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
$$Y=\sum_{n=1}^{\infty}\prod_{k=1}^{n}X_k$$
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?
Best Answer
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.