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?

The change of variables formula is (from the chain rule):
$$\begin{align}
f_{g(Z)}(y) & = f_Z(g^{-1}(y)) \cdot \lvert \mathcal D_y\; g^{-1}(y)\rvert & \textrm{where $g(z)$ is an invertable function}
\\[2ex]
f_Y(y) & = f_Z(1/y)\cdot\left\lvert\dfrac{\operatorname d y^{-1}}{\operatorname d y\quad}\right\rvert & \textrm{since $g(z)=1/z$}
\end{align}$$
