How to show that if $X \sim \Gamma(n/2, 1/2)$ and $Y \sim \Gamma(1/2, 1/2)$ then $X + Y \sim \Gamma((n+1)/2, 1/2)$?

Question

Since $$f_X(x) = \frac{(1/2)^{n/2}x^{n/2 - 1}e^{-x/2}}{\Gamma(n/2)}$$ and $$f_Y(y) = \frac{x^{-1/2}e^{-x/2}}{\sqrt{2} \Gamma(1/2)}$$ Using the convolution formula, we get that $$f_{X} * f_Y (z) = e^{-z/2}\frac{(1/2)^{(n+1)/2}}{\Gamma(\frac{n}{2})\Gamma(\frac{1}{2})}\int_{-\infty}^{\infty} (z-w)^{(n/2) - 1}w^{-1/2}\ dw$$ I am not sure how to evaluate the integral on the right.

Answer 1

You should let the lower limit become zero, then just proceed as normal. On the other hand, you can consider this "Laplace transform" approach.\

The Laplace transform of $X$ and $Y$ are given by $$ \mathcal{L}_{X}(s) = \left( 1-2s \right)^{-\frac{n}{2}},\\ \mathcal{L}_{Y}(s) = \left( 1-2s \right)^{-\frac{1}{2}}. $$ Hence, the Laplace transform of $X+Y$ is the product of $\mathcal{L}_{X}(s)$ and $\mathcal{L}_{Y}(s)$ given as above, or $$ \mathcal{L}_{X+Y}(s)=\mathcal{L}_{X}(s) \mathcal{L}_{Y}(s) =\left( 1-2s \right)^{-\frac{n+1}{2}}. $$ As you can observed, this $\mathcal{L}_{X+Y}(s)$ is also similar to that of $\text{Gamma}((n+1)/2,1/2)$ making $X+Y \sim \text{Gamma}((n+1)/2,1/2)$.