Prove that the difference of two chi square distributions is a chi square distribution, using the moment generating function

Question

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.

Answer 1

$$M_{X_3}(t) = M_{X_2}(t) M_{X_1}(t) \text{ from independence}$$ $$\text{ so we have that } \frac{M_{X_3}(t)}{M_{X_1}(t)} = M_{X_2}(t)$$ $$\text{and from the MGF of a chi square distribution (for $t< 1/2), $ we have that } \frac{(1+2t)^{-\frac{\nu_3}{2}}}{(1+2t)^{-\frac{\nu_1}{2}}} = (1+2t)^{-\frac{\nu_3-\nu_1}{2}}$$ $$\text{ which is precisely the MGF of } \chi^2_{\nu_3-\nu_1}$$

$$\text{So we have that } X_2 \sim \chi^2_{\nu_3-\nu_1}$$

Answer 2

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 $X_3$