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What is the expected value of the absolute standardized t-distribution - i.e.,: $E(|X|)$, where $X$ has the standardized t-distribution?

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    $\begingroup$ There are two close votes now, "needs details or clarity". I don't quite see what is unclear about this question. Could close-voters please comment on what clarification they would like to see? $\endgroup$ May 14 '20 at 19:27
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Per whuber's answer to Standardized Student's-t distribution, the density of the standardized $t$ distribution on $\nu$ degrees of freedom is

$$f(x) = \frac{\Gamma(\frac{\nu + 1}{2})}{\Gamma(\frac{\nu}{2})} \frac{1}{\sqrt{\pi(\nu-2)}} \left[1+\frac{x^2}{\nu-2}\right]^{-\frac{\nu+1}{2}}.$$

(Note that we need $\nu>2$ so we have two moments to standardize.)

Thus, your expectation is

$$ \begin{align*} E(|X|) & = 2\int_0^\infty xf(x)\,dx \\ & = \frac{\Gamma(\frac{\nu + 1}{2})}{\Gamma(\frac{\nu}{2})} \frac{2}{\sqrt{\pi(\nu-2)}} \int_0^\infty x\left[1+\frac{x^2}{\nu-2}\right]^{-\frac{\nu+1}{2}}\,dx \\ &= \frac{\Gamma(\frac{\nu + 1}{2})}{\Gamma(\frac{\nu}{2})} \frac{2}{\sqrt{\pi(\nu-2)}} \left(-\frac{\nu-2}{\nu-1}\right)\left(1+\frac{x^2}{\nu-2}\right)^{\frac{1-\nu}{2}}\bigg|_0^\infty \\ & = \frac{2}{\sqrt{\pi}}\frac{\Gamma(\frac{\nu + 1}{2})}{\Gamma(\frac{\nu}{2})} \frac{\sqrt{\nu-2}}{\nu-1} \end{align*} $$ by a rather simple integral evaluation.

I like sanity checking calculations like these using simulation, and it seems to check out:

> df <- 10
> nn <- 1e6
> 
> sims <- rt(nn,df)/(sqrt(df/(df-2)))# standardize by the variance
> mean(sims)
[1] -0.0006262779
> var(sims)
[1] 0.9995302
> 
> mean(abs(sims))
[1] 0.7732408
> 2/sqrt(pi)*gamma((df+1)/2)/gamma(df/2)*sqrt(df-2)/(df-1)
[1] 0.773398
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  • $\begingroup$ Thanks Stephan! $\endgroup$
    – user261912
    May 14 '20 at 10:36
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    $\begingroup$ +1. This works for any $\nu \gt 2;$ it is not necessary that $\nu \ge 3.$ $\endgroup$
    – whuber
    May 14 '20 at 20:35
  • $\begingroup$ @whuber: good point, I edited that into the post. $\endgroup$ May 15 '20 at 6:51

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