This is the question i'll show what i have done below,

Roll a standard 4-sided die (with sides labelled 1, 2, 3, 4) and let X be the number rolled. Then take a fair 8-sided die (with sides labelled 1, . . . , 8) and roll it repeatedly until you roll a number strictly larger than X. Let Y be the number of times you roll the 8-sided die, not including the last roll (i.e., the number of times you roll a number less than or equal to X). i. Write down the conditional distribution of Y given {X = x}, including parameter(s). ii. Find E (Y | X)

My answer,

Given the rolls of the second die are trials with "successes" defined to mean rolling higher than X conditional on {X = x}, the trails have a constant success probability P = 8-x/x per trail, each roll is independent, thus Y follows a geometric distribution with P = (8-x)/x

E(Y|X=x) = 1 - p / p = x / 8 - x E(Y|X) = X / 8 - X

Im struggling to find E(Y)

Can any one help?

  • 1
    $\begingroup$ $p=1-x/8$. $8-x/x =7$. $(8-x)/x > 1$ given $x = 1,2,3,4$. So both of them are not probability. So try to fix $p$ first. $\endgroup$ – user158565 May 31 '17 at 22:53
  • $\begingroup$ Add the self study tag. $\endgroup$ – Michael Chernick May 31 '17 at 23:11

On any roll of the 8 sided die the success probability is 1/2 that is getting 5,6,7 or 8 when X=4. So let n equal the number of roll before a success which is 1 less than the number of rolls to a success. P(n=0)=1/2 P(n=1)=(1/2)$^2$ , P(n=2)=(1/2)$^3$ and so on. To get E(Y|X=4) take y P(n=y) and sum over all y from 0 to infinity. This is conditional on X=4. If X=1 p=7/8. If X=2 p=3/4. If X=3 P=5/8. Repeat the process for 1, 2, and 3 and you get all the conditional probabilities and expectations. To get the unconditional probability take P(Y=y|X=x) P(X=x) summed for x=1,2,3 and 4. Note that P(X=x) =1/4 for each x between 1 and 4. Then take the sum y P(Y=y) for y between 0 and infinity.

  • $\begingroup$ You lost me at the first sentence: it appears to assume $X=4$. What if $X$ is not $4$? $\endgroup$ – whuber May 31 '17 at 22:55
  • $\begingroup$ Sorry I forgot about the conditioning on X. I will edit the answer. $\endgroup$ – Michael Chernick May 31 '17 at 22:59

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