Let $X,Y \sim \chi^2_n$ be two independent $\chi^2$-distributed random variables with $n$ degrees of freedom. What is the CDF/PDF of $$ Z = aX - b Y $$ with $a>0, b>0$ ?
P.S. This question answers the case $a=b=1$.
My attempt: Using the moment generating function (MGF), we can find $$ \mathcal{M}_Z(t) = \mathbb{E}[e^{tZ}] = \mathbb{E}[e^{taX}e^{-tbY}] = \mathcal{M}_X(at) \mathcal{M}_Y(-bt) $$ using independence, from which $$ \mathcal{M}_Z(t) = \left[(1 - 2at)(1 + 2bt) \right]^{-\frac{n}{2}} $$ Now, $\mathcal{M}_Z(-t) = \mathcal{L}_B[f_Z(z)]$, where $\mathcal{L}_B$ is bilateral Laplace transform and $f_Z(z)$ is the PDF of $Z$, so $$ f_Z(z) = \mathcal{L}_B^{-1}[\mathcal{M}_Z(-t)] = \int_{-\infty}^{+\infty} \left[(1 + 2at)(1 - 2bt) \right]^{-\frac{n}{2}} e^{-tz} dt $$ How to proceed from here? Is this way any good? Any help is appreciated.
