# Compound Poisson process with exponential distribution

Consider the following shock model. The count of shocks within a certain time $t$ is a Poisson process $N(t)$ with parameter $\lambda$, while every shock brings damage $Y_i$ to the subject, which is exponentially distributed with parameter $\mu$. The subject can at most withstand damages of $a$. So the subject has a lifespan of $T$. Now how to calculate $E[T]$?

My attempt: Since $T\ge 0$, so \begin{align} E[T]=&\int_0^{+\infty} P(T>t)\,dt\\ =&\int_0^{+\infty} P\left(\sum_{i=1}^{N(t)}Y_i\le a\right)\,dt\\ =&\int_0^{+\infty}\left( \sum_{n=1}^{+\infty} P\left(\sum_{i=1}^{n}Y_i\le a\right)\frac{(\lambda t)^n}{n!}e^{-\lambda t}\right)\,dt\\ =&\sum_{n=1}^{+\infty} P\left(\sum_{i=1}^{n}Y_i\le a\right)\int_0^{+\infty}\frac{(\lambda t)^n}{n!}e^{-\lambda t}\,dt\\ =&\frac{1}{\lambda}\sum_{n=1}^{+\infty} P\left(\sum_{i=1}^{n}Y_i\le a\right)\\ =&\frac{1}{\lambda}\sum_{n=1}^{+\infty}\int_0^a\left(\mu e^{-\mu \xi}\cdot \mathbf{1}_{\xi\ge 0}\right)^{*n}\,d\xi\\ =&\frac{1}{\lambda}\int_0^a \mathcal{F}^{-1}\left[\sum_{n=1}^{+\infty}\mathcal{F}\left[\mu e^{-\mu\xi}\cdot \mathbf{1}_{\xi\ge 0}\right]^n\right]\,d\xi\\ =&\frac{1}{\lambda}\int_0^a \mathcal{F}^{-1}\left[\sum_{n=1}^{+\infty}\left(\frac{i\mu}{s+i\mu}\right)^n\right]\,d\xi\\ =&\frac{1}{\lambda}\int_0^a \mathcal{F}^{-1}\left[\frac{i\mu}{s}\right]\,d\xi\\ =&\frac{1}{\lambda}\int_0^a \frac{\mu}{2}\mathrm{sign}(\xi)\,d\xi\\ =&\frac{\mu a}{2\lambda} \end{align}

But my answer is wrong, and the correct one is $(1+\mu a)/\lambda$. I don't know where I mess things up. Any hint will be appreciated, thank you.

EDIT: I am using Fourier transforms defined as $$\mathcal{F}[f]=E[e^{isT}]=\int_{-\infty}^{+\infty}f(t)e^{ist}\,dt$$ and $$\mathcal{F}^{-1}[f]=\frac{1}{2\pi}\int_{-\infty}^{+\infty}f(t)e^{-ist}\,dt$$

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To simplify the proof, note that $$T=\sum\limits_{k=1}^{+\infty}X_k\cdot\mathbf 1_{A_k},\quad\text{with}\quad A_k=[Y_1+\cdots+Y_{k-1}\leqslant a],$$ where $(X_k)_{k\geqslant1}$ is the sequence of interarrival times of the process $(N(t))_{t\geqslant0}$. Hence $\mathrm E(X_k)=\frac1\lambda$ for every $k\geqslant1$ and, by independence of the processes $(X_k)_{k\geqslant1}$ and $(Y_k)_{k\geqslant1}$, $$\mathrm E(T)=\frac1\lambda\sum\limits_{k=1}^{+\infty}\mathrm P(A_k).$$ To compute the sum of the last series, consider the Poisson process $(M(t))_{t\geqslant0}$ associated to the exponentially distributed interarrival times $(Y_k)_{k\geqslant1}$. For every $k\geqslant1$, $A_k=[M(a)\geqslant k-1]$, hence $$\mathrm E(T)=\frac1\lambda\sum\limits_{k=0}^{+\infty}\mathrm P(M(a)\geqslant k)=\frac1\lambda\cdot\mathrm E(1+M(a)).$$ The intensity of the Poisson process $(M(t))_{t\geqslant0}$ is the inverse of $\mathrm E(Y_k)$ and $\mathrm E(Y_k)=\frac1\mu$, hence $\mathrm E(M(a))=a\mu$ and, finally, $$\mathrm E(T)=\frac{a\mu}\lambda.$$

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I think

$$P\left(A\leq\sum_{i=1}^{n}Y_i\leq A+{\mathrm d}A\right)= \begin{array}{ll} \Bigg\{ & \begin{array}{ll} \mu ^n\frac{A^{n-1}}{(n-1)!}e^{-\mu A}\,{\rm d}A, & A\geq 0 \\ 0, & A<0 \\ \end{array} \end{array}$$

so the sum is evaluated to

$$\begin{split} \sum_{n=1}^{+\infty}P\left(\sum_{i=1}^{n}Y_i\leq a\right)&= \sum_{n=1}^{+\infty}\int_0^a \mu ^n\frac{A^{n-1}}{(n-1)!}e^{-\mu A}\, {\mathrm d}A\\ &=\int_0^a \sum_{n=1}^{+\infty}\mu ^n\frac{A^{n-1}}{(n-1)!}e^{-\mu A}\, {\mathrm d}A\\ &=\mu a \end{split}$$

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Possibly not a complete answer, but I have a qualm about this part: $$P\left(\sum_{i=1}^{N(t)}Y_i\le a\right) = \sum_{n=1}^{+\infty} P\left(\sum_{i=1}^{n}Y_i\le a\right)\frac{(\lambda t)^n}{n!}e^{-\lambda t}$$

I would instead write $$P\left(\sum_{i=1}^{N(t)}Y_i\le a\right) = \sum_{n=0}^{+\infty} P\left(\sum_{i=1}^{n}Y_i\le a\right)\frac{(\lambda t)^n}{n!}e^{-\lambda t}.$$

There is positive probability that $N(t)=0$. In the one term where we have $\displaystyle \sum_{i=1}^0$, I would construe that sum as being $0$. Then you're considering the probability that $0\le a$, which is $1$.

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Thank you, and now it's $(1+\mu a/2)\lambda^{-1}$, which seems closer. –  ziyuang Jun 12 '12 at 21:42
So it appears that the only remaining issue is in finding the inverse Fourier transform. –  Michael Hardy Jun 12 '12 at 22:34
But those two transforms are done by Mathematica, with FourierParameters -> {1, 1} –  ziyuang Jun 12 '12 at 22:55
I'll take a look at this further maybe tonight or tomorrow...... –  Michael Hardy Jun 13 '12 at 0:50