Conditional expectation student's t distribution I was looking at the student's t distribution and was interested in the following conditional expectation:
$$
E[X|X\geq t_v^{-1}(\alpha)] = \frac{g_v(t_v^{-1}(\alpha)}{1-\alpha}\left( \frac{v + (t_v^{-1}(\alpha))^2}{v-1} \right)
$$
where X $\sim g_v$ and where $g_v$ denotes the standard t density with degrees of freedom $v$, $t_v$ denotes the distribution function of t with d.o.f. $v$ and where $t_v^{-1}$ is the quantile function of the standard t distribution with d.o.f. $v$. So far I've considered the following steps using integration by parts:
\begin{align*}
E[X|X\geq t_v^{-1}(\alpha)] &= \int_{t_v^{-1}(a)}^{\infty} \frac{x \cdot g_v(x)}{Pr[X \geq t_v^{-1}(\alpha)]}dx
\end{align*}
Since $Pr[X \geq t_v^{-1}(\alpha)] = 1-\alpha$ the integral reduces to
$$
\frac{1}{1-\alpha}\int_{t_v^{-1}(a)}^{\infty} x \cdot g_v(x) dx
$$
Then by integration by parts 
$$
\frac{1}{1-\alpha}\int_{t_v^{-1}(a)}^{\infty} x \cdot g_v(x) dx  = \frac{1}{1-\alpha} \left[ x\cdot t_v(x)|_{t_v^{-1}(\alpha)}^{\infty} - \int_{t_v^{-1}(a)}^{\infty} t_v(x) dx \right].
$$
However I find it hard to simplify this expression. I have tried to substitue the expression for $t_v$ in the integral but it becomes a mess. Any help on how to continue?
 A: i think i have the answer. If you use a substitution of variables, the integral is quite easy to calculate.
Let us combine the constant terms in the density function of the t-distribution and call this C. Then
\begin{align*}\int_{t^{-1}_{\nu}(\alpha)}^{\infty} \frac{l\cdot g_{\nu}(l)}{\mathbb{P}(\tilde{L}\geq t^{-1}_{\nu}(\alpha))} dl & = \frac{C}{1-\alpha} \int_{t^{-1}_{\nu}(\alpha)}^{\infty} l \cdot \left(1 + \frac{l^{2}}{\nu}\right)^\frac{-(\nu+1)}{2} dl.
\end{align*}
Now we are going to use the substitution method. Let us write $t = \frac{l^{2}}{\nu}$. In that case, we find $\frac{dt}{dl} = \frac{\sqrt{\nu}}{2\sqrt{t}} \Leftrightarrow 2\sqrt{t}dl = \sqrt{\nu}dt$. Substituting this into the integral, we may write:
\begin{align*}
\frac{C}{1-\alpha} \int_{t^{-1}_{\nu}(\alpha)}^{\infty} l \cdot \left(1 + \frac{l^{2}}{\nu}\right)^\frac{-(\nu+1)}{2} dl  &=\frac{C}{1-\alpha} \int_{t^{-1}_{\nu}(\alpha)}^{\infty} \sqrt{\nu t}  \left(1 + t\right)^\frac{-(\nu+1)}{2} dl \\
&= \frac{C}{1-\alpha} \int_{t^{-1}_{\nu}(\alpha)}^{\infty} \frac{1}{2}\sqrt{\nu}  \left(1 + t\right)^\frac{-(\nu+1)}{2} 2\sqrt{t}  dl\\
 &=\frac{C}{1-\alpha} \int_{t^{-1}_{\nu}(\alpha)}^{\infty} \frac{1}{2}\nu \left(1 + t\right)^\frac{-(\nu+1)}{2} dt \\
&= \frac{1}{2} \frac{C\cdot \nu}{1-\alpha} \int_{t^{-1}_{\nu}(\alpha)}^{\infty}(1+t)^{-\frac{1}{2}(\nu+1)}dt  \\
&= \frac{1}{2} \frac{C\cdot \nu}{1-\alpha}\left[-\frac{2(1+t)^{-\frac{1}{2}(\nu-1)}}{\nu -1}\Big|_{t^{-1}_{\nu}(\alpha)}^{\infty} \right]
\end{align*}
Before we can substitute, we have to note that we still work with $t$, hence we have to write again $t=\frac{l^{2}}{\nu}$, as the boundaries are still in terms of the original specification. We therefore find 
\begin{align*}
\frac{1}{2} \frac{C\cdot \nu}{1-\alpha}\left[-\frac{2(1+t)^{-\frac{1}{2}(\nu-1)}}{\nu -1}\Big|_{t^{-1}_{\nu}(\alpha)}^{\infty} \right] =  \frac{C\cdot \nu}{1-\alpha}\left[-\frac{(1+\frac{l^{2}}{\nu})^{-\frac{1}{2}(\nu-1)}}{\nu -1}\Big|_{t^{-1}_{\nu}(\alpha)}^{\infty} \right]
\end{align*}
Note that since $\nu > 1 \Rightarrow -\frac{1}{2}(\nu-1) < 0$, we have that $\lim\limits_{l \rightarrow \infty} (1+\frac{l^2}{\nu})^{-\frac{1}{2}(\nu-1)} = 0$. Therefore, the only relevant part of the evaluation will be when we substitute $t^{-1}_{\nu}(\alpha)$. Hence the substitution gives:
\begin{align*}
\frac{C\cdot \nu}{1-\alpha}\left[-\frac{(1+\frac{l^{2}}{\nu})^{-\frac{1}{2}(\nu-1)}}{\nu -1}\Big|_{t^{-1}_{\nu}(\alpha)}^{\infty} \right] &=
\frac{C\cdot \nu}{1-\alpha}\left[0--\frac{(1+\frac{(t^{-1}_{\nu}(\alpha))^{2}}{\nu})^{-\frac{1}{2}(\nu-1)}}{\nu -1} \right] \\
&=\frac{C\cdot \nu}{1-\alpha}\left[\frac{(1+\frac{(t^{-1}_{\nu}(\alpha))^{2}}{\nu})^{-\frac{1}{2}(\nu-1)}}{\nu -1} \right] .
\end{align*}
Now, note that 
\begin{align*}
&g_{\nu}(t^{-1}_{\nu}(\alpha)) = C \left(1 + \frac{(t^{-1}_{\nu}(\alpha))^{2}}{\nu}\right)^{-\frac{1}{2} (\nu+1)} \\ 
\Rightarrow &\frac{C\cdot \nu}{1-\alpha}\left[\frac{(1+\frac{(t^{-1}_{\nu}(\alpha))^{2}}{\nu})^{-\frac{1}{2}(\nu-1)}}{\nu -1} \right] =\frac{\nu}{1-\alpha}\left[\frac{(1+\frac{(t^{-1}_{\nu}(\alpha))^{2}}{\nu}) g_{\nu}(t^{-1}_{\nu}(\alpha))}{\nu -1} \right]   .
\end{align*}
We can then write
\begin{align*}
\frac{\nu}{1-\alpha}\left[\frac{(1+\frac{(t^{-1}_{\nu}(\alpha))^{2}}{\nu}) g_{\nu}(t^{-1}_{\nu}(\alpha))}{\nu -1} \right]
  &= \frac{\nu}{1-\alpha} \left[\frac{g_{\nu}(t^{-1}_{\nu}(\alpha))}{\nu-1} \left(1+\frac{(t^{-1}_{\nu}(\alpha))^{2}}{\nu}\right)\right] \\
&=\frac{g_{\nu}(t^{-1}_{\nu}(\alpha))}{1-\alpha}\left[\frac{\nu }{\nu-1} + \frac{(t^{-1}_{\nu}(\alpha))^{2}}{\nu-1}\right] \\
&= \frac{g_{\nu}(t^{-1}_{\nu}(\alpha))}{1-\alpha}\left[\frac{\nu +(t^{-1}_{\nu}(\alpha))^{2} }{\nu-1} \right],
\end{align*}
which is what we wanted to show.
