Conditional distribution of $X_1 \mid X_1+X_2+X_3$ when $X_i$'s are exponential

Question

Suppose $X_1, X_2, X_3 \sim f_{\theta}$ independently , where $f_{\theta}(x)= \theta e^{-\theta x}$; if $x \ge 0$.

How can I find the conditional distribution of $X_1 \mid X_1+X_2+X_3$ ?

I mean , I can find the distribution $X_1+X_2+X_3$, but what will be the explicit form of the conditional pdf?

Suppose if I give it a try it is coming out to be:

$f_{X_1|X_1+X_2+X_3}(x_1|x_1+x_2+x_3) = \frac{f(x_1,x_2,x_3)}{g(s)}$ such that $s=x_1+x_2+x_3$ and $g$ is the pdf of $X_1+X_2+X_3$. So, I am not getting how to simplify this.

Can anyone help?

Answer 1

Let's set $X_1=X$, $X_2+X_3=Y$, $Z=X+Y$ and calculate the conditional density $f(x|x+y)$

Data:

$f_X(x)=\theta e^{-\theta x}$

$f_Y(y)=\theta^2 y e^{-\theta y}$

$f_Z(z)=\frac{\theta^3}{2} z^2 e^{-\theta z}$

$f_{XY}(x,y)=\theta^3 y e^{-\theta(x+y)}$

Now let's set

$\begin{cases} z=x+y \\ u=x \end{cases} \rightarrow \begin{cases} x=u \\ y=z-u \end{cases}$

without any calculation it is evident that $|J|=1$ and thus

$$f_{UZ}(u,z)=\theta^3(z-u)e^{-\theta z}\mathbb{1}_{(0;+\infty)}(z)\mathbb{1}_{(0;z)}(u)$$

Thus

$$f_{U|Z}(u|z)=2\frac{z-u}{z^2}\mathbb{1}_{(0;z)}(u)$$

$z>0$