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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)

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    $\begingroup$ For continuous r.v. $\mathbb P\{X<x\}=\mathbb P\{X\leq x\}$. $\endgroup$
    – Surb
    Commented Apr 30, 2018 at 15:40
  • $\begingroup$ Does the support being positive mean the image of $X$ is positive? $\endgroup$
    – user515599
    Commented Apr 30, 2018 at 15:48
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    $\begingroup$ The central issue that answers both of your questions is that for a continuous random variable $X$ and any fixed number $a$, $\mathbb P(X = a) = 0$. This is why $\mathbb P(X \leq c) = \mathbb P(X < c)$ -- after all, $\{X \leq c\} = \{X < c\} \cup \{X = c\}$ and the union is disjoint -- and why you don't have to worry about whether or not $X$ might be $0$. $\endgroup$
    – Aaron Montgomery
    Commented Apr 30, 2018 at 15:50

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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, $$\mathbb{P}\left(X \leq \dfrac{1}{x}\right) = \mathbb{P}\left(X < \dfrac{1}{x}\right)$$

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  • $\begingroup$ Thank you for addressing the continuity issue. What can we say about the distribution of $Y$? $\endgroup$
    – user515599
    Commented Apr 30, 2018 at 15:52
  • $\begingroup$ @whycantihavealongername Well, from your work, you have $F_Y(x) = 1 - F_X(1/x) = 1-(1-e^{-\lambda/x}) = e^{-\lambda / x}$. You can use this to find the PDF by taking the derivative if you need it. I don't know if there's a standard name for this distribution. $\endgroup$
    – Clarinetist
    Commented Apr 30, 2018 at 15:55
  • $\begingroup$ @whycantihavealongername I suppose you could call it an "inverse exponential distribution:" en.wikipedia.org/wiki/… $\endgroup$
    – Clarinetist
    Commented Apr 30, 2018 at 15:56

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