Intuition for relationship between variance of geometric and exponential distributions

Question

Given that the exponential distribution can be thought of as a continuous version of the geometric, is there an intuitive way to relate their variances?

I have some intuition for how the means of these distributions relate: a larger $\lambda$ corresponds to a larger $p$ (or, put intuitively, a faster rate of arrival in the exponential case corresponds to a higher chance of flipping 'heads' in the geometric case).

However, I have failed to come up with a way to understand how the variances relate.

Answer 1

There is a clear relationship between the geometric and exponential distributions.
If $X\sim\text{Exp}(\lambda)$, $F_X(x)=1-e^{-\lambda x}$, and $Y\sim\text{Geom}(p)$, $F_Y(y)=1-(1-p)^{\lfloor y \rfloor}$, where $\lfloor y \rfloor$ is the floor function, then $F_Y$ can be determined by $F_X$ with $\lambda=-\ln(1-p)$, that is $P(Y\le y)=P(X\le \lfloor y \rfloor)$. See B. J. Prochaska, "A Note on the Relationship Between the Geometric and Exponential distributions", The American Statistician, 27(1):7.

As to their variances, you can consider that \begin{align*} E[X]&=\frac{1}{\lambda},\qquad V[X]=\frac{1}{\lambda^2} =\frac{E[X]}{\lambda} \\ E[Y]&=\frac{1}{p},\qquad V[Y]=\frac{1-p}{p^2} \end{align*} i.e. the variance decreases as $\lambda$ or $p$ increases:

• if the rate of arrival is fast, arrivals "concentrate" around a short waiting time;
• if the chance of flipping heads is high, then the number of trials "concentrate" around a small number.

An example in R:

> set.seed(1234)
> e1 <- rexp(1000, 0.25)
> e2 <- rexp(1000, 0.75)
> round(range(e1),2)
[1]  0.00 29.07
> round(range(e2),2)
[1] 0.00 9.95
> g1 <- rgeom(1000, 0.25)
> g2 <- rgeom(1000, 0.75)
> range(g1)
[1]  0 22
> range(g2)
[1] 0 3