Variance of Poisson and Exponential distribution

Question

I have encountered a problem with the variance of Poisson and exponential distribution. Suppose there is an accident that follows a Poisson distribution with an occurrence rate of $\lambda$ per hour. The fuel needed for each accident follows an exponential distribution with mean $\mu$. I want to know the distribution of the fuel consumption per hour. The mean is quite straightforward, it should be just $\lambda * \mu$, but what about the variance(standard deviation)?

Answer 1

Let's formalize the problem. We have $N$~Pois$(\lambda)$, $X_i \overset{i.i.d}{\sim} Exp(1/\mu)$ and we are interested in $$S= \begin{cases} \sum_{i=1}^{N} X_i, & \text{if } N > 0 \\ 0, & \text{if } N = 0 \end{cases}$$

$\DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\E}{E} $

This is a case of compound Poisson distribution, and we can derive formulas for its variance and expectation.

Using law of total variance: $$\Var (S) = \E(\Var(S|N)) + \Var(\E(S|N)) \label{eq:pythagoras}.$$ Then we need derive $\Var(S|N)$ and $\E(S|N)$ : $$\E(S|N=n) = \E(\sum_{i=1}^N X_i|N=n) = \E(\sum_{i=1}^n X_i|N=n) = n \E(X_i|N=n) = n\E(X_i),$$ therefore $\E(S|N) = N \E (X_i) $. Similarly we get that $\Var(S|N) = N \Var(X_i) $.

Now let's substitute it to our formula for $\Var(S)$: $$\Var(S) = \E(N\Var(X_i)) + \Var(E(X_i) N).$$

Here $\Var(X_i)$ and $E(X_i)$ are just constants, therefore: $$\Var(S)= \E(N) \Var(X_i) + (\E (X_i) )^2 \Var(N).$$

In our problem $\E(X_i)=\mu, \Var(X_i)=\mu^2, \E(N)=\Var(N)=\lambda$.

Therefore $\Var(S) = 2 \mu^2 \lambda$.

Answer 2

Let $E_1,E_2,...\sim \exp(\frac{1}{\mu})$ be iid.

Fix $t\geq 0$, and let $A(t)\sim \text{Poisson}(\lambda t)$ denote the number of accidents that took plane on $[0,t]$. Assume $t$ is in hours and $A(t),E_1,E_2,...$ are independent.

You're trying to find the distribution of the random variable $$C(t)=E_1+\dots +E_{A(t)}$$

$C(t)$ is the amount of fuel consumed on $[0,t]$.

If we take $\mathcal{E}(t)\sim \text{Poisson}(t/\mu)$ then for $\tau \geq 0$ we have $C(t)\leq \tau$ if and only if $\mathcal{E}(\tau)\geq A(t)$ whose probability doesn't have a nice closed form, so from here I decided to try another approach.

By conditioning on $A(t)$ and using double expectation, we see that the moment generating function of $C(t)$ is $$\begin{eqnarray*}M_{C(t)}(s) &=& \mathbb{E}\Big(e^{C(t)s}\Big) &=& \mathbb{E}\left((1-\mu s)^{-A(t)}\right) &=& e^{-\lambda t}e^{\frac{\lambda t}{1-\mu s}}\end{eqnarray*}$$

But this means $$e^{-\lambda t}e^{\frac{\lambda t}{1-\mu s}}=\int_{0}^{\infty}f_{C(t)}(\tau)e^{s\tau}\mathrm{d}\tau$$

We can now say $f_{C(t)}(\tau)$ is $e^{-\lambda t}\mathcal{L}^{-1}\Big(e^{\frac{\lambda t}{1+\mu s}}\Big)(\tau)$ where $\mathcal{L}$ is the Laplace transform. (Here $\mathcal{L}$ maps a function of $\tau$ to a function of $s$).

This inverse Laplace transform doesn't seem to exist since $e^{\frac{\lambda t}{1+\mu s}}$ doesn't approach $0$ as $s$ gets large.

Does this mean the pdf of $C(t)$ isn't of exponential order?

Let's try another method.

We know that $C(t)|A(t)\sim \text{Gamma}(A(t),\frac{1}{\mu})$ (in shape$-$rate parameterization) and $A(t)\sim \text{Poisson}(\lambda t)$ so by using total law $f_{C(t)}(\tau)=\sum_{k=0}^{\infty}f_{C(t)|A(t)=k}(\tau|k)\mathbb{P}(A(t)=k)$ we have $$f_{C(t)}(\tau)=\frac{1}{\tau}\exp\Big(-\frac{\tau}{\mu}-\lambda t\Big) \sum_{k=1}^{\infty}\frac{1}{k!(k-1)!}\Big(\frac{\lambda t }{\mu}\tau\Big)^k+f_{C(t)|A(t)=0}(\tau|0)e^{-\lambda t}$$ But $f_{C(t)|A(t)=0}(\tau|0)=\delta(\tau)$ and, according to Wolfram alpha, $\sum_{k=1}^{\infty}\frac{1}{k!(k-1)!}\Big(\frac{\lambda t }{\mu}\tau\Big)^k=\sqrt{\frac{\lambda t}{\mu}\tau}\cdot \mathcal{I}_1\Big(2\sqrt{\frac{\lambda t}{\mu}\tau}\Big)$ where $\mathcal{I}_1$ is the modified Bessel function of the first kind. Finally, $$f_{C(t)}(\tau)=\frac{1}{\tau}\exp\Big(-\frac{\tau}{\mu}-\lambda t\Big)\sqrt{\frac{\lambda t}{\mu}\tau}\cdot \mathcal{I}_1\Bigg(2\sqrt{\frac{\lambda t}{\mu}\tau}\Bigg)+e^{-\lambda t}\delta(\tau)$$