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

Lecture 25

Homework 9 question 4 is subtle.

Consider this simplified version.

  1. Let W, X and Y be independent identically distributed (i.i.d) random variables.
  2. What is the probability that W<min(X,Y)?
  3. Method 1:
    1. Everything is symmetric, so each variable has 1/3 chance of being the smallest.
    2. P[W<min(X,Y)] = 1/3.
  4. Method 2:
    1. P[W<X] = P[W<Y] = 1/2 by symmetry.
    2. Therefore P[W<min(X,Y)] = P[W<X, W<Y] = P[W<X] P[W<Y] = 1/4
  5. However 1/4 != 1/3. Something is wrong.
  6. 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]
  7. However a few classes back I did this:
    1. Define Z=max(X,Y).
    2. The cdf FZ(z)=P[Z=max(X,Y)<=z] = P[X<=z, Y<=z] = P[X<=z] P[Y<=z] = FX(z)FY(z)
    3. 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.
  8. Here's another way to see that the 2 cases are different.
    1. If these are all continuous r.v., then P[W=X]=P[W=Y]=P[W=min(X,Y)]=0.
    2. However P[Z=min(X,Y)]=1 by the definition of Z.
  9. Agreed that this is all very confusing. Probability has some complicated parts.

Chapter 6 problems

  1. Problem 6.1, page 348.
  2. Problem 6.2.
  3. Problem 6.3.
  4. Problem 6.4.
  5. Problem 6.5.
  6. Problem 6.7.
  7. Problem 6.8.