Vtyjkyuk

Say I have $n$ independent random variable $X_1,...,X_n: Omega rightarrow mathbbR$ on a probability space $Omega$ with the same density $p:mathbbRrightarrow [0,infty)$. Could someone explain to me in the most measure-theoretic terms (i.e. assuming I know plenty of measure theory but not much probability theory terminology) how you can explicitly write down the density of the joint probability distribution $mathbbR^2rightarrow [0,infty)$ of $X_1$ and $t= sum_j=1^nX_j$ using only $p$? For instance I have an example where $p(x) = xe^-x$, and I'm supposed to get $$frac1Gamma ( 2n-2)x_1(t-x_1)^2n-3e^-t$$ for the density of the joint distribution.

What if you just push the measure from $mathbbR^n$ to $mathbbR^n$ via $$L: (x_1,...,x_n-1, x_n)mapsto (x_1,...,x_n-1,sum_i=1^n x_i)$$ and then project down $ mathbbR^nrightarrow mathbbR^2$ via

$$pi :(x_1,...,x_n-1, t)mapsto (x_1,t).$$

Then the density function is easy to track: Since $L$ has an inverse, the density $p$ goes to $pcirc L^-1$ and then since $pi$ is just a projection, the density can be pushed down just by integrating over the fibers of points.

You will need the following identity,

$$fracC^2k+1(2k+1)! = int_A_k(C)big(C-sum_i=1^k x_ibig)cdot x_1x_2...x_k~~ dx_1dx_2...dx_k,$$

where $$A_k(C):= (x_1,...,x_k)in mathbbR^k:~~~~sum_i=1^k x_ileq C,~~textand~~ x_igeq 0~~textfor all ~i~.$$

This can be proved by induction:
$$int_A_k(C)big(C-sum_i=1^k x_ibig)cdot x_1x_2...x_k~~ dx_1dx_2...dx_k =$$ $$int_0^C ~x_k bigg( ~int_A_k-1(C-x_k)big((C-x_k)- sum_i=1^k-1 x_ibig)cdot x_1x_2...x_k-1~~ dx_1dx_2...bigg) ~dx_k=$$ $$int_0^C ~x_kfrac(C-x_k)^2k-1(2k-1)!dx_k.$$

Then just integrate by parts.