I have the following question:

Suppose $X_1, X_2$ iid $\sim f_X(x)=\theta e^{-\theta x}1\{x\ge 0\}$ and $S=X_1+X_2$. What is $f_{X_1,X_2}(x_1,x_2|S=s)$?

The solution from our exercise class to this problem was $$f_{X_1,X_2}(x_1,x_2|S=s)=\dfrac{f_{X_1,X_2,S}(x_1,x_2,s)}{f_S(s)}=\dfrac{\theta^2 e^{-\theta s}}{\theta^2 s e^{-\theta s}}= \dfrac{1}{s} \quad \text{,where} \quad s=x_1+x_2$$

However, to me it is not clear how to understand this "where $s=x_1+x_2$". Shouldn't it rather be something like: $$f_{X_1,X_2}(x_1,x_2|S=s)= \dfrac{\delta(s=x_1+x_2)}{s}$$ where $\delta$ is a dirac delta function? Or how can we otherwise recover $f_{X_1,X_2}(x_1,x_2)$ from $f_{X_1,X_2,S}(x_1,x_2,s)$ by integrating out $s$?

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    $\begingroup$ 1) Please add the tag self-study. 2) some hints: You are conditioning on $s=S=X_1+X_2$ so no need for the delta function. 3) Use that $X_1+X_2=S$ has a gamma distribution (with shape parameter 2) and the known result that $X_1/S$ has a beta distribution with params 1,1; that is, it is uniform, so $X_1$ given $S=s$ is uniform on $(0,s)$ $\endgroup$ – kjetil b halvorsen Jan 25 at 15:33
  • $\begingroup$ Thanks for your answer. Isn't that exactly what you get with the delta function that $X_1|S=s$ has uniform distribution on $(0,s)$? I also don't see how $f_{X_1,X_2}(x_1,x_2|S=s)$ is integrating to $1$ (in $x_1,x_2$) otherwise, as it is only nonzero on a set of L-measure zero right? $\endgroup$ – Andreas Schlaginhaufen Jan 25 at 16:27
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    $\begingroup$ Well, yes, you can write it that way, if you want. But what is wrong with just saying that $s=x_1+x_2$? $\endgroup$ – kjetil b halvorsen Jan 25 at 16:32
  • $\begingroup$ Well I don't understand how this can be a density. What is the dominating measure? $\endgroup$ – Andreas Schlaginhaufen Jan 25 at 17:39
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    $\begingroup$ The dominating measure is the Lebesgue measure on the line $x_1+x_2=s$. $\endgroup$ – Xi'an Jan 25 at 18:20

One way of thinking about your problem is connecting it to the Poisson process. Start a (homogeneous) Poisson process at time zero. Then $X_1$ is the waiting time to the first event (point), and $S=X_1+X_2$ is the waiting time to the second event. So we know there is exactly one event in the interval $(0,S)$, and the result follows from the uniform distribution of the event, given the endpoint of the interval.

You can add the algebra.


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