Stochastic processes - claim arrival process

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April 3rd 2011, 01:49 PM
allandmunja

Stochastic processes - claim arrival process

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

Well, I have tried to calculate expected value of $\nu$ first and then $S_{\nu}$ has the distribution equal to $Exp(\lambda )$ convolved ${\bf E}\nu$ 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?
April 3rd 2011, 03:54 PM
theodds

It will be easier to derive the distribution of $S_\nu$ if you consider the distribution of $S_\nu | \nu$ and $\nu$ . With that, you can get the joint distribution rather easily, and then it is just a matter of integrating out $\nu$ . I worked it out in a few minutes; it really isn't so bad if you attack it from that angle.
April 8th 2011, 10:59 AM
allandmunja

I've worked on this for some time... Well, if I'm correct the joint density function will be $f_{S_{\nu}|\nu =n}(x)=\sum_{n=1}^{\infty}\lambda e^{-\lambda x}\frac{(\lambda u)^{n-1}}{(n-1)!}\cdot (1-p)^{n -1}$ . Than the distribution function I am looking for becomes $P(S_{\nu}\le x)=\int_{0}^{x}\sum_{n=1}^{\infty}\lambda e^{-\lambda u}\frac{(\lambda u)^{n-1}}{(n-1)!}\cdot (1-p)^{n -1}du$ . Am I right and this is the final answer or something else can be done?
April 8th 2011, 03:53 PM
theodds

The first expression you give is the marginal density of $S_\nu$ if I recall correctly; you are summing out $\nu$ 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.
April 8th 2011, 04:45 PM
matheagle

Why dont you try using moment generating functions where you use the double expected value.
where you first condition on the number of terms.
April 9th 2011, 02:27 AM
allandmunja

$P(S_{\nu}\le x)=\int_{0}^{x}\lambda e^{-\lambda u}\sum_{n=0}^{\infty}\frac{(\lambda u(1-p))^{n}}{(n)!} du=\int_{0}^{x}\lambda e^{-\lambda u}\cdot e^{\lambda u-\lambda u p}=\int_{0}^{x}\lambda e^{-\lambda u p}=\frac{1}{p}P(X\le x)$ , where $X\sim Exp(\lambda p)$
Am I right?
April 9th 2011, 02:45 AM
allandmunja

@matheagle, did I understood correctly: you are suggesting taking ${\bf E}[{\bf E}[e^{S_{\nu}}]|\nu ]$ ? I'll try that...
April 9th 2011, 02:47 AM
allandmunja

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April 9th 2011, 06:55 AM
matheagle

Quote:

Originally Posted by allandmunja

@matheagle, did I understood correctly: you are suggesting taking ${\bf E}[{\bf E}[e^{S_{\nu}}]|\nu ]$ ? 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
April 9th 2011, 07:22 AM
theodds

Quote:

Originally Posted by allandmunja

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

I just noticed that you forgot to put a $p$ in the pmf of $\nu$ ; as is, the pmf of $\nu$ does not sum to 1. But yes, this is correct, and after multiplying your answer by the factor of $p$ that was left out, you will be done :)
April 9th 2011, 08:29 AM
allandmunja

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