Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

(About the first step)

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 $$\mathcal{F}^{-1}[f]=\frac{1}{2\pi}\int_{-\infty}^{+\infty}f(t)e^{-ist}\,dt$$

share|cite|improve this question
up vote 1 down vote accepted

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, $$ \mathrm E(T)=\frac{a\mu}\lambda. $$

share|cite|improve this answer

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

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

share|cite|improve this answer

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 $1$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.