# Sums of Chi-Square Random Variables

Let $X_1\dots X_n$ be independent random variables with moment generating functions (mgf) $M_i(t) (i = 1 \dots n)$.

The chi-squared random variable with ν degrees of freedom has mgf $M(t) = (1 − 2t)^{ −ν/2}$. When $ν$ is a positive integer, the random variable is the sum of $ν$ independent squared standard normal random variables, i.e.$\sum_{i=1}^{v} Z_i^2$.

(i) Let the $X_i$ be chi-squared random variables with $ν_i$ degrees of freedom respectively. Using the result "$Y = \sum_{i=1}^{n} a_iX_i$ has mgf $M_Y (t) = \prod_{i=1}^nM_i(a_it)$", show that $Y =\sum_{i=1}^n X_i$ is chi-squared with $\sum_{i=1}^n ν_i$ degrees of freedom. Assuming the $ν_i$ are positive integers, give intuition for this result by describing a scenario by which $Y$ might be generated.

HI, would someone please solve for me? It is not my homework, just one of the problems from the past exam. I've tried substituting $M_Y (t) = \prod_{i=1}^n (1 − 2t)^{−ν/2}=(1 − 2t)^{−nν/2}$ and what now?

• You've done the case where all the $\nu_i$ are the same, but the problem has a different $\nu_i$ for each $X_i$. But otherwise you are 95% of the way there... look at what you get for $M_Y$ and compare it to the MGF of the distribution that you want to arrive at. Oct 30 '17 at 4:28

Just take the product of the mgfs directly: you have that $M_{X_i}(t) = (1-2t)^{-{\frac{\nu}{2}}}$, so
\begin{align} M_{Y}(t) &= \prod_{i=1}^n(1-2t)^{-{\frac{\nu_i}{2}}} \\ &= (1-2t)^{-{\frac{\sum_{i=1}^n\nu_i}{2}}} \end{align}