How is $1/T$ distributed if $T$ follows a Student's $t$-distribution?

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    $\begingroup$ With 1 df, it is a $t$-distribution *with 1 df :) $\endgroup$ – Elvis Jan 6 '12 at 14:19
  • $\begingroup$ What if $n > 2$? $\endgroup$ – petrichor Jan 6 '12 at 14:23
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    $\begingroup$ Have you tried deriving it from the definition of the Student t distribution? The calculations are simple. $\endgroup$ – whuber Jan 6 '12 at 14:24
  • $\begingroup$ I tried but couldn't figure out. I need to check how functions of r.v.'s are considered. $\endgroup$ – petrichor Jan 6 '12 at 14:27
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    $\begingroup$ To generalize the first comment by @Elvis ever-so-slightly: Note that, for general df $\nu$, if $T \sim t_\nu$, then $T^2 \sim F_{1,\nu}$ and so $T^{-2} \sim F_{\nu,1}$. Elvis' first comment is then recovered immediately via the symmetry of the $t$-distribution. $\endgroup$ – cardinal Jan 6 '12 at 15:48

One can show that if $X$ has density $f(t)$, then $Y = 1/X$ has density $g(t) = {1\over t^2} f\left( {1\over t} \right)$ (for $t\ne0$).

The density of $T$ with $k$ degrees of freedom is $$\frac{1}{\sqrt{k\pi}}\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k}{2})}\frac{1}{(1+\frac{t^2}{k})^{\frac{k+1}{2}}}$$ so the density of $1 \over T$ is $$\frac{1}{\sqrt{k\pi}}\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k}{2})}\frac{1}{t^2(1+\frac{1}{kt^2})^{\frac{k+1}{2}}}.$$

Note that for $k = 1$ this is the same density (in this case $t$ is the quotient of two iid centered gaussian variables).

Here is the allure of this density for $k=1, 2, 30$. As whuber says in the comments when $k>1$ it is bimodal, and all moments diverge.

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Edit How do you show with elementary tools $g(t) = {1\over t^2} f\left( {1\over t} \right)$ ?

One possible solution is to first verify (draw a graph) that : $$\mathbb P(Y\le t) = \left\{\begin{array}{ll} \mathbb P\left(X \ge {1\over t} \right) + \mathbb P(X \le 0) & \mbox{ if } t> 0 \\ \mathbb P(X \le 0) & \mbox{ if } t= 0 \\ \mathbb P(X\le 0) - \mathbb P\left(X \le {1\over t} \right) & \mbox{ if } t< 0 \\ \end{array}\right.$$

Using this you will easily find how to express the cdf of $Y$ in terms of the cdf of $X$ ; derive this expression to obtain the density.

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    $\begingroup$ (+1) Note that the reciprocal distribution is bimodal when $k\gt 1$. All the moments diverge (for any $k$). $\endgroup$ – whuber Jan 6 '12 at 15:28
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    $\begingroup$ You’re right, I’ll add a graph, it’s a matter of five minutes. $\endgroup$ – Elvis Jan 6 '12 at 15:30
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    $\begingroup$ A nit pick: the bimodality kicks in as soon as $k$ exceeds 1; e.g., it's there for $k=1.01$. (Student's t is defined for all real $k\gt 0$, not just positive integral $k$.) $\endgroup$ – whuber Jan 6 '12 at 15:48
  • $\begingroup$ @whuber yes! and the limit in 0 would deserve some special attention when $k<1<2$. No time now... $\endgroup$ – Elvis Jan 6 '12 at 16:01
  • $\begingroup$ That limit is $0$ because the original Student t distribution decays faster than $t^{-2}$ as $t\to\pm\infty$ for $1\lt k$. $\endgroup$ – whuber Jan 6 '12 at 16:28

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