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

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