find the probability density function of $Z=-2\sum_{i=1}^{n} \ln F(x_i)$

Question

Let $x_1, x_2, x_3...,x_n$ be an independent and identically distributed random sample with CDF $F(X_i)$ where $i=1,2,...,n$. Find the probability density of $Z=-2\sum_{i=1}^{n} \ln F(x_i)$

my attempt:

i used a $$w=-2\ln F(x_i)$$ and then $$w/2=\ln F(x_i)$$ so $${e^{w/2}}=F(x_i)$$

but now i dont know what to do with the $F(x_i)$ and i tried to use the cumulative function technique, but i did not get a specific result

Answer 1

I assume that you mean that $Z=-2\sum_{i=1}^n\log(F(X_i))$, where $F$ is the distribution function of $X$ and that is why you couldn't get any further. The answer is that $Z \sim \chi_{2n}^2$.

Indeed, let us denote $F\equiv F_X$. Now, let us denote $Y_i:=-\log(F_X(X_i))$. Then, the distribution of $Y$ is: \begin{gather*} F_{Y_i}(y)=P(Y_i\leq y) = P(-\log(F_X(X_i))\leq y) = P(F_X(X_i)\geq e^{-y})= P(X\geq F^{-1}_X(e^{-y})) \\= 1 - F_X(F^{-1}_X(e^{-y})) = 1 - e^{-y},\ y>0 \end{gather*} This implies that $Y_i\sim \operatorname{exp}(1)$, and they are all independent, so $Z/2 = \sum_{i=1}^n Y_i\sim\gamma(a=1,p=n)$ and finally $Z\sim \chi_{2n}^2$. And well, hence, the density of $Z$ will be that of a chi-squared with $2n$ degrees of freedom.

Answer 2

Use the fact that if $Y\sim \mathcal U(0, 1)$, then $-\log(Y)\sim \exp(1)$. Using the summation property of Gamma distribution, the sum is just another Gamma distribution, and then you can have the density.