If $X$ follows an $\operatorname{Exp}(\theta)$, does $1/X$ follow an $\operatorname{Exp}(1/ \theta)$?

Question

I heard a teacher say that if $$X \sim \operatorname{Exp}(\theta)$$ then $$\frac{1}{X} \sim \operatorname{Exp}\left(\frac{1}{\theta}\right)$$ I don't trust this teacher because he has given us wrong answers before.

I tried to obtain this myself:

If $y = g(x) = \frac{1}{x}$, then $$f_Y(y) = f_X(g^{-1}(y)) \cdot \left|\frac{d}{dy}g^{-1}(y)\right|$$ $$f_Y(y) = \theta \cdot e^{-\theta/y} \cdot \frac{1}{y^2}$$ where $1/y \in (0, +\infty)$.

I don't see an $\operatorname{Exp}(1/\theta)$ here. Am I doing something wrong?

Answer 1

You are right and your transformation is correct!

But without doing any calculation, what is true is that $X\sim \operatorname{Exp}(\theta)=\Gamma(1;\theta)$

And this means that $\frac{1}{X}\sim \text{Inv-Gamma}$

See "Related distributions" in this link

Answer 2

If $1/X \sim \operatorname{Exponential}(1/\theta)$ as claimed, then $\theta = 1$ would suggest $1/X$ is exponential with rate $1$ if $X$ is exponential with rate $1$. This of course is absurd, since for example $$\Pr[X > 1] = e^{-1}$$ but $$\Pr[1/X > 1] = \Pr[X < 1] = 1 - e^{-1}.$$

Answer 3

Let $Y= \frac1X$

• $\mathbb P(X \ge x)= \exp(-\theta x)$ for $x >0$
• $\mathbb P(Y \le \frac1x)= \exp(-\theta x)$
• $\mathbb P(Y \le y)= \exp(-\theta / y)$ which is certainly not the CDF of an exponential distribution

The density for $Y$ is then $\frac{\theta}{y^2} \exp(-\theta / y)$ which, as tommik says, is called an inverse gamma distribution

Answer 4

Three things to note for finding CDF and PDF of functions of rvs:

1. If the function is decreasing, the inequality sign flips
2. It's often easier to find CDF first.
3. Be careful if support of the function changes, e.g. if $Y= \log X$, you can only consider $X>0$ even if $X$ is defined on $X \leq 0$. In your case, for $X>0$: $$ F_Y(y) = P(Y \leq y) = P(\frac{1}{X} \leq y) = P(X \geq \frac{1}{y}) = \int_{\frac{1}{y}}^{\infty} \lambda e^{- \lambda x} dx = e^{-\frac{\lambda}{y}} $$ Now use the definition of PDF (derivative of CDF): $$ f_{Y}(y) = \frac{\lambda}{y^2}e^{-\frac{\lambda}{y}}, y > 0 $$