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Studying for a test in course about stochastic processes, here's a test question that I can't fully understand:

An insurance company insures its policyholders against damages of a particular kind. Damages of this kind are reported by the policyholders according to a Poisson process with intensity λ = 10 (per month). For each reported such damage, the company pays out a random amount which is exponentially distributed with mean 5 (thousand dollars). The different amounts are independent of each other and of the number of damages. Let X be the total amount paid out by the company for this kind of damages during one month. Compute the moment generating function for X.

And here is the solution:

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I don't understand the last part of the solution. Specifically why the need to take the log of φZ. Perhaps anyone here can share some insight.

Sycorax
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    please [edit] your post to give it a more informative title. – Sycorax Aug 04 '21 at 20:04
  • Please add the [tag:self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. – kjetil b halvorsen Aug 04 '21 at 22:53
  • @kjetilbhalvorsen, excuse me? I was very clear that I didn't understand the last part of the solution, which clearly means that I understand everything up to that point, i.e., it’s obvious what I do and do not understand. Also, how is this considered homework? It's a question from a previous exam with its solution, a solution that I didn't fully understand. I was clearly asking a specific question, which I got explained, not asking for someone else to do some sort of homework for me. –  Aug 05 '21 at 08:35
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    The message that the moderator @kjetilbhalvorsen posted is a standard copy and paste message, so not all of its aspects may apply to your particular situation. So the part on "posting your homework & hoping someone will do it for you" doesn't really apply to you. As you are new here, any questions of a routine nature i.e. from a textbook or past exam should be tagged as self-study, even if it's an enquiry concerning conceptual difficulties with a solution. – microhaus Aug 05 '21 at 15:49

1 Answers1

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First, note that:

$\theta_X(t) = \mathbb{E}_N\theta_Z^N(t)$

(the expected value of the MGF of $Z$ raised to the $N^{th}$ power), as the sum of $N$ independent variates with MGF $\theta_Z(t)$ has MGF $\theta_Z^N(t)$.

Onwards! Expanding this expression gives us:

$\theta_X(t) = \sum_N \theta_Z^N(t)p(N;\lambda) = \sum_N e^{N \log \theta_Z(t)}p(N;\lambda)$

Since the definition of the MGF of $N$ is $\theta_N(t) = \sum_N e^{Nt}p(N;\lambda)$, we can clearly just replace $t$ in the expression for $\theta_N(t)$ with $\log \theta_Z(t)$ to get the MGF of $X$.

Note that this is just a pattern recognition thing; we know the expression for $\theta_N(t)$, and we can see that the expression for $\theta_X(t)$ will have the same form, just with $t$ replaced by $\log \theta_Z(t)$.

jbowman
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