Conditional expectations related to count of event occurring kth time

[psie]

Homework Statement

Independent repetitions of an experiment are performed. $A$ is an event that occurs with probability $p$, $0<p<1$. Let $T_k$ be the number of the performance at which $A$ occurs the $k$th time, $k=1,2,\ldots$. Compute
(a) $E(T_3\mid T_1=5)$,
(b) $E(T_1\mid T_3=5)$.

Relevant Equations

I think the relevant distributions here are the geometric and negative binomial distributions, i.e. the number of trials until first success and the number of trials with $n$ successes respectively.

I am stuck at obtaining the joint pmf of $T_3$ and $T_1$. It is clear I think that $T_1\in\text{Ge}(p)$, where the pmf of $T_1$ is given by $p(1-p)^{k-1}$, $k=1,2,\ldots$. Now, the negative binomial distribution counts the number of trials with $n$ successes and with success probability $p$. So I would say $T_2$ and $T_3$ are $\text{NBin}(2,p)$ and $\text{NBin}(3,p)$ respectively. I also know of the fact that $\text{NBin}(n,p)$ is just a sum of $n$ geometrically independent distributed random variables with success probability $p$.

Yet I do not know how to find the joint pmf of $T_1$ and $T_3$. If I can find that, then I have the conditional pmf of $T_3$ given $T_1$ and vice versa. Then I think I'd be able to solve both (a) and (b).

Also, this exercise does appear in a chapter on order statistics, and I notice that $T_1\leq T_2\leq T_3$, so I'm curious if one could solve it using one of those methods as well (I think this is the way it's intended to be solved).

 

[psie]

Ok, I think I've worked out an answer. I am not 100% sure about my answer. Also, I am not using order statistics at all (perhaps there is some slick way of doing it using order statistics).

Anyway, for a), if we assume $T_3\in \text{NBin}(3,p)$, then $T_3=X_1+X_2+X_3$ where $X_i\in\text{Ge}(p)$ are independent with expectation $1/p$. I would guess e.g. $X_1=T_1$ and so \begin{align*}E(T_3\mid T_1=5)&=E(X_1+X_2+X_3\mid X_1=5)\\ &=E(5+X_2+X_3)\\ &=5+\frac2p.\end{align*} For b), we observe that $E(X_1\mid X_1+\ldots+X_n=x)=\frac{x}{n}$ if $X_1,\ldots,X_n$ are iid (for proof, see here). Then simply $$E(T_1\mid T_3=5)=E(X_1\mid X_1+X_2+X_3=5)=\frac53.$$

 

[nuuskur]

b) is correct. It is enough to note that $n E(X_i | S_n) = S_n$ for all $i$. It's a routine check that sum of iid geometric variables is negative binomial.

side note: when making calculations, one should take care with the distribution of NB: do we count the number of trials until $k$ successes or the number of failures until $k$ successes.