# Integral involving Student's t distribution

Is there a reference that has the value of the following integral: $$F(a) =\int_{-\infty}^{a}xt_{\nu}(x)dx = \int_{-\infty}^{a}\frac{x}{s\sqrt{\nu\pi}}\frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\Gamma\left(\frac{\nu}{2}\right)}\left(1+\frac{1}{\nu}\left(\frac{x-m}{s}\right)^2\right)^{-\frac{\nu+1}{2}} d x$$ where $$t_{\nu}(x)$$ is the pdf for Student's t distribution.

• I think your $F(a)$ is the mean of the truncated t-distribution with lower limited = $-\infty$ and upper limited = $a$. en.wikipedia.org/wiki/Truncated_distribution Except that yours isn't normalized. – const-ae Aug 28 '20 at 15:27
• This is straightforward to evaluate in terms of the Student $t$ CDF and elementary functions, so why do you need a reference? – whuber Aug 28 '20 at 16:05
• Yes, perhaps it is straightforward, but it's not a one-line long derivation. – Alex Aug 28 '20 at 16:40

You're right: this is not (quite) a one-liner. But I can do a thorough job in just two lines by building on what you already know.

Let's make some standard preliminary observations to simplify the work and establish the notation.

Your $$t_\nu$$ is a location-scale family based on the Student $$t$$ distribution with $$\nu$$ degrees of freedom with a location parameter $$m$$ (which I will write as $$\mu$$ to make it clear it's a parameter and not a variable) and scale parameter $$s$$ (which I will write as $$\sigma$$). That is tantamount to a change of units of measurement of the original Student $$t$$ variable $$Z$$ to a new variable $$X=\sigma Z + \mu.$$ Thus, your integral becomes a linear combination of integrals with integrands $$zt_\nu(z)$$ and $$t_\nu(z)$$ (where from now on I will use "$$t_\nu$$" in the narrower sense of the standard Student $$t$$ density). The latter integrates, by definition, to the (standard) student $$t$$ CDF, which I will write $$T_\nu(z) = \int^z t_\nu(x)\,\mathrm{d}x.$$

This leaves us with the (genuine) problem of integrating $$xt_\nu(x)$$ for the standard $$t$$ density $$t_\nu.$$ The answer is contained in one of your previous questions, where you note

$$\frac{d}{dx}t_{\nu}=-\frac{\nu+1}{\nu+x^2}\,xt_{\nu}(x).$$

This can be rewritten (using the product rule of differentiation) in the form

\begin{aligned}\frac{d}{dx} \left(\frac{x^2+\nu}{1-\nu}\,t_\nu(x)\right) &= \frac{2x}{1-\nu}\,t_\nu(x) + \frac{x^2+\nu}{1-\nu}\frac{d}{dx}t_\nu(x) = \left(\frac{2}{1-\nu} - \frac{1+\nu}{1-\nu}\right) x t_\nu(x) \\ &= xt_\nu(x). \end{aligned}

In other words, you already know an antiderivative of $$xt_\nu(x).$$ The Fundamental Theorem of Calculus states this is the indefinite integral. All we need to compute is the constant of integration. But, provided $$\nu \gt 1,$$ the integrand $$xt_\nu(x)$$ decreases sufficiently rapidly as $$a\to -\infty$$ that the limiting value of its integral must be $$0.$$ Thus, the constant of integration is $$0.$$

Putting these results together gives for $$\nu \gt 1$$

\begin{aligned} \frac{1}{\sigma}\int^a x t_\nu\left(\frac{x-\mu}{\sigma}\right)\mathrm{d}x &= \sigma \int^{(a-\mu)/\sigma} z t_\nu(z)\,\mathrm{d}z + \mu \int^{(a-\mu)/\sigma} t_\nu(z)\,\mathrm{d}z \\ &= \sigma \left(\frac{\left(\frac{a-\mu}{\sigma}\right)^2+\nu}{1-\nu}\right) t_\nu\left(\frac{a-\mu}{\sigma}\right) + \mu T_\nu\left(\frac{a-\mu}{\sigma}\right). \end{aligned}

When $$\nu \le 1,$$ you may verify the integral does not converge.

As a quick check, I computed this integral (as a function of $$a$$) both numerically and using the final formula. Here is a plot of the two computations, one in black and the other in red. They agree.