The reciprocal of $t$-distributed random variable How is $1/T$ distributed if $T$ follows a Student's $t$-distribution?
 A: 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.

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.
