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I'm trying to find an online derivation of why the mean of the Chi-Distribution with $n$ degrees of freedom is $$E(\chi_n)= \sqrt2 \frac{ \Gamma(\frac{n+1}{2})}{\Gamma(\frac{n}{2})}$$

Is it too involved to find online? Does anyone have any idea how I could derive it using something much more well-known (and available) online like the chi-squared distribution?

I've looked everywhere online, and even looked through some of the books referenced on Wikipedia and WolframAlpha's pages on the Chi Distribution. None of them list how to derive it, they just have the final answer.

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    $\begingroup$ Find it as $E(\chi_n)=E(\sqrt{\chi_n^2})=\int \sqrt{x}f_{\chi^2_n}(x)\,dx=\cdots$, i.e. using the pdf $f_{\chi^2_n}$ of chi-squared distribution. $\endgroup$ – StubbornAtom Feb 13 at 18:17
  • $\begingroup$ Hmm that might actually work! But how do you reach that form? Where did you get that $\sqrt{x}$ from? $\endgroup$ – QuantumHoneybees Feb 14 at 4:28
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    $\begingroup$ It is just an application of this theorem. You have two options: find $E(\chi_n)$ from the pdf of $\chi_n$, or find it using the pdf of $\chi^2_n$ as indicated in my last comment. $\endgroup$ – StubbornAtom Feb 14 at 7:31

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