My book says the following

Suppose $X_{i} \sim$ iid $Exp(1,\eta)$

Where $Exp(\theta,\eta)$ is the shifted exponential ie has density

$$\frac{1}{\theta}e^\frac{-(x-\eta)}{\theta}$$ for $x \ge \eta$ and zero otherwise.

Then my book states

$\bar X-\eta$ $\sim$ $Gamma(\frac{1}{n},n)$

But I do not see why this is true.

I tried using mgf but it is hard to work with sample mean in mgf.

So why is it true?

  • $\begingroup$ What expression for the the gamma pdf are you using? $\endgroup$ – Sycorax Apr 8 at 5:14
  • $\begingroup$ k Shape- $\theta$ Scale $\endgroup$ – Quality Apr 8 at 5:17

If we expand your final expression, we have:$$\bar{X}-\eta=\frac{1}{n}\sum ( X_i - \eta )$$

$X_i-\eta$ is classic exponential RV with parameter $\theta = 1$. By denition Sum of $n$ iid exp RVs (with parameter $\theta$) is gamma distributed, i.e. $G(n,\theta)$, so, the sum $\sum (X_i-\eta)\sim G(n, 1)$. And gamma Rv has scaling property, i.e. if $Y$ is Gamma with $(k,\theta)$, then $cY\sim G(k,c\theta)$. Applying here $\bar{X}-\eta\sim G(n,1/n)$. I think the parameter order is swapped in your book, I’m following wiki convention, where you can also find sum and scale properties.


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