Given a random variable $Y = Exp(\lambda)$, what is the mean and variance of $G=\dfrac{1}{Y}$ ?

I look at the Inverse Gamma Distribution, but the mean and variance are only defined for $\alpha>1$ and $\alpha>2$ respectively...

  • $\begingroup$ This is a special case of stats.stackexchange.com/questions/299722, so the answer there applies directly and tells you the mean and variance are both infinite. $\endgroup$
    – whuber
    Commented Jul 7, 2022 at 15:33

4 Answers 4


Given that the inverse exponential distribution has $\alpha = 1$, you have stumbled upon the fact that the mean of the inverse exponential is $\infty$. And therefore, the variance of the inverse exponential is undefined.

If $G$ is inverse exponentially distributed, $E(G^r)$ exists and is finite for $r < 1$, and $= \infty$ for $r = 1$.

  • $\begingroup$ This is linked with my question in here $\endgroup$ Commented Aug 14, 2016 at 11:23

I'll show the calculation for the mean of an Exponential distribution so it will recall you the approach. Then, I'll go for the inverse Exponential with the same approach.

Given $f_Y(y) = \lambda e^{-\lambda y}$

$E[Y] = \int_0^\infty{yf_Y(y) dy}$

$ = \int_0^\infty{y \lambda e^{-\lambda y} dy}$

$ = \lambda \int_0^\infty{y e^{-\lambda y} dy}$

Integrating by part (ignore the $\lambda$ in front of the integral for the moment),

$u = y, dv=e^{-\lambda y} dy$

$du = dy, v = \frac{-1}{\lambda}e^{-\lambda y}$

$ = y \frac{-1}{\lambda}e^{-\lambda y} - \int_0^\infty{ \frac{-1}{\lambda}e^{-\lambda y} dy}$

$ = y \frac{-1}{\lambda}e^{-\lambda y} + \frac{1}{\lambda} \int_0^\infty{ e^{-\lambda y} dy}$

$ = y \frac{-1}{\lambda}e^{-\lambda y} - \frac{1}{\lambda^2} e^{-\lambda y}$

Multiply by the $\lambda$ in front of the integral,

$ = - y e^{-\lambda y} - \frac{1}{\lambda} e^{-\lambda y}$

Evaluate for $0$ and $\infty$,

$ = (0 - 0) - \frac{1}{\lambda} (0 - 1)$

$ = \lambda^{-1}$

Which is a known results.

For $G = \frac{1}{Y}$, the same logic apply.

$E[G] = E[\frac{1}{Y}]= \int_0^\infty{\frac{1}{y} f_Y(y) dy}$

$ = \int_0^\infty{\frac{1}{y} \lambda e^{-\lambda y} dy}$

$ = \lambda \int_0^\infty{\frac{1}{y} e^{-\lambda y} dy}$

The main difference is that for an integration by parts,

$u = y^{-1}$


$du = -1y^{-2}$

so it doesn't help us for $G = \frac{1}{y}$. I think the integral is undefined here. Wolfram alpha tell me it doesn't converge.


So the mean doesn't exist for the inverse Exponential, or, equivalently, for the inverse Gamma with $\alpha=1$. The reason is similar for the variance and $\alpha \gt 2$.

  • 2
    $\begingroup$ Note that (as Whuber commented on another answer) $\exp(-\lambda y)$ is bounded away from $0$ for $y$ near $0$, and $\int_0^\epsilon \frac{1}{y}\;dy$ diverges for any $\epsilon > 0$, so the integral for $E[G]$ does indeed diverge. $\endgroup$
    – user46207
    Commented Aug 12, 2016 at 16:05

After a quick simulation (in R), it seems that the mean does not exist : enter image description here

rates <- c(1,0.5,2,10)

par(mfrow = c(2,2))
for(rate in rates)
  plot(cumsum(1/rexp(n, rate))/seq(1,n),type='l',main = paste0("Rate = ",rate),
       xlab = "Sample size", ylab = "Empirical Mean")

For the sake of comparison, here is what happens with a genuine exponential random variable.

enter image description here

  • 5
    $\begingroup$ The mean cannot exist because the exponential has positive density in any neighborhood of zero. $\endgroup$
    – whuber
    Commented Aug 12, 2016 at 15:45
  • $\begingroup$ @whuber indeed, this is what I tried to stress : the empirical mean does not converge for the inverse of an exponential law, whereas it does for an exponential law. $\endgroup$
    – RUser4512
    Commented Aug 12, 2016 at 15:51
  • 5
    $\begingroup$ Yes, but (1) from the fact I quoted, the conclusion of no expectation is immediately obvious and (2) no amount of simulation can do any more than suggest that an expectation might be undefined. For instance, if one were to truncate the exponential at a lower limit of $10^{-1000}$, its inverse would indeed have a finite expectation, but your simulations would not look any different. Therefore the simple observation (1) would appear to be much more informative and reliable than the simulations. $\endgroup$
    – whuber
    Commented Aug 12, 2016 at 15:56
  • $\begingroup$ See stats.stackexchange.com/questions/299722/… $\endgroup$ Commented Oct 9, 2019 at 7:28

You actually can bypass this expectation by introducing a small number $\varepsilon$, so that $$ \frac{E[G]}{\lambda} = \int_0^{\infty}{ \frac{1}{t} e^{-\lambda t} dt } = \lim\limits_{\varepsilon\to 0} \int_0^{\frac{1}{\varepsilon}}{ \frac{1}{t+\varepsilon} e^{-\lambda t} dt } = \lim\limits_{\varepsilon\to 0}~\big\{e^{\lambda \varepsilon} \Gamma(0; \lambda \varepsilon) - e^{\lambda \varepsilon} \Gamma(0; \lambda (\varepsilon^{-1}+\varepsilon)) \big\} $$ where $\Gamma(\cdot,\cdot)$ is the upper incomplete Gamma function.

With the above equation, you can do whatever you want with $\lambda$.

For example, if you choose a $\lambda$ satisfying $\lambda \epsilon \to 1$, thus $\lambda \to \infty$, we obtain that $$ E[G] \to e \lambda \Gamma(0,1). $$

In this case, $E[G]$ does exist as $\lambda \to \infty$.


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