Given $P(X=k|T=t)=\frac{t^ke^{-t}}{k!}$ for $k=0,1,2,...$ and $P(T<t)=1-e^{-t}$ for $t>0$, what is $P(X=k)$.
n.b.: This is not homework; it's just review.
I believe one approach is to do this integral: $$\int_t P(X=x|T=t)f_T(t)dt$$
Which yields $$ \frac{-2^{-k-1}}{k!} \Gamma(k+1,2t) $$ evaluated from zero to infinity. It seems as though this answer is a little too complicated, however.
I would appreciate any help. Thanks.
Note $F_T(t) = \Pr[T \le t] = \Pr[T < t] = 1-e^{-t}$, so $$f_T(t) = \frac{d}{dt}[F_T(t)] = e^{-t}, \quad t > 0.$$ Then $$\begin{align*} \Pr[X = k] &= \int_{t=0}^\infty \Pr[X = k \mid T = t]f_T(t) \, dt \\ &= \int_{t=0}^\infty e^{-t} \frac{t^k}{k!} e^{-t} \, dt \\ &= \frac{1}{k!} \int_{t=0}^\infty t^k e^{-2t} \, dt \\ &= \frac{\Gamma(k+1)}{k! \, 2^{k+1}} \int_{t=0}^\infty \frac{2^{k+1} t^k e^{-2t}}{\Gamma(k+1)} \, dt \\ &= \frac{\Gamma(k+1)}{k! \, 2^{k+1}} \\ &= 2^{-(k+1)}, \quad k \in 0, 1, 2, \ldots. \end{align*}$$ Notice that I used the fact that the integral of a ${\rm Gamma}(k+1, 2)$ density (using a shape and rate parametrization) over its support is equal to $1$; and then noting that for nonnegative integers $k$, $\Gamma(k+1) = k!$. The resulting unconditional distribution for $X$ is obviously geometric with parameter $p = 1/2$.