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In my probability class I got a problem about conditional distributions to solve.

Let $\lambda, a, b >0$. Random variable $N$ has Poisson distribution ($Po(\lambda) $). Conditioning on $N$, random variable $T$ has distribution $Exp(an+b)$. I need to calculate unconditional distribution of random variable $T$ and conditional distribution of random variable $N$ given $T$.

If $N$ is Poisson distributed, then $P(N=n)= \frac{\lambda^n e^{-\lambda}}{n!}$ for $n=0,1,2,...$ and exponential distribution has density $f(t)= \lambda e^{-\lambda t} $ for $t>0,$ and $0$ otherwise.

I started like this: let $0 \leq c \leq d $, then $P(c < T < d)= \sum_{n=0}^{\infty} P(N=n) P(c < T

I know I can exchange the integral and the sum but I don't know how to calculate this sum anyway.

If someone has some idea of how to sum this, I would be really thankful.

2 Answers 2

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Work with densities instead. Let $f_T$ be the density function of $T$ and $f_{T\mid N}$ be the density of $T\mid N$. The latter is naturally the density of an exponential r.v. Let $\Omega$ be the probability space. To get the unconditional density of $T$, use that

$$ f_T(t) = \int_\Omega f_{T\mid N}(t\mid n) \,\mathrm dP(N = n) $$ where the integral is a Riemann-Stieltjes integral, i.e. \begin{align*} f_T(t) &= \sum_{n = 0}^\infty f_{T\mid N}(t\mid n) P(N = n) \\&= \sum_{n = 0}^\infty (an+b)e^{-(an+b)t} \frac{\lambda^n e^{-\lambda}}{n!}. \end{align*} By expanding a bit and pulling constants out of the sum we see that the sums are power series of the exponential function. Hence, \begin{align*} f_T(t) &= e^{-(\lambda+bt)} \biggl(a\sum_{n = 0}^\infty \frac{\lambda^n e^{-ant}}{(n-1)!} + b\sum_{n = 0}^\infty \frac{\lambda^ne^{-ant}}{n!} \biggr) \\&= e^{-(\lambda+bt)} \left(a\, e^{-at+\lambda \exp\{-at\}}+ be^{\lambda \exp\{-at\}}\right) \end{align*}

To find $N\mid T$, use Baye's theorem, $$ P( N = n\mid T\leq t) = \frac{P(T\leq t \mid N = n)P(N = n)}{P(T\leq t)} $$

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    Thank you very much for your help. I also thought about doing it with densities. And I think about calculating density of $f_N|T$ it is sufficient to use formula $f_{N|T}(n|t)=\frac{f_{N,T}(n,t)}/{f_T(t)}$ since we can represent 'density' of discrete variable $N$ as density relative to counting measure on $\mathbb{N}$. – 2017-02-12
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You don't need to exchange the integral and the sum. Just calculate the inegral: $$(an+b)\int_c^de^{-(an+b)t}\,dt=-\left.e^{-(an+b)t}\right\vert_c^d=e^{-(an+b)c}-e^{-(an+b)d}.$$ Then the sum becomes: \begin{align} \sum\limits_{n=0}^{\infty} \frac{\lambda^n e^{-\lambda}}{n!}(e^{-(an+b)c}-e^{-(an+b)d})&=\sum\limits_{n=0}^{\infty} \frac{\lambda^n e^{-\lambda}}{n!}e^{-(an+b)c}-\sum\limits_{n=0}^{\infty} \frac{\lambda^n e^{-\lambda}}{n!}e^{-(an+b)d}\\ &=e^{-(\lambda+bc)}\sum\limits_{n=0}^{\infty} \frac{(\lambda e^{-ac})^n}{n!}-e^{-(\lambda+bd)}\sum\limits_{n=0}^{\infty} \frac{(\lambda e^{-ad})^n}{n!}\\ &=e^{-(\lambda+bc)}e^{\lambda e^{-ac}}-e^{-(\lambda+bd)}e^{\lambda e^{-ad}}. \end{align} In the last step we're using the series expansion $e^x=\sum\limits_{n=0}^\infty\frac{x^n}{n!}$.

Also, to get the distribution, it's enough to calculate $P(T

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    Thank you very very much. It helped a lot. But now that I have distribution, I would need it's density to calculate distribution of $N|T$, because I would do that with formula $f_{N|T} (n|d) = \frac{f_{N,T}(n,d)}{f_T(d)}$. I only have to do derivative of $P(T – 2017-02-11
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    Yes, to get the density of $T$ you just take that derivative. However I'm not sure if you can use that formula as $N$ is discrete and doesn't have a density. Perhaps to get the distribution of $N|T$ it's easiest to use the Bayes' theorem, Therkel said at the end of his answer. – 2017-02-11
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    Well, I think we can write also 'densitiy' of N - relative to counting measure $m_0$ (along natural numbers). Thanks for help. – 2017-02-12