# Knowing $E(Y|X)$, compute $E(Y)$

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 $$1,\ldots,8$$) 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 $$E(Y | X)$$

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 $$P = (8-x)/x$$.

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

I'm struggling to find $$E(Y)$$.

• $p=1-x/8$. $8-x/x =7$. $(8-x)/x > 1$ given $x = 1,2,3,4$. So both of them are not probability. So try to fix $p$ first. May 31, 2017 at 22:53
• Add the self study tag. May 31, 2017 at 23:11
• Small detail. I believe you need to use $p=\frac{8-x}{8}$ as succes probability instead of $p=\frac{8-x}{x}$. Nov 22, 2022 at 8:17

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 rolls before a success which is 1 less than the number of rolls to a success. Then $$P(n=0)=1/2$$, $$P(n=1)=(1/2)^2$$, $$P(n=2)=(1/2)^3$$ 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 will 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.
• You lost me at the first sentence: it appears to assume $X=4$. What if $X$ is not $4$?