What is the distribution of $S_n$ [closed]

Let $x_1 ,\ldots ,x_n $ be a random sample of size $n$ taken from the generalized inverted exponential distribution with pdf $$f(x|\lambda , \theta )=\frac{\lambda}{\theta}x^{-2}\exp \left(-\frac{\lambda}{x}\right)\left[1-\exp \left(-\frac{\lambda}{x}\right)\right]^{\frac{1}{\theta}-1}, \; x>0,\; \lambda ,\theta >0 .$$ What is the distribution of $$S_n =-\sum_{i=1}^{n}\ln \left(1-\exp \left(-\frac{\lambda}{x_i }\right)\right)?$$ What is $E(S_n)$?

Thanks in advance.

$$f_X(x)=\frac{\lambda}{\theta}\frac{1}{x^2}e^{-\lambda/x}[1-e^{-\lambda/x}]^{1/\theta-1}$$

set

$$Y=-\ln(1-e^{-\lambda/X})$$

that is

$$x=-\frac{\lambda}{\ln(1-e^{-y})}$$

$$x'=\lambda\frac{e^{-y}}{(1-e^{-y})\ln^2(1-e^{-y})}$$

the use the formula

$$f_Y(y)=f_X(g^{-1}(y))\left|\frac{d}{dy}g^{-1}(y)\right|$$

substitute, simplify the expression and get the result.

your random variable $S_n$ is the sum of $n$ independent exponential thus it is a Gamma

$$S_n\sim\text{Gamma}[n;\theta]$$

with $\theta$ as scale parameter, thus its expectation is

$$\mathbb{E}[S_n]=n\theta$$

and variance

$$\mathbb{V}[S_n]=n\theta^2$$