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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.

Anthony Lahmann
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TheProofIsTrivium
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    You forgot about $w >0$ and $z-w > 0$ this will give your integration to be between $0$ and $z$. Then substitute $w = zs$ and then $s$ is from $0$ to $1$. Then you can take $z$ in front of the integral and it will become constant, known as Beta function. You do not need to calculate it, as long as you know what you need to get. Does things in front of that beta integral seems familiar? – Dominik Kutek Feb 21 '20 at 01:06
  • @DominikKutek Thanks for your response. I end up with $z^{(n-1)/2}\int_{0}^{1} (1-w)^{(n/2 - 1)}w^{-1/2}\ dw$. Does that look right? How do I turn this into something using the Gamma function? – TheProofIsTrivium Feb 21 '20 at 01:47
  • Yes, you can by induction and by parts compute that integral, but note it isn't neccessary. Let $J$ be equal that integral You probably ended up with $\frac{1}{\Gamma(\frac{n}{2})\Gamma(\frac{1}{2})} \cdot z^{\frac{n-1}{2}} \cdot e^{-\frac{1}{2}z} (\frac{1}{2})^{\frac{n+1}{2}} \cdot J$. When you multiply by $\frac{\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n+1}{2})}$ you get part of Gamma density and something like $\frac{\Gamma(\frac{n+1}{2}) \cdot J}{\Gamma(\frac{n}{2})\Gamma(\frac{1}{2})}$ and since it is a density, the last term is $1$ hence $J$ is fraction of those gammas – Dominik Kutek Feb 21 '20 at 01:52

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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 $X+Y \sim \text{Gamma}((n+1)/2,1/2)$.

Kyle
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