Chi-squared distribution related to the mean of an exponential sample

Question

I'm trying to solve the following problem :

Let $T$ be the units of time until some device fails. This random variable has the following distribution density:

$f(t,\phi) = \frac{1}{\phi}e^{-\frac{t}{\phi}},t>0,\phi>0$

I estimated $\phi$ using the sample mean with the maximum likelihood estimator based on a sample $\{T_1,T_2,T3,\dots,T_n\}.$

Well, now it asks me to prove that the random variable $U=\frac{2\sum_{k=1}^nT_k}{\phi}$ has a chi squared distribution with $2n$ degrees of freedom? I don't have any idea of how I might prove this. Can someone offer some indications / a solution?

Not necessarily complete, but it would be really helpful to contain as many steps of the proof as possible! Or at least enough to help me understand how to solve it.

Thanks!

Answer 1

The exponential distribution with rate $\lambda = 1/\varphi = 1$ has a chi-squared distribution with 2 degrees of freedom. (See Wikipedia on 'exponential distribution' or your text.) Accordingly, the distribution of the sum of $n$ independent exponential random variables is related to a chi-squared distribution with $2n$ degrees of freedom. You can prove this using moment generating functions.

Just to illustrate: Below is a histogram resulting from many samples of size $n = 5$ from $\mathsf{Exp}(\lambda = 1/3)$ and the corresponding values of $U$. The density of $\mathsf{Chisq}(df = 10)$ is superimposed on this histogram. (Of course, this figure is not a proof, but it is an indication that the distribution of $U$ is as stated in the problem.)

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Note: In case it is of any interest, here is the code used to make the figure in R statistical software:

m = 10^6;  n = 5;  phi=3
t = replicate( m, sum(rexp(n, 1/phi)) )
u = 2*t/phi
mean(u);  var(u)
## 10.0039    # aprx mean of CHISQ(10);  exact = 10
## 19.96353   # aprx variance of CHISQ(10)  exact = 20
hist(u, prob=T, col="skyblue2")
curve(dchisq(x, 2*n), add=T, lwd=2, col="maroon")