Chi-Squared 2df = 2nλ/λˆ MGF Needed?

Question

Question: Assume that $T_i (i = 1, ..., n)$ are independently identically distributed with exponential distribution with hazard $\lambda$. Assuming there is no censoring, prove that:

$$\frac{2nλ}{\hat \lambda} ∼ χ^2_{2n} $$ where $\hat \lambda$ is the MLE of $\lambda$.

Notes(adding all I can to help):

$$\hat \lambda = \frac{n}{\sum_{i=1}^n T_i}$$

I know this has something to do with the moment generating function, but I just can't get it to come out right.

Also, I believe that(if I remember correctly):

MGF of $\sum_{i=1}^n T_i$ is the same as MGF of $T_i$ raised to the $n^{\text{th}}$ power.

Any ideas?

Answer 1

I figured it out!!!!

Here's the answer in case anyone was interested:

Things I needed to know:

Expoential distribution is scalar. MGF of a sum of $n$ identical distributions is the MGF of one to the $n^{\text{th}}$ power.