ECSE-2500, Engineering Probability, Spring 2010, Rensselaer Polytechnic Institute

# Lecture 25

# Homework 9 question 4 is subtle.

Consider this simplified version.

- Let W, X and Y be independent identically distributed (i.i.d) random variables.
- What is the probability that W<min(X,Y)?
*Method 1:*- Everything is symmetric, so each variable has 1/3 chance of being the smallest.
- P[W<min(X,Y)] = 1/3.

*Method 2:*- P[W<X] = P[W<Y] = 1/2 by symmetry.
- Therefore P[W<min(X,Y)] = P[W<X, W<Y] = P[W<X] P[W<Y] = 1/4

- However 1/4 != 1/3. Something is wrong.
- The problem is that W<X and W<Y are dependent since a small W makes both more probable. Therefore P[W<X, W<Y] != P[W<X] P[W<Y]
- However a few classes back I did this:
- Define Z=max(X,Y).
- The cdf F
_{Z}(z)=P[Z=max(X,Y)<=z] = P[X<=z, Y<=z] = P[X<=z] P[Y<=z] = F_{X}(z)F_{Y}(z) - Why can I factor the cdf here? Since Z is defined as max(X,Y), and X and Y are independent, P<=X and P<=Y are independent.

- Here's another way to see that the 2 cases are different.
- If these are all continuous r.v., then P[W=X]=P[W=Y]=P[W=min(X,Y)]=0.
- However P[Z=min(X,Y)]=1 by the definition of Z.

- Agreed that this is all very confusing. Probability has some complicated parts.

# Chapter 6 problems

- Problem 6.1, page 348.
- Problem 6.2.
- Problem 6.3.
- Problem 6.4.
- Problem 6.5.
- Problem 6.7.
- Problem 6.8.