Integrating the product of Poisson and exponential pdf

Question

So I'll spare the background as to why, but I'm trying to integrate the following:

$$\int_0^{\infty} \frac{e^{-(\lambda+\mu)t}(\lambda t)^n}{n!} dt$$

If you parameterize a Poisson w/ $\lambda$ and an exponential w/ $\mu$ and multiply their pdf's, you get the above. I just can't seem to do the integration by parts.

Is this a two step IBP? Is there an easier way to solve than actually integrating (product of random variables, etc.)?

Answer 1

If $T \sim {\rm Exponential}(\mu)$ and $N \mid T \sim {\rm Poisson}(\lambda T)$, where $\mu$ is a rate parameter, not scale--so in particular, ${\rm E}[T] = 1/\mu$ (an unfortunate choice of parameter name), then the marginal distribution of $N$ is given by $$\Pr[N = n] = \int_{t = 0}^\infty \Pr[N = n \mid T = t] f_T(t) \, dt = \int_{t=0}^\infty \mu e^{-t(\lambda + \mu)} \frac{(\lambda t)^n}{n!} \, dt.$$ To evaluate this integral, we note that the integrand is akin to a gamma density with shape parameter $a = n+1$ and rate parameter $b = \lambda + \mu$, but missing a factor of $b^a$. Explicitly, we have $$\Pr[N = n] = \frac{\mu \lambda^n}{(\lambda + \mu)^{n+1}} \int_{t = 0}^\infty e^{-t(\lambda + \mu)} \frac{(\lambda + \mu)^{n+1} t^n}{\Gamma(n+1)} \, dt = \frac{\lambda^n \mu}{(\lambda + \mu)^{n+1}}.$$ Taking the sum of this for $n = 0, 1, 2, \ldots$, we indeed get $$\sum_{n=0}^\infty \frac{\lambda^n \mu}{(\lambda + \mu)^{n+1}} = \frac{\mu}{\lambda + \mu} \sum_{n=0}^\infty \biggl(\frac{\lambda}{\lambda + \mu}\biggr)^n = \frac{\mu/(\lambda + \mu)}{1 - \lambda/(\lambda + \mu)} = 1.$$ We also see that the unconditional/marginal distribution of $N$ is $N \sim {\rm Geometric}(p = \tfrac{\mu}{\lambda + \mu})$.

Answer 2

You can use the Laplace Transform. The transform of $t^n$ is $$\frac{n!}{s^{n+1}}$$ So you get, after some algebra, the quantity must equal $$\lambda^n {n!\over n!}\frac{1}{((\lambda +\mu))^{n+1}}=\frac{\lambda^n}{(\lambda+\mu)^{n+1}}$$ Can someone check this please? And evidently you've made a mistake somewhere because that is clearly not a probability function.

Answer 3

I'm not sure if I understand $\mu$ and $\lambda$, but if they're just constants we have a $n$ step integration by parts in $t$:

$$ \frac{\lambda^n}{n!} \int_0^\infty e^{-(\lambda + \mu) t } t^n dt =\frac{\lambda^n}{n!} \left ( -\frac{ e^{- ( \lambda + \mu ) t } t^n }{\lambda + \mu } \Big |_0^\infty + \frac{n}{\lambda + \mu } \int_0^\infty e^{- ( \lambda + \mu ) t } t^{n-1} dt \right ) = \frac{n \lambda^n }{(\lambda + \mu) n! } \int_0^\infty dte^{- ( \lambda + \mu ) t } t^{n-1} $$

Just rinse and repeat.