I am attempting to algorithmically estimate the burstiness of a dataset and have found two comparable metrics.
The coefficient of variation is the ratio of the standard deviation to the mean. The index of dispersion is the ratio of the variance to the mean.
Why would a person use one over the other? What are the pros and cons of each?
Note that the coefficient of variation (CV) is always dimensionless and is scale invariant. On the other hand, the index of dispersion (ID) is not scale invariant and is dimensionless only when it applies to a dimensionless variable such as a count, as is the case in practice. Both CV and ID are for non-negative variables, but they are used in different contexts.
The sample and theoretical CVs provide nice indications about continuous distributions and samples. The exponential distribution has unit CV and can be seen as a reference within some families of distributions. The gamma, the Weibull and the Generalised Pareto (GP) families embed distributions with arbitrary CVs, and there is a one-to-one relation between their shape parameter and the CV. In the three families, $\text{CV}<1$ indicates a tail which is thinner than exponential, while $\text{CV}>1$ is for a tail thicker than exponential, and even is an heavy tail in the case of GP.
The sample and theoretical IDs are most often used for discrete variables with non-negative integer values such as counts. The reference distribution with $\text{ID} = 1$ is now the Poisson distribution, notably in the family made of the three distributions. Binomial, Poisson and Negative Binomial. The binomial is underdispersed ($\text{ID} < 1$) and the Negative Binomial is overdispersed ($\text{ID} > 1$). The ID is often used in the theory of point processes where the Poisson distribution plays a major role.
An interesting relationship between the two notions is provided by the renewal process: given as sequence of i.i.d. positive r.vs $X_i$ usually representing lifetimes, the interest is on the sum $S_n := X_1 + X_2 + \dots +X_n$ for large $n$, and on the number $N_t$ of renewals $S_n$ falling in the interval $(0,\,t)$. When the $X_i$ are exponential, $N_t$ is Poisson. Under quite general assumptions the ID of $N_t$ tends for large $t$ to the square of the CV of $X$ so $N_t$ is overdispersed when the CV of $X$ is $> 1$.
