# Thread: Stochastic processes - claim arrival process

1. ## Stochastic processes - claim arrival process

Hey guys! I am working with stochastic processes and came round the flowing problem:
$\displaystyle \tau_1,\tau_2,...\sim Exp(\lambda)$ are iid and $\displaystyle \nu$ is such independent random variable that $\displaystyle P(\nu =n)=(1-p)^{n-1},n\ge 1$. What is the distribution of $\displaystyle S_{\nu}=\tau_1+...+\tau_{\nu}$.

Well, I have tried to calculate expected value of $\displaystyle \nu$ first and then $\displaystyle S_{\nu}$ has the distribution equal to $\displaystyle Exp(\lambda )$ convolved $\displaystyle {\bf E}\nu$ times. Bot this somehow doesn't feel right. However, this looks awfully similar to compound Poison process but claim arrival times are not Poison distributed... Could anyone give a hint on this problem?

2. It will be easier to derive the distribution of $\displaystyle S_\nu$ if you consider the distribution of $\displaystyle S_\nu | \nu$ and $\displaystyle \nu$. With that, you can get the joint distribution rather easily, and then it is just a matter of integrating out $\displaystyle \nu$. I worked it out in a few minutes; it really isn't so bad if you attack it from that angle.

3. I've worked on this for some time... Well, if I'm correct the joint density function will be $\displaystyle f_{S_{\nu}|\nu =n}(x)=\sum_{n=1}^{\infty}\lambda e^{-\lambda x}\frac{(\lambda u)^{n-1}}{(n-1)!}\cdot (1-p)^{n -1}$. Than the distribution function I am looking for becomes $\displaystyle P(S_{\nu}\le x)=\int_{0}^{x}\sum_{n=1}^{\infty}\lambda e^{-\lambda u}\frac{(\lambda u)^{n-1}}{(n-1)!}\cdot (1-p)^{n -1}du$. Am I right and this is the final answer or something else can be done?

4. The first expression you give is the marginal density of $\displaystyle S_\nu$ if I recall correctly; you are summing out $\displaystyle \nu$ so it is no longer conditional on it.

Yes, there is more you can do. You can simplify the infinite summation. Move the constants outside the summation and, hopefully, you will recognize the well-known form of the summation.

5. Why dont you try using moment generating functions where you use the double expected value.
where you first condition on the number of terms.

6. $\displaystyle P(S_{\nu}\le x)=\int_{0}^{x}\lambda e^{-\lambda u}\sum_{n=0}^{\infty}\frac{(\lambda u(1-p))^{n}}{(n)!} du=\int_{0}^{x}\lambda e^{-\lambda u}\cdot e^{\lambda u-\lambda u p}=\int_{0}^{x}\lambda e^{-\lambda u p}=\frac{1}{p}P(X\le x)$, where $\displaystyle X\sim Exp(\lambda p)$
Am I right?

7. @matheagle, did I understood correctly: you are suggesting taking $\displaystyle {\bf E}[{\bf E}[e^{S_{\nu}}]|\nu ]$? I'll try that...

8. -

9. Originally Posted by allandmunja
@matheagle, did I understood correctly: you are suggesting taking $\displaystyle {\bf E}[{\bf E}[e^{S_{\nu}}]|\nu ]$? I'll try that...
that should work because the second expectation is just a geometric series
But I dont know how you write your parameter in the exponential

10. Originally Posted by allandmunja
$\displaystyle P(S_{\nu}\le x)=\int_{0}^{x}\lambda e^{-\lambda u}\sum_{n=0}^{\infty}\frac{(\lambda u(1-p))^{n}}{(n)!} du=\int_{0}^{x}\lambda e^{-\lambda u}\cdot e^{\lambda u-\lambda u p}=\int_{0}^{x}\lambda e^{-\lambda u p}=\frac{1}{p}P(X\le x)$, where $\displaystyle X\sim Exp(\lambda p)$
Am I right?
I just noticed that you forgot to put a $\displaystyle p$ in the pmf of $\displaystyle \nu$; as is, the pmf of $\displaystyle \nu$ does not sum to 1. But yes, this is correct, and after multiplying your answer by the factor of $\displaystyle p$ that was left out, you will be done

11. Thank you guys! I'm still calculating matheagle's solution, but it appears result is the same. Thanks again!