My problem is to calculate $E[\max(S-5000, 0)]$ where $$S = \sum_{i=1}^{N} X_i,$$ $N$ is a random variable with geometric distribution, parametrized as follows: $$P(N=n) = \frac{\beta^n}{(1+\beta)^{n+1}}, ~~~~~ n=\{0,1,2, ...\}$$ and ${X_1, X_2...}$ are all independent and exponentially distibuted with density: $$f_X(x) = \lambda e^{-\lambda x}.$$

I came up with an idea that I can find the density function of $S$ which is (according to my knowledge): $$f_S(s) = \sum_{n=0}^{\infty} f_{X_1 + X_2 + ... +X_n}(s) \cdot P(N=n).$$

However, I encountered a problem - $N$ may have values $\{0,1,2,...\}$ (natural numbers WITH zero) but I can't calculate the density of the sum of zero random variables (the number of random variables to be sumed is zero when $n=0$).

Is it possible to calculate such density? It is there any other way to calculate $E[\max(S-5000, 0)]$?

  • $\begingroup$ en.wikipedia.org/wiki/Wald's_equation Note that your question about $N=0$ is basically answered by the convention for the empty sum: when $N=0$ you have $S=0$. Wald's equation is consistent with this convention. $\endgroup$ – Ian Jun 2 '16 at 17:53
  • $\begingroup$ Though Wald's equation itself won't actually help you with the maximum part of the problem, hmm... $\endgroup$ – Ian Jun 2 '16 at 18:14
  • $\begingroup$ But if I consider that $f_S(s) = 0$ for $n=0$ and calculate the rest of the series and after that I calculate the integral $$\int_{0}^{\infty}f_S(s)ds$$ it turns out that it doesn't equal 1... $\endgroup$ – Jerry Jun 2 '16 at 18:15
  • $\begingroup$ Ah, I managed to write $1-\beta$ instead of $1+\beta$. My apologies. $\endgroup$ – Math1000 Jun 2 '16 at 20:15

First note that the sum of $n$ i.i.d. exponential random variables has an Erlang distribution, that is, for any $n\geqslant1$, the sum $\sum_{i=1}^n X_i$ has probability density

$$\frac{(\lambda t)^{n-1}}{(n-1)!}\lambda e^{-\lambda t}\mathsf 1_{(0,\infty)}(t). $$ (How sum of exponential variables is a gamma variable) It follows then that the distribution of $S$ is given by $\nu_S = \nu_S^d + \nu_S^c$, where for each Borel set $B$, $$\nu_S^d(B) =\mu(B\cap\{0\})\mathbb P(N=0) $$ where $\mu$ is counting measure, and $$\nu_S^c(B)=\int_B f_S\ \mathsf d\lambda$$ where $\lambda$ is Lebesgue measure. We compute $$\nu_S^d(B) =\mu(B\cap\{0\})(1+\beta)^{-1} $$ and \begin{align} f_S(t) &= \sum_{n=0}^\infty f_{S\mid N=n}(t)\mathbb P(N=n)\\ &= \sum_{n=1}^\infty \frac{(\lambda t)^{n-1}}{(n-1)!}\lambda e^{-\lambda t}\frac{\beta^n}{(1+\beta)^{n+1}}\\ &= \frac{\beta \lambda e^{-\lambda t}}{(1+\beta)^2}\sum_{n=0}^\infty\frac{\left(\frac{\beta\lambda t}{1+\beta}\right)^n}{n!}\\ &=\frac{\beta\lambda}{(1+\beta)^2} \exp\left(-\frac{\lambda t}{1+\beta}\right)\mathsf 1_{(0,\infty)}(t). \end{align} It follows that \begin{align} \mathbb E[(S-5000)^+] &= \int_{5000}^\infty (t-5000)\frac{\beta\lambda}{(1+\beta)^2} e^{-\frac{\lambda t}{1+\beta}}\ \mathsf dt\\ &= \frac\beta\lambda \exp\left(-\frac{5000\beta\lambda}{1+\beta}\right). \end{align}

  • $\begingroup$ That is exactly the form of $f_S(t)$ that I managed to derive on my own. But notice that it is not a valid probability density function, since $$\int_{0}^{\infty}\frac{\beta\lambda}{(1+\beta)^2} \exp\left(-\frac{\lambda t}{1+\beta}\right) dt = \\ \frac{\beta}{1+\beta} \int_{0}^{\infty}\frac{\lambda}{1+\beta} \exp\left(-\frac{\lambda t}{1+\beta}\right) dt = \\ = \frac{\beta}{1+\beta} $$ I used the fact that $f(t) = \frac{\lambda}{1+\beta} \exp\left(-\frac{\lambda t}{1+\beta}\right)$ is the density of exponential distribution, so it's integral from $0$ to $\infty$ equals 1. $\endgroup$ – Jerry Jun 2 '16 at 21:04
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    $\begingroup$ @Jerry You are forgetting to add the probability that $S=0$. $\endgroup$ – Math1000 Jun 3 '16 at 11:28

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