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?
Best Answer
Just take the product of the mgfs directly: you have that $M_{X_i}(t) = (1-2t)^{-{\frac{\nu}{2}}}$, so
$$ \begin{align} M_{Y}(t) &= \prod_{i=1}^n(1-2t)^{-{\frac{\nu_i}{2}}} \\ &= (1-2t)^{-{\frac{\sum_{i=1}^n\nu_i}{2}}} \end{align} $$
For the second part, recall that the chi-squared distribution is the sum of squares of normal random variables. What happens when you add these sums together?