Distribution of $\frac{1}{X}$ if $X\sim N(0,1)$ Given a random variable $X$ that has a normal ditribution with mean $\mu$ and standard deviation $\sigma$, what is the distribution of $\frac{1}{X}$?
I guess that it like a normal distribution applying some kind of transformation to the mean and std, but I do not see how to proceed. Any hints or suggestions to see this?
 A: Since the random variable $X \sim \text{N}(0,1)$ is symmetric around zero, the random variable $Y = 1/X$ will also be symmetric around zero.  Let's have a look at its distribution on the positive side.  For all $y > 0$ we have:
$$\begin{equation} \begin{aligned}
\mathbb{P}(0 < Y \leqslant y)
&= \mathbb{P}(X \geqslant 1/y) \\[6pt]
&= 1- \mathbb{P}(X < 1/y) \\[6pt]
&= 1- \Phi(1/y), \\[6pt]
\end{aligned} \end{equation}$$
where $\Phi$ is the standard normal distribution function.  Differentiating and applying the chain rule gives the following density over the positive argument values $y>0$:
$$\begin{equation} \begin{aligned}
f_Y(y)
&= \frac{d}{dy} \mathbb{P}(0 < Y \leqslant y) \\[6pt]
&= -\frac{d}{dy} \Phi(1/y) \\[6pt]
&= \frac{1}{y^2} \cdot \text{N}(1/y|0,1) \\[6pt]
&= \frac{1}{\sqrt{2 \pi} y^2} \cdot \exp \Big( - \frac{1}{2 y^2} \Big). \\[6pt]
\end{aligned} \end{equation}$$
Using the symmetry of the distribution of $Y$ means that this same expression for the density holds over the values $y<0$.  Thus, we can write the density as:
$$f_Y(y) = \begin{cases}
\frac{1}{\sqrt{2 \pi} y^2} \cdot \exp \Big( - \frac{1}{2 y^2} \Big) & & \text{if } y \neq 0, \\[6pt]
0 & & \text{if } y = 0. \\[6pt]
\end{cases}$$
This distribution is actually quite unlike a normal distribution.  In particular, it is still symmetric, but it is bimodal.  Note also that the density above is continuous, since the value at the argument $y=0$ is equal to the limit of the density from above and below.
