Take into account an infinity sequence of independent tosses of an unbiased die. Let $X$ be the number of tosses necessary to obtain a five and $Y$ the number of tosses necessary in order to obtain a six. Determine:
(a) $\textbf{E}(X)$
(b) $\textbf{E}(X \mid Y = 1)$
(c) $\textbf{E}(X \mid Y=2)$
MY ATTEMPT
I am particularly having trouble to calculate the conditional expectation. As I far as I know, it should be something like \begin{align*} \textbf{E}(X\mid Y = y) = \sum_{x} x\cdot p_{X\mid Y}(x,y) = \sum_{x} \frac{x\cdot p_{X,Y}(x,y)}{p_{Y}(y)} \end{align*}
But I don't know how to proceed. Could someone pave the way towards the right solution? Thanks in advance.
Here is a solution for (b). You should be able to get (c) similarly.
Let $Z_1,Z_2,...,Z_n,...$ be a countable sequence of iid random variables valued in $\{1,...,6\}$ (the dice throws). Then $X=\min\{t\ge 0: Z_t = 5\}$ and \[ E[X|Y=1] = \sum_{k=1}^{\infty} k \frac{P(X=k,Y=1)}{P(Y=1)} \] and the prpobability of the denomiator is $1/6$.
For $k=1$, the probability of the event $\{X=1,Y=1\}=\{Z_1=5,Z_1=6\}$ is 0 since $Z_1$ takes only one value.
For $k\ge 2$, the event $\{X=k\}$ can be rewritten as $\{Z_1\ne 5, ..., Z_{k-1}\ne 5, Z_k=5\}$ and the intersection $\{X=k,Y=1\}$ can be rewritten as $\{Z_1 =6, Z_2 \ne 5,..., Z_{k-1}\ne 5, Z_k=6\}$. By independence, the probability of this event is $\frac 1 6(\frac 5 6)^{k-2}\frac 1 6$ for $k\ge 2$. Hence \[ E[X|Y=1] = \sum_{k=2}^{\infty} k (1/6) (5/6)^{k-2} = \frac 6 5 \sum_{k=2}^{\infty} k (1/6) (5/6)^{k-1} \] and $\sum_{k=2}^{\infty} k (1/6) (5/6)^{k-1} = E[X]$ already computed in (a).