This was a part question to what I posted here and it went like this:
Suppose $Y_1,Y_2,...$ are independently and identically distributed random variables with common distribution function $F$. For any positive integers $\alpha$ and $\beta$, derive the distribution of $$V=-\beta\sum_{i=1}^\alpha\log_e{F\left(Y_i\right)}$$
I proceeded with the following assumption (and as suggested by Oliver Diaz and @kimchilover in the tagged question), that $Y_i$'s are continuous random variables. So, $$X_i=F\left(Y_i\right)\sim U\left[0,1\right]$$ I defined $Z_i=\log_e{X_i}$. Thus, $Z_i\in\left(-\infty,0\right]$. The pdf of $Z_i$ comes out as follows: $$\begin{align}f_{Z_i}{\left(z_i\right)}&=f_{X_i}{\left(e^{z_i}\right)}\cdot \frac{d}{dz_i}e^{z_i}&,z_i\leq0\\ &=e^{z_i}&,z_i\leq0 \end{align}$$ I then proceeded to find the pdf of $A_i=-\beta Z_i$. (Thus, $A_i\in\left[0,\infty\right)$) $$\begin{align} f_{A_i}{\left(a_i\right)}&=f_{Z_i}{\left(-\frac{a_i}{\beta}\right)}\cdot\frac{d}{da_i}\left(-\frac{a_i}{\beta}\right)&,a_i\geq0\\ &=-\frac{e^{-a_i/\beta}}{\beta}&,a_i\geq0 \end{align}$$ But this is not a valid pdf. I'm stuck here thinking as to where I went wrong. My plan of action was this: I would find the pdf of $-\beta Z_i$, and then, I'll try to find the distribution of $V=-\beta\sum_{i=1}^\alpha Z_i$ using the mgf of $A_i$.
Can somebody point out the mistake in my process? Also, any hints and alternative methods are very much welcome.
