I'm trying to work on the following problem:
Given n=8 random variables $(X_1, ..., X_8)$ where $X_i$ ~ $U(0,1)$.
Let $Y=ln(\frac{1}{X_{1}\cdot\ldots\cdot X_{8}})$
Prove that $P(Y\leq3)=F_V(6)$ where $V$~$\chi^2(16)$.
This is what I tried:
$P(Y\leq3)=P(ln(\frac{1}{X_{1}\cdot\ldots\cdot X_{8}})\leq3)=P((\frac{1}{X_{1}\cdot\ldots\cdot X_{8}})\leq e^3)=P(e^{-3} \leq X_1*...*X_8)=1-P\left(e^{-3}\geq X_{1}\cdot\ldots\cdot X_{8}\right)$
Also we know that $\chi^2(n)\equiv\Gamma(n/2,1/2)$, so using $F_{X}\left(x\right)=\begin{cases} 1-\sum_{k=0}^{n-1}\frac{\left(\lambda\cdot x\right)^{k}\cdot e^{-\lambda\cdot x}}{k} & x>0\\ 0 & else \end{cases}$
We get that $F_{V}\left(t\right)=1-\sum_{k=0}^{7}\frac{\left(\frac{1}{2}\cdot t\right)^{k}\cdot e^{-\frac{1}{2}\cdot t}}{k}$
so $F_{V}\left(6\right)=1-\sum_{k=0}^{8}\frac{3^{k}\cdot e^{-3}}{k}$
But I couldn't understand why both terms are equal?
$\sum_{k=0}^{8}\frac{3^{k}\cdot e^{-3}}{k}=P\left(e^{-3}\geq X_{1}\cdot\ldots\cdot X_{8}\right)$