Consider the following result in terms of Beta functions $B(\cdot, \cdot)$.

$$ \mathbb E \left( \frac{Y_i^2}{\sigma^2} \right) = \frac{1}{r^2}\frac{n!}{(i-1)!(n-i)!} [B(n-i+1, i) -2B(n-i+r+1, i) + B(n-i+2r+1, i)], $$

I read a paper saying that by rewrite Beta functions using Gamma functions, i.e., $B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$, the above result becomes

$$ \mathbb E \left( \frac{Y_i^2}{\sigma^2} \right) = \frac{1}{r^2} \frac{\Gamma(n+1)\Gamma(n-i+2r+1)}{\Gamma(n-i+1)\Gamma(n+2r+1)}. $$

I could not get this claimed result myself. I am wondering whether I made any errors along the way. Could anyone help me, please? Thank you!

Here is my workings.

\begin{align} \mathbb E \left( \frac{Y_i^2}{\sigma^2} \right) &= \frac{1}{r^2}\frac{n!}{(i-1)!(n-i)!} [B(n-i+1, i) -2B(n-i+r+1, i) + B(n-i+2r+1, i)] \\[10pt] &= \frac{1}{r^2}\frac{\Gamma(n+1)}{\Gamma(i)\Gamma(n-i+1)} \left[\frac{\Gamma(n-i+1)\Gamma(i)}{\Gamma(n+1)} -2\frac{\Gamma(n-i+r+1)\Gamma(i)}{\Gamma(n+r+1)} + \frac{\Gamma(n-i+2r+1)\Gamma(i)}{\Gamma(n+2r+1)}\right] \\[10pt] &=\frac{1}{r^2} \left[ 1-2\frac{\Gamma(n+1)\Gamma(n-i+r+1)}{\Gamma(n-i+1)\Gamma(n+r+1)} + \frac{\Gamma(n+1)\Gamma(n-i+2r+1)}{\Gamma(n-i+1)\Gamma(n+2r+1)} \right] \\[10pt] &\neq \frac{1}{r^2} \frac{\Gamma(n+1)\Gamma(n-i+2r+1)}{\Gamma(n-i+1)\Gamma(n+2r+1)}, \end{align}


$$ 1-2\frac{\Gamma(n+1)\Gamma(n-i+r+1)}{\Gamma(n-i+1)\Gamma(n+r+1)} = 0. (**) $$

I do not see why $(**)$ is true.


Something's definitely off. Take e.g. $n=3,r=2,i=1$. I get $2/35$ for your first formula in terms of Beta functions, and $3/28$ for the alleged answer. Perhaps there's a typo somewhere, either in what you wrote or what the authors wrote.


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