# Mean of inverse exponential distribution

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...

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 $r = 1$.

• This is linked with my question in here – Diogo Santos Aug 14 '16 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,

$u = y^{-1}$

and

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

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.

http://www.wolframalpha.com/input/?i=integrate+from+0+to+infinity+(1%2Fx)+exp(-x)+dx

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 $\alpha \gt 2$.

• 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. – Strants Aug 12 '16 at 16:05

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

n<-1000
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.

• The mean cannot exist because the exponential has positive density in any neighborhood of zero. – whuber Aug 12 '16 at 15:45
• @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. – RUser4512 Aug 12 '16 at 15:51
• 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. – whuber Aug 12 '16 at 15:56
• – kjetil b halvorsen Oct 9 '19 at 7:28