! 5 ;! 6 ;! 7 g f! >: 1 ;! 2 ;! 3 ;! 4 ;! 5 ;! 6 g ; f! 1 ;! 2 ;! 3 ;! 4 ;! 5 ;! 7 g ; f! 1 ;! 2 ;! 3 ;! 6 ;! 7 g >; f! 1 ;! 2 ;! 4 ;! 5 ;! 6 ;! 7 g ; f! 3 ;! 4 ;! 5 ;! 6 ;! 7 g b) What is the smallest -algebra G such that jxj is a random variable??; ; f! 1 ;! 2 g ; f! 3 ;! 4 ;! 5 g ; f! 6 ;! 7 g f! 1 ;! 2 ;! 3 ;! 4 ;! 5 g ; f! 1 ;! 2 ;! 6 ;! 7 g ; f! 3 ;! 4 ;! 5 ;! 6 ;! 7 g c) Is jxj a F random variable? Justify your answer. Yes since 8? 2 F if x < 0 >< f! f! 2 : jx (!)j xg = 1 ;! 2 g 2 F if 0 x < 1 f! 1 ;! 2 ;! 3 ;! 4 ;! 5 g 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! 7 g =2 G: 6