How to Estimate Uniform Distribution Parameters by Moments

The method of moments estimates distribution parameters by matching theoretical moments to moments calculated from the data. For a uniform distribution, the formula depends on which endpoint or endpoints are unknown. The common one-parameter model is uniform on $[0,\theta]$; if both endpoints are unknown, you need two moments.

The estimates below use the ordinary method-of-moments convention: the $k$th sample raw moment is $n^{-1}\sum_{i=1}^n X_i^k$. This means the variance-like quantity in the two-parameter formula uses divisor $n$, not $n-1$.

If the distribution is uniform on $[0,\theta]$

For $X\sim\operatorname{Uniform}(0,\theta)$, the population mean is

The first sample raw moment is the sample mean, $\bar X=n^{-1}\sum_{i=1}^n X_i$. Match it to the population mean and solve:

For example, with observations $1, 2, 3, 6$, the sample mean is $3$, so the method-of-moments estimate is $\hat\theta=6$. This estimate is not based on the largest observation; it comes from matching means. If the known lower endpoint is instead $c$, so that $X\sim\operatorname{Uniform}(c,c+\theta)$, then $\mathbb{E}[X]=c+\theta/2$ and $\hat\theta=2(\bar X-c)$.

If both endpoints are unknown

Suppose $X\sim\operatorname{Uniform}(a,b)$, with $a<b$. Its first two raw moments are

Define the first and second sample raw moments as

The moment equations are $m_1=(a+b)/2$ and $m_2=(a^2+ab+b^2)/3$. It is convenient to use the identity $\operatorname{Var}(X)=\mathbb{E}[X^2]-\mathbb{E}[X]^2=(b-a)^2/12$. The corresponding sample moment difference is

Thus the estimated midpoint is $\bar X$, and the estimated width is $\sqrt{12(m_2-m_1^2)}$. Solving for the endpoints gives

Equivalently, if $v_n=n^{-1}\sum_{i=1}^n(X_i-\bar X)^2$, then $\hat a=\bar X-\sqrt{3v_n}$ and $\hat b=\bar X+\sqrt{3v_n}$.

Worked example

Take the data $1,2,4,5$. Their mean is $\bar X=3$, and their second raw moment is

Therefore $m_2-\bar X^2=11.5-9=2.5$. The estimated endpoints are

The moment estimate is approximately $\operatorname{Uniform}(0.26,5.74)$. If you are given the usual unbiased sample variance $s^2$ (with divisor $n-1$), first convert it: $v_n=((n-1)/n)s^2$. Do not substitute $s^2$ directly for $v_n$.

A symmetric uniform distribution

A different model is $X\sim\operatorname{Uniform}(-\theta,\theta)$. Its mean is always zero, so the first moment cannot identify $\theta$. Use the second moment, $\mathbb{E}[X^2]=\theta^2/3$, instead. Matching it to $m_2=n^{-1}\sum X_i^2$ gives $\hat\theta=\sqrt{3m_2}$, with $\theta\geq 0$.

Common mistakes

• Assuming there is one formula for every uniform model. The answer changes depending on whether the interval is $[0,\theta]$, $[c,c+\theta]$, $[-\theta,\theta]$, or $[a,b]$.
• Using the sample maximum for the method of moments. The maximum is central to the maximum-likelihood estimate in some uniform models, but it is not the method-of-moments calculation. For $[0,\theta]$, the moment estimate is $2\bar X$.
• Using divisor $n-1$ without adjustment. Standard sample raw moments use divisor $n$. If starting from the unbiased sample variance, multiply it by $(n-1)/n$ before applying the two-endpoint formula.
• Treating the result as a maximum-likelihood estimate. With two unknown endpoints, the moment estimates need not place every observation between the estimated endpoints. They are still the algebraic moment estimates, but a fitted uniform density with those endpoints would assign zero density to any observation outside the interval. Maximum likelihood instead uses the sample minimum and maximum (with parameter-space details depending on the convention).

If all observations are identical, the two-endpoint formulas give equal endpoints. This is a degenerate boundary case rather than a nonzero-width continuous uniform distribution.