# 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?

• Please see “Reciprocal normal distribution” in wiki.
– Ed V
Feb 23, 2020 at 3:05

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{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}

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{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}

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. It has zero mean and infinite variance. 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.

• Great, thank you! Feb 23, 2020 at 4:13