# Compute mean value of the exponential of product of 2 exponential variables

I have to compute $$E[e^{vS}]$$, where $$v$$ and $$S$$ are two independent RVs exponentially distributed with different parameters, $$\lambda_1, \lambda_2$$.

According to this question I started computing $$Y = vS$$, but then it is hard to continue with the integral and $$e^Y$$

How can I solve it?

EDIT 1

In particular, I have to compute the mean completion time $$C$$ of processes in a queuing system with priorities: $$E[C] = (E[e^{vS}]-1)*({E[D]+ \frac{1}{v}})$$ so $$v$$ is an interarrival time, $$S$$ is a service time and $$D$$ is the duration of the process with higher priority.

EDIT 2 The paper that I am studying is this, in I am trying to solve $$E[C]$$ and $$E[C^2 ]$$ on page 8. In the start pages, there is the therminolgy used

• Did you mean $C=(e^{vS}-1)(D+1/v),\,E[C]=E[(e^{vS}-1)(D+1/v)]$?
– J.G.
Oct 16, 2019 at 10:12
• No, $C$ in the paper that I am studying is $C = S + \sum_{i=0}^{N} S'(i) + \sum_{i=0}^{N} D(i)$, where $S, S'$ and $D$ are times. Oct 16, 2019 at 10:17
• So where does $v$ get involved? To be honest, I can't see why multiplying exponential variables would be part of a model of how long anything takes.
– J.G.
Oct 16, 2019 at 10:19
• I am trying to study paper that talks about this stuff. The paper is dropbox.com/s/wx25iludgchit14/… Oct 16, 2019 at 10:26
• Yes, added. We can also talk through a chat.. Oct 16, 2019 at 10:28

Your mean is$$\int_0^\infty dv\int_0^\infty dS\lambda_1\lambda_2\exp(vS-\lambda_1v-\lambda_2S)=\int_0^\infty dv\int_0^\infty dS\lambda_1\lambda_2\exp((S-\lambda_1)(v-\lambda_2)-\lambda_1\lambda_2).$$The region $$S>\lambda_1,\,v>\lambda_2$$ makes an infinite contribution $$\lambda_1\lambda_2\exp(-\lambda_1\lambda_2)\int_0^\infty da\int_0^\infty db\exp(ab)$$ with $$a:=S-\lambda_1,\,b:=\lambda_2$$, so the final result is $$\infty$$.

• But I have to compute a mean time of a process, that should be finite. I edited the question with more information. Oct 16, 2019 at 10:09

Making a start:$$\mu=\int_{0}^{\infty}\int_{0}^{\infty}e^{xy}\lambda_{1}\lambda_{2}e^{-\lambda_{1}x}e^{-\lambda_{2}y}dxdy=\lambda_{1}\lambda_{2}\int_{0}^{\infty}e^{-\lambda_{2}y}\int_{0}^{\infty}e^{x\left(y-\lambda_{1}\right)}dxdy$$

Now observe that $$\int_{0}^{\infty}e^{x\left(y-\lambda_{1}\right)}dx=\infty$$ if $$y>\lambda_{1}$$ and consequently $$\mu=\infty$$.

• But I have to compute a mean time of a process, that should be finite. I edited the question with more information. Oct 16, 2019 at 10:10