Distribution of $Y$ if $Y=\dfrac1X$ and $F_X(x) = \left(1 - \frac1x\right)\mathsf1_{[1,\infty)}(x)$ Let $X$ be a random variable with cumulative distribution function
$$F_X(x) = \left(1 - \frac1x\right)\mathsf1_{[1,\infty)}(x).$$
Let $Y=\dfrac1X$. What is the distribution of $Y$?
 A: First, observe that $1<x<\infty$ implies that $0<\frac1x<1$, so $\mathbb P(Y\in(0,1))=1$. Now for $0<y<1$ we compute
\begin{align}
F_Y(y)&=\mathbb P(Y\leqslant y)\\
&= \mathbb P\left(\frac1X\leqslant y\right)\\
&= \mathbb P\left(X\geqslant \frac 1y\right)\\
&= 1 - \mathbb P\left(X\leqslant \frac1y\right)\\
&= 1 - F_X\left(\frac1y\right)\\
&= (1 - (1 - y))\mathsf 1_{(0,1)}(y)\\
&= y\mathsf 1_{(0,1)}(y).
\end{align}
It follows that $Y$ is uniformly distributed over $(0,1)$. Since $F_Y$ is absolutely continuous, it is differentiable a.e. and $F_Y'$ is the probability density function of $Y$:
$$f_Y(y) = \frac{\mathsf d}{\mathsf d y} F_Y(y) = \mathsf 1_{(0,1)}(y)$$
(where $\mathsf 1_A$ denotes the indicator function of a set $A$).
Recall that any random variable $W$ has a cumulative distribution function $F_W:\mathbb R\to[0,1]$ defined by $F_W(w) = \mathbb P(W\leqslant w)$. When $W$ is a continuous random variable, it has a probability density function, that is, a function $f_W$ which satisfies $F_W' = f_W$ and for $w\in\mathbb R$,
$$F_W(w) = \int_{-\infty}^w f_W(t)\ \mathsf dt. $$
The distribution function and the density are two different (but related) things. Note that not all random variables are continuous, so they may not have a density.
