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I'm given an exercise to find a distribution of a r.v. $S_N$ constructed as follows: $$S_N = \sum_{k=1}^{N+1} X_k$$ where $N \sim Geom(p), X_k \sim Exp(\lambda)$ for $p \in(0,1), \lambda >0$.

I recall from Non-Life Insurance course the following property of Laplace transform/PGF: $$L_{S_N}(t) = g_N(L_X(t))$$ Since $L_X(t) = \frac{\lambda}{\lambda+t}$ and $g_N(t) = \frac{p}{1-(1-p)t}$ , plugging in one expression into another I obtained: $$L_{S_N}(t) = p+(1-p)\frac{p\lambda}{p\lambda+t}$$

So here is my actual question. Is it possible to somehow recover a cdf from the given Laplace transform of the corresponding distribution ? I found a script where it's done but simply as a "property" but I would like to know some more detail and the theory behind it.

Any hint would be highly appreciated :)

john_jerome
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  • Use the law of total probability conditioning on N and you should be able to arrive at the conclusion. – user52227 Dec 10 '16 at 13:17
  • Yeah, that's the most straightforward approach and I tried it in the first place. The problem arises when you need to deal with a cdf of a sum of $n$ exponentially distributed variables (which has Erlang distribution). And that cdf isn't nice at all – john_jerome Dec 10 '16 at 13:21
  • There is no problem with that, do some calculations with the Erlang and several things will get simplified. – user52227 Dec 10 '16 at 13:30
  • I would rather use the characteristic function. So to use the inversion and find the cdf. As in the case of Laplace transform, there are several approaches in complex analysis which are for more comlicated than with the case of Laplace transform. – kolobokish Dec 10 '16 at 13:51
  • @user52227 I'm stuck with the following expression that can be obtained after using the mentioned lat ow total probability conditioning on N: $$F_{S_N}(x)=\sum_{n=1}^\infty P(S_n \le x)P(N=n)$$ where the first probability is the cdf of Erlang. Since we have an infinite sum then, what would be the next step to simplify the expression ? – john_jerome Dec 10 '16 at 21:11
  • https://math.stackexchange.com/q/634158/321264 – StubbornAtom May 11 '21 at 18:56

1 Answers1

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Fixed $n$, the R.V. $S_{n+1}$ has Erlang density, i.e., \begin{equation} \nonumber f_{S_{n+1}} (t) = \lambda^{n+1} \frac{t^n}{n!} e^{-\lambda t} \end{equation} which implies \begin{equation} \nonumber \Pr(S_{n+1}\le t) = \int_0^t \lambda^{n+1} \frac{z^n}{n!} e^{-\lambda z} \mbox{d}z. \end{equation} Therefore $$ \Pr(S_N\le t) = \Pr\left ( \bigcup_{n\ge 0} \left\{ S_N\le t \cap N=n\right\} \right)\\ = \sum_{n\ge 0} \Pr\left ( S_{N}\le t \cap N=n \right) \\ = \sum_{n\ge 0} \Pr\left ( S_{N}\le t| N=n\right) \Pr\left (N=n \right)\\ = \sum_{n\ge 0} \Pr\left ( S_{n+1}\le t\right) \Pr\left (N=n \right)\\ = \sum_{n\ge 0} \int_0^t \lambda^{n+1} \frac{z^n}{n!} e^{-\lambda z} \mbox{d}z \,p^n (1-p)\\ = \lambda (1-p) \int_0^t \sum_{n\ge 0} \frac{(p \lambda z)^n}{n!} e^{-\lambda z} \mbox{d}z \\ = \lambda (1-p) \int_0^t e^{p \lambda z} e^{-\lambda z} \mbox{d}z \\ = \lambda (1-p) \int_0^t e^{- \lambda(1-p) z} \mbox{d}z\\ = 1- e^{- \lambda(1-p) t}, $$ for which we recognize that $S_n$ is a random variable with the exponential law of parameter $\lambda(1-p)$. Note that I could invert the integral and summation because the involved terms are all positive.

The same conclusion can be obtained with a characteristic function argument.

user52227
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  • Great, thanks. I tried to work with the explicit cdf for Erlang distribution instead of the integral over pdf and that's the point I got stuck on. – john_jerome Dec 12 '16 at 17:49