Distribution of the random variable $W=\frac{2}{3} \sum_{i=1}^{20}|X_i-2|$ is

Question

Let $X_1, \ldots, X_{20}$ be independent and identically distributed random variables with the common probability density function

$$ f(x) = \frac{1}{6} e^{-\frac{|x-2|}{3}}, \qquad -\infty < x < \infty. $$

Then distribution of the random variable $W=\frac{2}{3}\sum_{i=1}^{20}|X_i-2|$ is?

I proceeded this way:

$$ f(x) = \frac{1}{3} e^{-\frac{|x-2|}{3}}, \qquad x > 0. $$

$Y=|X-2|$ follow $\mathrm{Gamma}(1,\frac{1}{3})$, and we know if $X$ follows $\mathrm{Gamma}( \alpha , \lambda )$ then $2\lambda X$ follows $\chi^2(2\alpha)$.

This means $\frac{2}{3} Y$ follows $\chi^2(2)$.

So my answer is $\chi^2(40)$. But answer to this question is $\chi^2(20)$.

Answer 1

Such things are mechanized nowadays. For example, the code of Mathematica 13.1

k = ProbabilityDistribution[1/6*Exp[-RealAbs[x - 2]/3], {x, -Infinity, Infinity}];
Probability[2/3*Sum[RealAbs[z[j] - 2], {j, 1, 5}] < x, 
Table[z[j] \[Distributed] k, {j, 1, 5}]]

produces $$\frac{1}{384} e^{-\frac{x}{2}} \left(-x^4-8 x^3-48 x^2-192 x+384 e^{x/2}-384\right), x>0, $$ and $0$ otherwise for the CDF. It takes approximately half an hour. If you have many time, try it with $20$ instead of $5$.