I have problems with this exercise.

Let $X_1, X_2, \ldots $ r.v. independent and equally distributed exponential with parameter $\lambda > 1$. Verify if random variable


has finite mean and calculate the characteristic function.

First of all, I need to calculate

$$E[Y]=E\left[\sum_{n=1}^{\infty}\prod_{k=1}^{n}X_k\right]=\sum_{n=1}^{\infty}E\left[\prod_{k=1}^{n}X_k\right],$$ this is true becase the r.v. are positive.

$$\sum_{n=1}^{\infty}E\left[\prod_{k=1}^{n}X_k\right]=\sum_{n=1}^{\infty}\prod_{k=1}^{n}E\left[X_1\right],$$ this is true since r.v. are independent and equally distributed.

$$\sum_{n=1}^{\infty}\prod_{k=1}^{n}E\left[X_1\right]=\sum_{n=1}^{\infty}\prod_{k=1}^{n}1/\lambda=\sum_{n=1}^{\infty}(1/\lambda)^n=\frac{1}{\lambda-1},$$ since r.v. is exponential distributed.

Am I right? Or there is a wrong step? If I am right how I can calculate the characteristic function with that expected value?

Any help?

  • 4
    $\begingroup$ Some thoughts: $$Y=X_1+X_1X_2+X_1X_2X_3+\dots\\=X_1(1+\color{red}{X_2+X_2X_3+\dots})\\=X_1(1+\color{red}{Y'})\hspace{3cm}\tag1$$where $Y'$ is distributed exactly like $Y$, and $Y'$ is independent of $X_1$. If you take take the characteristic function of both sides (and use this for the ch.f. of an independent product), you get$$\phi_Y(t)=e^{it}\int_0^\infty \phi_Y(tx) (\lambda e^{\lambda x})\,dx\tag2$$(1) seems promising, (2) maybe less so. $\endgroup$ Commented Jan 21, 2022 at 19:27
  • $\begingroup$ Yes, that was the correct answer... there is no closed form. Thanks! $\endgroup$ Commented Feb 8, 2022 at 17:14

1 Answer 1


Everything is correct except for your last line. Instead:

$$\sum_{n=1}^{\infty}\prod_{k=1}^{n}E\left[X_1\right]=\sum_{n=1}^{\infty}\prod_{k=1}^{n}1/\lambda \color{red}=\sum_{n=1}^{\infty} \frac{1}{{\lambda^n}}= \frac{1}{\lambda - 1}$$

where in the last step we used the formula for a geometric series.

  • $\begingroup$ Oh, I see my mistake. I think I confused the multiplication and the sum symbols. Now I think for the characteristic function I have to integrate the expected value, right? $\endgroup$ Commented Jan 21, 2022 at 16:29
  • 1
    $\begingroup$ I don't see how one would use an expected value to calculate the characteristic function. $\endgroup$
    – Gregory
    Commented Jan 21, 2022 at 16:33
  • $\begingroup$ I agree with @Gregory. It's not obvious to me how one may use this mean to get the characteristic function. $\endgroup$ Commented Jan 21, 2022 at 16:36
  • $\begingroup$ Yes, I think it is independent from the expected value... but I got stuck when I try to calculate with $E[e^{itY}]$ $\endgroup$ Commented Jan 21, 2022 at 17:21

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