I am reading through modern mathematical statistics with applications by Devore and Berk. I came across the following "exercise left to the reader" and I couldn't figure it out easily.
Question:
Use moment generating functions to show that if $$X_3 = X_1+X_2, $$
with $$X_1 \sim \chi ^2_{\nu_1},$$ $$X_3 \sim \chi ^2_{\nu_3},$$ and
$$\nu_3 > \nu_1$$ and $X_1, X_2$ are independent, then $X_2 \sim \chi^2_{\nu_3-\nu_1}$
Attempt:
Using the MGF we would have
$$ M_{X_2}(t) = M_{X_3-X_1}(t) = M_{X_3}(t)M_{X_1}(-t) = (1+2t)^{-\nu_3/2}(1-2t)^{-\nu_1/2} .$$
I do not see an easy way to deal with $(1+2t)^{-\nu_3/2}(1-2t)^{-\nu_1/2} $ since the exponents are not the same.
Best Answer
Prove it in the following way:
$X_1 \sim \chi_{v_3}^2$
suppose that
$X_2 \sim \chi_{v_3-v_1}^2$
thus for independence you know that
$$M_{X_1}(t)M_{X_2}(t)=(1-2t)^{-\frac{v_1}{2}}(1-2t)^{-\frac{v_3-v_1}{2}}=(1-2t)^{-\frac{v_3}{2}}$$
that is exactly the MGF of the distribution of $X_3$