Variance Bounds from Bounded Supports

Why a Bounded Range Limits Variability

If a random variable cannot leave a finite interval, its values cannot be arbitrarily far from their mean. This simple restriction gives a precise upper bound on its variance. The bound depends only on the interval width, and it is sharp: an appropriate distribution concentrated at the two endpoints attains it.

A more informative inequality retains the location of the mean within the interval. We will derive that bound first, then maximize it over possible means. Keeping these two steps separate makes the equality conditions transparent.

Since \(a\le X\le b\) almost surely, the mean also satisfies \(a\le\mu\le b\). The first bound below uses both endpoints and the value of \(\mu\); the second removes the mean to give a bound depending only on the interval length.

The Mean-Sensitive Variance Bound

Proof. For every \(x\in[a,b]\), both \(x-a\) and \(b-x\) are nonnegative. Therefore \((X-a)(b-X)\ge0\) almost surely. Expanding this product and taking expectations gives

Because \(X\) has a finite second moment, \(\operatorname{Var}(X)=E[X^2]-\mu^2\). Subtracting \(\mu^2\) from the last inequality yields

To bound the product, write \(d=b-a\). Direct expansion gives

Combining the inequalities proves \(\operatorname{Var}(X)\le(b-a)^2/4\). When \(a=b\), the almost-sure condition forces \(X=a\) almost surely, so \(\mu=a\) and the variance is zero; the same displayed bounds also give zero. \(\square\)

The mean-sensitive bound is strongest when the mean is near the center of the interval. If \(\mu\) is close to either endpoint, the product \((b-\mu)(\mu-a)\) is small. In particular, if \(\mu=a\) or \(\mu=b\), the variance must be zero.

When Are the Bounds Sharp?

There are two distinct equality questions. The mean-sensitive inequality can be sharp for a given mean anywhere in \([a,b]\). The range-only inequality is sharp only when the mean is at the midpoint as well as when the mean-sensitive inequality is sharp.

Assume first that \(a<b\). Equality in the first inequality of the proof holds exactly when \((X-a)(b-X)=0\) almost surely. To see this, the random variable \((X-a)(b-X)\) is nonnegative and bounded; its expectation is zero if and only if it is zero almost surely. Within \([a,b]\), its zeros are precisely \(a\) and \(b\). Thus equality in the mean-sensitive bound holds exactly when \(X\) is supported on the two endpoints, allowing one endpoint to have probability zero.

More explicitly, if \(P(X=b)=p\) and \(P(X=a)=1-p\), then

This formula is valid under the stated condition \(a<b\). For every \(\mu\in[a,b]\), it gives a probability \(p\in[0,1]\), and the resulting endpoint-supported distribution attains the mean-sensitive bound. Indeed, its variance is

If \(a=b\), this parameterization is not used: \(X=a\) almost surely, both bounds are zero, and any endpoint-probability description is unnecessary.

For the range-only bound, the identity

shows that its maximum over \(\mu\in[a,b]\) is attained only at \(\mu=(a+b)/2\), provided \(a<b\). Combining this with the equality condition just established, the range bound is attained exactly by a distribution assigning probability \(1/2\) to each endpoint. When \(a=b\), it is attained by the only possible distribution, the constant \(X=a\).

Comparing Continuous and Interior-Supported Distributions

The extremal distribution is not generally continuous, nor does it need to put positive probability throughout the interval. The following examples show how the same bound applies across different types of distributions and how interior values can make it strict.

Interpretation and a Common Pitfall

The range-only bound gives an immediate scale statement: a random variable confined to an interval of width \(b-a\) has standard deviation at most half that width. This does not say that every distribution on the interval is nearly as variable as the extremal endpoint distribution. The uniform and three-point examples have variance strictly below the maximum.

Common mistake: applying the endpoint probability formula \(p=(\mu-a)/(b-a)\) without checking that \(a<b\). If \(a=b\), the denominator is zero, and the expression is undefined. The correction is to handle this case directly: \(a\le X\le a\) almost surely forces \(X=a\) almost surely, hence \(\operatorname{Var}(X)=0\). For \(a<b\), the formula is valid, and it constructs an endpoint distribution with any feasible mean.

A second useful check is to distinguish equality in the two bounds. Endpoint support alone gives equality in the mean-sensitive bound, but it does not by itself give equality in the range-only bound. The mean must also be the interval midpoint, which for an endpoint-supported distribution requires equal endpoint probabilities.