Exact distribution of MLE exponential distribution

Question

Let $y_1, \dots,y_n$ be i.i.d. random variables from $Exp(\theta)$, where $\theta$ is scale parameter. I've found the MLE $$\hat \theta=\frac{\sum^{n}_{i=1}y_i}{n}$$

Now I need to find the exact distribution of $\hat \theta$.

I know that if $y\sim Exp(\theta)$ than $\sum^{n}_{i=1}y_i\sim Ga(n,\theta)$, so I would say $\hat \theta\sim Ga(n,n\theta)$. Is that correct?

Edit

The right solution is $\hat \theta\sim Ga(n,\frac{\theta}{n})$.

Answer 1

1) If $X\sim\mathcal{E}xp(\theta)$ where $f_X(x)=\theta e^{-\theta x}$, then $\hat{\theta}=1/\bar{Y}_n$. As such $\sum Y_i \sim Gam(n, \theta)$, then $1/\sum Y_i \sim \text{Inv-Gam}(n, 1/\theta)$. So, your MLE is $n$ times Inverse-Gamma random variable. You can derive its analytical expression by using $f_Y(y)=f_X(g^{-1}(y))|(g^{-1}(y))'|$.

2) If $X\sim\mathcal{E}xp(\theta)$ where $f_X(x)=1/\theta e^{-x/\theta }$, then your answer is correct.