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Suppose $X_1,X_2,\ldots,X_n$ are iid with density $f(x;\theta) = \theta e^{-\theta x}, x,\theta > 0$.
Let $T = \sum_{i=1}^n X_i$.
Then $T$ has density $f(t;\theta) = \frac{1}{\Gamma(n)}\theta^n t^{n-1}e^{-\theta t}, t>0.$
Show that the conditional density of $X_1$ given $T=t$ is
$$f(x_1|t) = \frac{n-1}{t}(1-\frac{x_1}{t})^{n-2}, 0<x_1<t<\infty. $$ How would I do this?
I know that
$$f(x_1|t) = \frac{f_{X_1,T}(x_1,t)}{f_T(t)} \\ = \frac{f_{X_1,t}(x_1,x_1+x_2+\ldots+x_n)}{f_T(t)}$$

I'm not sure how to proceed, since $T$ and $X_1$ are not independent.

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  • $\begingroup$ Hint: Try to find f(t|x1) $\endgroup$ – Vikash B Mar 14 '18 at 9:25
  • $\begingroup$ I'm not sure how to do that either (I would just do what I did in the OP). I think the main problem is that the distributions are continuous $\endgroup$ – Mr. Bromwich I Mar 14 '18 at 9:31
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We have that: $$f_{X_1|T}(x_1|t)=\frac{f_{T|X_1}(t|x_1)\,f_{X_1}(x_1)}{f_T(t)}$$

Let $T'$ = $T - X_1$

So we have, $$f_{T|X_1}(t|x_1)=f_{T'}(t-x_1)$$

$T'$ is a gamma r.v with shape $n-1$

Now you have everything needed to to calculate: $f_{X_1|T}(x_1|t)$

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  • $\begingroup$ That worked! The solution used a weird limiting argument. $\endgroup$ – Mr. Bromwich I Mar 14 '18 at 10:24
  • $\begingroup$ Does your line after "So we have," follow because $X_1 + \ldots + X_n = t | X_1 = x_1$ simplify to just $X_2 + \ldots + X_n = t-x_1$? $\endgroup$ – Mr. Bromwich I Mar 14 '18 at 10:25
  • $\begingroup$ @Mr.BromwichI, correct. This is a pretty standard way you can overcome independence requirements. $\endgroup$ – Vikash B Mar 14 '18 at 11:43

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