Law of Total Expectation

  • Thread starter adnaps1
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  • #1
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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.
 

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  • #2
Stephen Tashi
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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).
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) ?
 
  • #3
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Yes, you're right. Thank you very much.
 

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