finding density of $1/Z$ when $Z$ is a standard random normal variable If $Z$ is a standard normal r.v., we know that its density is
$$f_Z(z)=\frac{1}{\sqrt{2\pi}}e^{-z^2/2},$$
where $-\infty \leq z\leq \infty$. I want to find what $f_{1/Z}(z)$ is.
I let $Y=1/Z$, so $Z=1/Y$, and then I used the above formula to obtain 
$$f_{1/Z}(z)=\frac{1}{\sqrt{2\pi}}e^{-1/2y^2},$$
but my solution is wrong. I don't know what else to do. Some help?

By the properties of a continuous density, 
$$
\mathbb P(0 < Y \le b) = \int_0^b f_Y(y) \, dy. 
$$
Therefore, for $Y = 1/Z$, 
$$
\mathbb P(0 < Y \le y) = \mathbb P(Z \ge 1/y) = \int_{1/y}^{\infty} f_Z(z) \, dz = 1/2 - \int_0^{1/y} f_Z(z) \, dz. 
$$
Differentiating with respect to $y$ to recover $Y$'s density, 
$$
f_Y(y) = \frac d{dy} \left( 1/2 - \int_0^{1/y} f_Z(z) \, dz \right) = -f_Y(1/y) \cdot \frac{-1}{y^2} = \frac{f_Y(1/y)}{y^2}. 
$$
All that is left to do now is plugging in. Note that we have only taken care of the case where $z \ge 0$ ; I leave the case $y \le 0$, it can be done similarly (hint : consider $\mathbb P(1/Z \le y)$ ; note that I considered $\mathbb P(0 < 1/Z \le y)$ because the density of $Y$ satisfies $f_Y(y) = \frac{d}{dy} \mathbb P(Y \le y)$). 
Hope that helps,
