Find the $Var(X+Y)$, $f (x,y) = cx^{α−1}(y − x)^{β−1}e^{−y}$

Question

Let X and U have the joint p.d.f. $$ f_{X,Y} (x,y) = cx^{α−1}(y − x)^{β−1}e^{−y}, 0<x<y<\infty $$ Find the Var(X+Y).

V(X+Y) = V(X) + V(Y) + cov(X,Y)

V(X) = E(x^2)-(Ex)^2

To fine the marginal function of X, fx, $$ f_x(x) = \int_0^y cx^{α−1}(y − x)^{β−1}e^{−y} dy $$ $$ = cx^{α−1} \int_0^y (y − x)^{β−1}e^{−y} dy$$

I struggle from now.

Answer 1

Your integral bounds are incorrect, Actually, $$ f_X(x){=\int_x^\infty cx^{\alpha-1}(y-x)^{\beta-1}e^{-y}dy \\= cx^{\alpha-1}\int_x^\infty (y-x)^{\beta-1}e^{-(y-x)-x}dy \\= cx^{\alpha-1}\int_0^\infty u^{\beta-1}e^{-u-x}dy \\= cx^{\alpha-1}e^{-x}\int_0^\infty u^{\beta-1}e^{-u}dy \\= c\Gamma(\beta)x^{\alpha-1}e^{-x} } $$where $\Gamma(z)$ is the Gamma function of $z$.

Remark

From $\int_0^\infty f_X(x)dx=1$, the value of $c$ is equal to $\frac{1}{\Gamma(\alpha)\Gamma(\beta)}$, therefore the PDF of $X$ is equal to $$ f_X(x)=\frac{1}{\Gamma(\alpha)}x^{\alpha-1}e^{-x}. $$

Answer 2

In terms of $Z=Y-X$,$$E(f(X)g(Z))=c\int_0^\infty dx f(x)x^{\alpha-1}e^{-x}\int_0^\infty dz g(z)z^{\beta-1}e^{-z}.$$In particular, $X,\,Z$ are independent Gamma-distributed variables, in agreement with @MostafaAyaz's computation of $X$'s marginal distribution. You can then finish as follows:

• Take $f=g=1$ to prove $c=\frac{1}{\Gamma(\alpha)\Gamma(\beta)}$.
• Compute $E(X),\,E(X^2),\,E(Z),\,E(Z^2)$, and hence $V(X),\,V(Z)$. (The above link will tell you the Gamma distribution's variance, but it's worth deriving it if it's unfamiliar, in which case you'll need to know $c$ from the previous step.)
• Since $X,\,Z$ are uncorrelated, $V(X+Y)=V(2X+Z)=4V(X)+V(Z)$.