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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

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  • $\begingroup$ Did you mean $C=(e^{vS}-1)(D+1/v),\,E[C]=E[(e^{vS}-1)(D+1/v)]$? $\endgroup$
    – J.G.
    Oct 16, 2019 at 10:12
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    $\begingroup$ 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. $\endgroup$
    – linofex
    Oct 16, 2019 at 10:17
  • $\begingroup$ 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. $\endgroup$
    – J.G.
    Oct 16, 2019 at 10:19
  • $\begingroup$ I am trying to study paper that talks about this stuff. The paper is dropbox.com/s/wx25iludgchit14/… $\endgroup$
    – linofex
    Oct 16, 2019 at 10:26
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    $\begingroup$ Yes, added. We can also talk through a chat.. $\endgroup$
    – linofex
    Oct 16, 2019 at 10:28

2 Answers 2

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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 $\infty$.

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    $\begingroup$ But I have to compute a mean time of a process, that should be finite. I edited the question with more information. $\endgroup$
    – linofex
    Oct 16, 2019 at 10:09
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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 $\mu=\infty$.

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

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