To find the distribution of the random variable based on uniform distribution

Question

Let $X_1,X_2,...,X_n$ be iid $U(-5,5)$ random variables. Then the distribution of the random variable $Y=-2\sum\limits_{i=1}^{10}\log(|X_i|/5)$ is

(A) $\chi_{10}^2$ (B) $10\chi_{2}^2$ (C) $\chi_{20}^2$ (D) $\frac{1}{2}\chi_{20}^2$

My Steps:

$$P(Y \leq y) = P(-2\sum\limits_{i=1}^{10}\log(|X_i|/5) \leq x) = P(\sum\limits_{i=1}^{10}(|X_i|/5) \geq 10e^{\frac{-x}{2}}$$

Now, the PDF of $\chi^2$ distribution with parameter $2$ is given by

$$f(x) = \frac{1}{2}e^{-x/2}$$

So, the answer should option (B)

Did I solve this correctly ?

EDIT: A sum of logs is a product inside the log and then it would become $20\log |X_i|$ .. so distribution would be $\chi_{20}^2$ ?

Answer 1

The first question you need to ask yourself is, if $X_i \sim \operatorname{Uniform}(-5,5)$, then what is the distribution of $Y_i = -2 \log (|X_i|/5)$? In other words, look at a single $X_i$ first. What is the support of $Y_i$?

Here is a hint: $$\Pr[Y_i \le y] = \Pr[-2 \log (|X_i|/5) \le y] = \Pr[|X_i| \ge 5 e^{-y/2}] = 2 \Pr[ X_i \ge 5e^{-y/2} ],$$ the last equality arising from the symmetry of the distribution of $X_i$.