Why does the Poisson distribution have variance equal to its mean?

## The claim For X ~ Poisson(λ), both `E[X] = λ` and `Var(X) = λ`. The standard deviation is therefore `sqrt(λ)`. This property is called **equidispersion**, and it is a consequence of the family having only one parameter: unlike the normal family, where the centre and the spread are set independently, a Poisson count that averages 100 *must* have a standard deviation of 10. ## Derivation from the mass function Start with `P(X = k) = e^(-λ) λ^k / k!`. The mean: `E[X] = Σ_{k≥0} k · e^(-λ) λ^k / k!`. The k = 0 term vanishes, and for k ≥ 1 the k cancels one factorial factor, leaving `λ · Σ_{k≥1} e^(-λ) λ^(k-1)/(k-1)! = λ · 1 = λ`, because the remaining sum is the full Poisson mass function summing to 1. The same trick twice gives the second factorial moment: `E[X(X-1)] = Σ_{k≥2} k(k-1) · e^(-λ) λ^k / k! = λ² · Σ_{k≥2} e^(-λ) λ^(k-2)/(k-2)! = λ²`. So `E[X²] = E[X(X-1)] + E[X] = λ² + λ`, and `Var(X) = E[X²] - (E[X])² = λ² + λ - λ² = λ`. ## Derivation from the binomial limit A second, more intuitive route: a binomial count has mean np and variance np(1-p). Hold np = λ fixed while letting p shrink toward 0. The mean stays λ, and the variance np(1-p) → np = λ because the (1-p) factor tends to 1. The binomial's variance is always *below* its mean by the factor (1-p); the Poisson is the boundary case where that discount disappears. This also explains why a Poisson count never has variance below its mean while a binomial always does. ## What equidispersion buys you It gives a free diagnostic. The **variance-to-mean ratio**, sometimes called the index of dispersion, is 1 for a Poisson count. Compute the sample mean and the sample variance of your observed counts and compare: - **Ratio near 1** — consistent with Poisson; the model's spread is credible. - **Ratio well above 1 (overdispersion)** — the counts vary more than a single rate can explain. The usual causes are clustering (one event triggers others, so arrivals come in bursts) or heterogeneity (the underlying rate itself differs across the days, users or regions being pooled, adding rate-variation on top of the Poisson noise). Consequence: Poisson-based intervals are too narrow and extreme days look far more surprising than they are. - **Ratio well below 1 (underdispersion)** — the counts are *more regular* than random arrivals would be. This happens with scheduled or capacity-capped events: a queue that processes at most 10 items an hour cannot produce a Poisson tail. Here Poisson overstates the variability. With counts averaging 20 per day and a sample variance of 95, the ratio is about 4.75 — decisively not Poisson. The right reading is not "the mean is wrong" but "one rate does not describe these days". ## A caution on the diagnostic Sample variance is itself noisy. With only 10 or 15 observed windows, a ratio of 1.8 is unremarkable; you need either many windows or a large ratio before calling overdispersion. Also compare the mean and variance on the **same** window definition — mixing hourly counts with daily counts guarantees nonsense. ## Relative variability shrinks as λ grows Because the standard deviation is sqrt(λ) while the mean is λ, the coefficient of variation is `sqrt(λ)/λ = 1/sqrt(λ)`. A process averaging 4 events has a standard deviation of 2 — a 50 percent relative swing. One averaging 400 has a standard deviation of 20, only 5 percent. This is why low-count segments look wildly unstable while high-count aggregates look smooth, even when the underlying rate is identical. Ranking small regions by raw event counts is treacherous for exactly this reason: the noisiest-looking ones are usually just the smallest. ## When you need variance above the mean If counts really are overdispersed, a one-parameter family cannot represent them. A two-parameter discrete family such as the negative binomial has variance strictly greater than its mean and can absorb the extra spread, which is why it is the standard fallback whenever the variance-to-mean check fails. ## Common errors Saying the variance is λ² (that is the square of the mean, not the variance), claiming the standard deviation is λ, believing the mean and variance can be tuned separately, or observing variance ≫ mean and continuing to use Poisson because "counts are always Poisson".