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?

$$P\{ Y \le y\} = P\left\{\frac{1}{Z}\le y\right\} = P\left\{Z \ge \frac{1}{y}\right\} = \int_{1/y}^\infty f_Z(z)\mathrm{d}z$$ 
$$\implies f_Y(y) = \frac{\partial}{\partial y} P\{Y \le y\} = \frac{1}{y^2} f_Z(1/y)$$

Alternatively, recall that densities of transformed random variables are in the ratio of the Jacobian of the transformation. $$z = \frac{1}{y} \implies |J| = \left| \frac{\mathrm{d} u}{\mathrm{d} y}\right| = \frac{1}{y^2},$$ and $$f_Y (y) = |J| f_Z(1/y) = \frac{1}{y^2} f_Z(1/y)$$

The reason this is happening is essentially because the density of the random variable is not a probability. Roughly, the probability of $Z$ lying in a region of width $\mathrm{d}z$ around $z$ is $f_Z(z) \mathrm{d}z$, and that should equal the probability of the transformed realisation, which is $f_Y(y) \mathrm{d}y$. Equating these gives you the relatoin between $f_Z$ and $f_Y$ 
