Roll an unbiased die successively and call $X$ and $Y$ denote the
rolls necessary to obtain a 6 and a 5 , respectively. Compute
$E(X|Y=5)$. In general, can you compute $E(X|Y=y)$?
Try
The event $X|Y=5$ is also geometric but we have cases.
before the fifth roll notice that $X$ is still geometric with parameter $p=1/6$ since neither 5 or a 6 have occured. but, after the fifth roll, a 6 has shown up so we only have 5 number to pick from and thus after thus we have a geometric rv with parameter $p' = \frac{1}{5}$. Therefore
$$ E(X|Y=5) = 6 + 5 = 11 $$
is this correct?
Best Answer
The key fact here is that Geometric Distributions are memory-less.
$E(X|Y=5) = E(Y=5) + E(X) = 5 + \frac{1}{p_X} = 5+6 = 11$.
So, yes your reasoning is correct.