Let $X\sim \text{Exp}(\lambda)$ be an exponentially distributed random variable.
That is it has the probability density function
$f(x)=\lambda e^{-\lambda x}1_{[0,\infty)}(x)$ and cumulative distribution function
$$F_X(x)=\int_ {0}^x\lambda e^{-\lambda x}=[-e^{-\lambda x}]_{0}^x=1-e^{-\lambda x}$$
Let $Y:= \frac{1}{X}$. We have for $F_Y(x)$:
$$F_Y(x)=\mathbb{P}({Y \leq x})=\mathbb{P}\left(\frac{1}{X}\leq x\right )=\mathbb{P}\left(X\geq\frac{1}{x}\right )=1-\mathbb{P}\left(X < \frac{1}{x} \right)=^!1-F_X\left(\frac{1}{x}\right)$$
The problem at $!$ is that definition of cumulative density function requires a non strict inequality so I don't know why this holds.
Aside from that, how does one get the distribution of $Y$ from the cumulative density function/probability density function? How do we deal with $Y$ at $X=0$ (does $X$ attain $0$ as a value even)
Best Answer
Observe that $$\left(-\infty, \dfrac{1}{x}\right) \cup \left\{\dfrac{1}{x} \right\}=\left(-\infty, \dfrac{1}{x} \right]$$ and that the sets $\left(-\infty, \dfrac{1}{x}\right)$, $\left\{\dfrac{1}{x} \right\}$ are disjoint.
Therefore, on the left-hand side, $$\mathbb{P}\left[\left(-\infty, \dfrac{1}{x}\right) \cup \left\{\dfrac{1}{x} \right\}\right]=\mathbb{P}\left[\left(-\infty, \dfrac{1}{x}\right) \right] + \mathbb{P}\left(\left\{\dfrac{1}{x} \right\}\right) = \mathbb{P}\left(X < \dfrac{1}{x}\right)+0$$ because $$\mathbb{P}\left(\left\{\dfrac{1}{x} \right\}\right) = 0$$ due to $X$ being a continuous random variable, and on the right-hand side, we simply have $$\mathbb{P}\left[\left(-\infty, \dfrac{1}{x} \right] \right] = \mathbb{P}\left(X \leq \dfrac{1}{x}\right)$$ Hence, equating the two sides, $$\mathbb{P}\left(X \leq \dfrac{1}{x}\right) = \mathbb{P}\left(X < \dfrac{1}{x}\right)$$