Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Law of Total Expectation

  1. Feb 3, 2013 #1
    Suppose a box initially contains 1 red marble and 1 black marble and that, at each time n = 1, 2, ..., we randomly select a marble from the box and replace it with one additional marble of the same color. Let X_n denote the number of red marbles in the box at time n (note that X_0 = 1). What is E(X_3)?

    In solving this problem, I would like to calculate E(X_3 | X_2 = 1). If there is 1 red marble at time 2 (X_2 = 1), that means the first 2 selections resulted in black marbles. So at the time of the third selection, there are 3 black marbles in the box and 1 red marble. Therefore, E(X_3 | X_2 = 1) = 1(3/4) + 2(1/4) = 5/4 (that is, 1 with probability 3/4 and 2 with probability 1/4).

    However, if I would like to calculate E(X_3 | X_2 = 1) differently, using the law of total expectation, I can write E(X_3 | X_2 = 1) = E(X_3 | X_2 = 1, X_1 = 1) P(X_1 = 1). (There are no other values of X_1 to condition on, because if we know there is 1 red marble at time n = 2, there cannot be 2 red marbles at time n = 1.) However, this simplifies to [1(3/4) + 2(1/4)](1/2) = 5/8.

    Why am I getting different results? I think the problem has something to do with the following: When I condition on X_1 = 1, I already know X_1 = 1 with probability 1 because X_2 = 1; however, then I say P(X_1 = 1) = 1/2, which is also true.
     
  2. jcsd
  3. Feb 3, 2013 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    Shouldn't that be E(X_3 | X_2 = 1) = E(X_3 | X_2 =1, X_1=1) P(X_1=1 | X_2 = 1) ?
     
  4. Feb 3, 2013 #3
    Yes, you're right. Thank you very much.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Law of Total Expectation
Loading...