# Expected value of the absolute standardized t distribution

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

• 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? May 14, 2020 at 19:27

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.)

\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

• Thanks Stephan! May 14, 2020 at 10:36
• +1. This works for any $\nu \gt 2;$ it is not necessary that $\nu \ge 3.$
– whuber
May 14, 2020 at 20:35
• @whuber: good point, I edited that into the post. May 15, 2020 at 6:51