Normal Mean And Variance By Moment Matching

Why Match Moments for a Normal Model?

A normal model is described by two parameters: its mean, which determines its center, and its variance, which determines its spread. The method of moments estimates these parameters by equating selected population moments to their sample counterparts. For the normal model, the first raw moment identifies the center, while the second central moment identifies the variance.

The calculations are direct, but one detail matters: the sample central second moment used for moment matching divides by $n$, not $n-1$. That convention gives a method-of-moments estimate; it does not make the estimate unbiased. Keeping these two ideas distinct is important when interpreting a result.

Definition: Suppose $X_1,\ldots,X_n$ are an independent random sample from $N(\mu,\sigma^2)$, where $\mu$ is the population mean and $\sigma^2>0$ is the population variance. The method-of-moments estimates obtained by matching the first raw sample moment and second central sample moment are $$ \widehat{\mu}_{\mathrm{MM}}=\bar X=\frac{1}{n}\sum_{i=1}^{n}X_i, \qquad \widehat{\sigma^2}_{\mathrm{MM}}=\frac{1}{n}\sum_{i=1}^{n}(X_i-\bar X)^2. $$ Here $\bar X$ is the sample mean. The corresponding estimate of the standard deviation is $\widehat{\sigma}_{\mathrm{MM}}=\sqrt{\widehat{\sigma^2}_{\mathrm{MM}}}$.

Population and Sample Moments

For $X\sim N(\mu,\sigma^2)$, the first raw population moment is $E[X]=\mu$. The second central population moment is $E[(X-\mu)^2]=\sigma^2$. A raw moment is taken about zero; a central moment is taken about the population mean. The sample analogues are the sample mean and the average squared deviation from the sample mean.

The sample central moment uses $\bar X$ in place of the unknown $\mu$. Thus the two moment equations are

\bar X=\mu, \qquad \frac{1}{n}\sum_{i=1}^{n}(X_i-\bar X)^2=\sigma^2.

Solving these equations for the unknown parameters gives the estimates in the definition. In actual data, substitute the observed values for $X_1,\ldots,X_n$. The normal assumption supplies the model and the population moment relationships; these moment-matching calculations themselves do not require likelihood maximization.

Worked Example: Estimating Both Parameters From Observations

A hypothetical laboratory records five measurement errors, in the same measurement unit: $3, 5, 7, 9, 11$. Treat them as an independent sample from $N(\mu,\sigma^2)$. Find both method-of-moments estimates, including the estimate of the standard deviation.

Method and first moment. The method is to match the first raw sample moment to $E[X]=\mu$. The sum of the observations is $3+5+7+9+11=35$, so

\widehat{\mu}_{\mathrm{MM}}=\bar x=\frac{35}{5}=7.

Second central moment. The deviations from the observed mean are $-4,-2,0,2,4$. Their squared deviations are $16,4,0,4,16$, with total $40$. Therefore,

\widehat{\sigma^2}_{\mathrm{MM}} =\frac{16+4+0+4+16}{5} =\frac{40}{5}=8.

The estimated standard deviation is $\widehat{\sigma}_{\mathrm{MM}}=\sqrt{8}\approx 2.828$, rounded to three decimal places. The variance estimate is in squared measurement units, while the standard deviation estimate is in the original measurement unit.

Check. The deviations sum to zero, as they must when measured from the sample mean. The squared-deviation average is $8$, and its square root is approximately $2.828$. A common mistake would be to divide the sum $40$ by $n-1=4$, producing $10$; that computes the unbiased sample variance, not the moment-matching estimate.

Why the Raw-Moment Equations Give the Same Variance Estimate

The focus here is matching the first raw moment and second central moment. It is useful to see why the same normal variance estimate also results if one instead matches the first two raw moments. For any random variable with a finite second moment,

E[(X-\mu)^2]=E[X^2]-\mu^2.

To verify this identity, expand the square and use $E[X]=\mu$:

E[(X-\mu)^2] =E[X^2-2\mu X+\mu^2] =E[X^2]-2\mu E[X]+\mu^2 =E[X^2]-\mu^2.

For a normal population, this says $\sigma^2=E[X^2]-\mu^2$. Matching the first raw sample moment gives $\widehat{\mu}_{\mathrm{MM}}=\bar X$. Matching the second raw moment gives $\widehat{E[X^2]}=\frac{1}{n}\sum X_i^2$. Subtracting the square of the first estimate yields

\widehat{\sigma^2}_{\mathrm{MM}} =\frac{1}{n}\sum_{i=1}^{n}X_i^2-\bar X^2 =\frac{1}{n}\sum_{i=1}^{n}(X_i-\bar X)^2.

The final equality follows by expanding the squared deviations and using $\sum X_i=n\bar X$. Therefore, either choice of the second moment equation leads to the same variance estimate when the first equation is also matched.

Worked Example: Using Summary Totals

A hypothetical analyst has a sample of $n=5$ recorded values. The available summaries are $\sum x_i=30$ and $\sum x_i^2=218$. Assuming an independent sample from $N(\mu,\sigma^2)$, calculate the method-of-moments estimates using the raw-moment form, then verify the variance using the central-moment identity.

Method. The first raw sample moment is the sample mean:

\bar x=\frac{\sum x_i}{n}=\frac{30}{5}=6.

The second raw sample moment is

\frac{1}{n}\sum x_i^2=\frac{218}{5}=43.6.

Subtract the square of the sample mean to obtain the variance estimate:

\widehat{\sigma^2}_{\mathrm{MM}} =43.6-6^2 =43.6-36 =7.6.

Verification. The central-moment calculation can be obtained without listing the individual observations:

\frac{1}{5}\sum_{i=1}^{5}(x_i-\bar x)^2 =\frac{1}{5}\sum x_i^2-\bar x^2 =43.6-36 =7.6.

Thus the estimates are $\widehat{\mu}_{\mathrm{MM}}=6$ and $\widehat{\sigma^2}_{\mathrm{MM}}=7.6$. The estimated standard deviation, if required, is $\sqrt{7.6}\approx2.757$, rounded to three decimal places. The common error is to report $43.6$ as the variance; it is the second raw sample moment and still includes the contribution from the nonzero center.

Check. The computed variance is nonnegative, as a sum of squared deviations must be. The raw-moment subtraction and central-moment identity agree.

The Denominator and the Meaning of the Estimate

For observed values $x_1,\ldots,x_n$, define the sample central second moment used here by

m_2=\frac{1}{n}\sum_{i=1}^{n}(x_i-\bar x)^2.

A different familiar quantity is the unbiased sample variance,

s^2=\frac{1}{n-1}\sum_{i=1}^{n}(x_i-\bar x)^2, \qquad n>1.

These are related by $s^2=\frac{n}{n-1}m_2$. The factor $n-1$ is used in the unbiased sample variance because estimating the mean from the same observations affects the expected sum of squared deviations. For method-of-moments estimation by matching the sample central moment, use $m_2$, which divides by $n$. Do not switch denominators partway through the calculation.

Worked Example: Avoiding the Unbiased-Variance Substitution

Five hypothetical sensor readings are $2, 4, 6, 8, 10$, modeled as an independent sample from $N(\mu,\sigma^2)$. Find the method-of-moments estimates, and compare the variance estimate with the unbiased sample variance.

Method and mean. The observations sum to $30$, so

\bar x=\frac{30}{5}=6.

The deviations are $-4,-2,0,2,4$, whose squared values sum to $16+4+0+4+16=40$. The method-of-moments variance estimate is

\widehat{\sigma^2}_{\mathrm{MM}} =\frac{40}{5}=8.

The unbiased sample variance is instead $s^2=40/4=10$. The two values answer different questions: the moment-matching estimate is $8$, while $10$ is the unbiased sample variance. The method-of-moments standard deviation estimate is $\sqrt{8}\approx2.828$, rounded to three decimal places.

Check. The mean is the midpoint of the symmetric observations, and the squared deviations total $40$. Dividing by $n=5$ gives the required moment estimate. Dividing by $n-1=4$ is the specific denominator error to avoid when the question asks for method of moments.

Conditions, Interpretation, and Exam Traps

The normal model assumes each observation has the same distribution $N(\mu,\sigma^2)$, and an independent sample means the observations are mutually independent. Under this model, $\mu$ and $\sigma^2$ are the parameters being estimated. The moment formulas also produce numerical summaries for data that are not normal, but without the stated model they should not automatically be interpreted as estimates of normal-distribution parameters.

For a nonconstant observed sample, the method-of-moments variance estimate is positive. If all observations are identical, it is zero. That result is the direct consequence of the sample central moment; it does not establish that the assumed population variance is zero. In particular, a zero estimate from a small or degenerate sample does not prove that the underlying population is degenerate.

Do not confuse raw and central moments. The second raw moment is $\frac{1}{n}\sum x_i^2$, not the variance estimate. Subtract $\bar x^2$, or calculate squared deviations from $\bar x$.
Do not substitute $n-1$ automatically. The method-of-moments central sample moment divides by $n$. The unbiased sample variance is a distinct statistic.
Keep variance and standard deviation separate. The estimate of $\sigma^2$ is a squared quantity. Take its square root only when an estimate of $\sigma$ is requested.
Check the scale of the result. If each observation is recorded in a given unit, the mean and standard deviation use that unit, while the variance uses its square.

This method is aligned with the SOA VEE Mathematical Statistics syllabus: identify the model’s parameter functions, select the corresponding sample moments, and solve the matching equations. Writing the equations before substituting numbers helps prevent confusion between the second raw moment, the central moment, and the unbiased sample variance.

Check Your Understanding

For each problem, use the normal method-of-moments equations and keep the variance denominator consistent with moment matching.

A hypothetical sample from $N(\mu,\sigma^2)$ is $2, 5, 8, 9$. Calculate the method-of-moments estimates of $\mu$ and $\sigma^2$. (Answer: $\widehat{\mu}_{\mathrm{MM}}=6$, $\widehat{\sigma^2}_{\mathrm{MM}}=7.5$)
For five hypothetical observations, $\sum x_i=30$ and $\sum x_i^2=218$. Find the estimates of $\mu$ and $\sigma^2$ by raw-moment matching. (Answer: $\widehat{\mu}_{\mathrm{MM}}=6$, $\widehat{\sigma^2}_{\mathrm{MM}}=7.6$)
A sample of six hypothetical values has $\sum x_i=42$ and $\sum x_i^2=330$. Compute the method-of-moments estimates and the estimate of the standard deviation to three decimal places. (Answer: $\widehat{\mu}_{\mathrm{MM}}=7$, $\widehat{\sigma^2}_{\mathrm{MM}}=6$, $\widehat{\sigma}_{\mathrm{MM}}=2.449$)
Five hypothetical observations are $1, 4, 4, 7, 9$. Calculate the method-of-moments variance estimate and the unbiased sample variance. State which one is the moment estimate. (Answer: method-of-moments variance $=7.6$; unbiased sample variance $=9.5$)
For $n=10$ hypothetical observations, $\sum x_i=25$ and $\sum x_i^2=90$. Find the method-of-moments estimates of the mean, variance, and standard deviation; round the standard deviation to three decimal places. (Answer: $\widehat{\mu}_{\mathrm{MM}}=2.5$, $\widehat{\sigma^2}_{\mathrm{MM}}=2.75$, $\widehat{\sigma}_{\mathrm{MM}}=1.658$)

What You'll Learn