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?

  • 1
    $\begingroup$ Since the PDF is proportional to a power of $x^2 + \nu$ and the term $x$ is proportional to its derivative, the mindless substitution $t = x^2 + \nu$ works immediately. $\endgroup$
    – whuber
    Commented Oct 4, 2017 at 13:12
  • $\begingroup$ I am not entirely sure what you mean by 'proportional to a power of $x^2 + v$'. So you are basically saying I needn't use integration by parts? $\endgroup$
    – Pim
    Commented Oct 4, 2017 at 15:00
  • $\begingroup$ See the PDF at en.wikipedia.org/wiki/Student%27s_t-distribution. $\endgroup$
    – whuber
    Commented Oct 4, 2017 at 15:19

1 Answer 1


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.


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