if $X$ is a random variable what is the distribution of $W(x)=1/x$? If $X$ is a random variable, what is $W(x) = 1/x$?
if $X$ was say normal distribution $\frac{1}{\sqrt{2\pi\sigma}} e^{-(x-\mu)^2/(2\sigma^2)}$
Would that mean $W(x)$ is distributed as $\dfrac{1}{\frac{1}{\sqrt{2\pi\sigma}} e^{-(x-\mu)^2/(2\sigma^2)}} \text{?}$
or 
Would that mean W(x) is distributed as ${\frac{1}{\sqrt{2\pi\sigma}}e^{-\frac{(\frac{1}{x}-\mu)^2}{2\sigma^2}}}$
Or is $W(x)=1/x$ to be interpreted as a conditional probability of $W$ given $X=x$? Or $f_{W\mid X}(W\mid X=x)$. I feel like this is right because X is used to denote distributions while x is used to denote realised variables. 
 A: If $X$ is a random variable and $g \colon \mathbb R \to \mathbb R$ is a Borel measurable function, then $Y = g(X)$ is another random variable with cumulative distribution function $F_Y(t) = P(g(X) \le t)$.
Let's assume that $X$ is a normal variable with mean $\mu$ and variance $\sigma^2$. Now $$F_X(t) = P(X \le t) = \frac 12 \left( 1 + \textrm{erf} \left( \frac{x- \mu}{\sigma \sqrt 2} \right) \right),$$
so $$F_Y(t) = P\left(\frac 1 X \le t\right).$$
To expand it, you have to distinguish two cases: $X \le 0$ or $X > 0$. Anyway, $1/X$ has no expected value. According to What is the name of this theorem, and are there any caveats?, it is equal to
$$\int_{-\infty}^\infty \frac{e^{-x^2 / \sqrt{2}}}{\sqrt{2 \pi}x} \,\textrm{d} x.$$
However, the integral isn't absolutely convergent. 
A: I think you're asking "what is the pdf of 1/X given the pdf of X".  Is that right? 
It is not the reciprocal as that would violate the criteria of the total area must be 1.  Something which might be relevant is that whatever the pdf is of 1/X is that E(1/X) will equal the integral (over the relevant domain) of 1/x times the pdf of X.  See if this helps: https://imai.princeton.edu/teaching/files/Expectation.pdf
See theorem 14.
