We have $X_n=1$ if it's heads (T) on the $n$-toss and $X_n=0$ otherwise ($n=1,...$). We also have $T_1=$ number of coin tosses needed to observe the first T and $T_2=$ total number of coin tosses to observe the second T. Find $\mathbb{E}[T_2|T_1]$.
I've found the joint law that is $\mathbb{P}(T_1=k,T_2=n)=\mathbb{P}(T_1=k)\mathbb{P}(T_2=n|T_1=k)=\frac{1}{2^n}$, but now I'm stuck.
Can you help me? Thanks in advance.
In the expression $E(T_2 | T_1)$, $T_1$ is no longer a random variable. Therefore, the question reduces to "if I've already observed one head after $T_1$ flips, how many more flips should I expect for the second head?" Or, phrased differently, how many flips until I see a head? Thus, \begin{equation*} E(T_2 | T_1) = T_1 + E(Y) \end{equation*}
where $Y$ is the number of flips I need to wait to see a head. As lulu mentioned in the comments, finding $E(Y)$ is the same problem as finding $E(T_1)$. Hence, \begin{equation*} E(T_2 | T_1) = T_1 + E(T_1) = T_1 + 2 \end{equation*}
Personally, I prefer to introduce the new notation, $Y$, to make it clear there is a step in between the final statement. The notation is a bit tricky here, as $T_1$ is not a random quantity in the term $E(T_2 | T_1)$. Others may cringe at introducing more notation.
It's helpful to note that the answer to $E(X|Y)$ will always be written in terms of $Y$ (assuming we aren't dealing with degenerate random variables).