Estimating a Normal Mean With Known Variance

Why the Known-Variance Case Matters

A normal model is often used when measurements vary around a common location. In some estimation problems, the variance is treated as known from the model or supplied as part of the problem, while the mean remains unknown. The task is then to find the value of the mean that makes the observed sample most likely.

The key result is that the maximum likelihood estimate (MLE) of the mean is the sample average. The derivation matters: it shows how the sample size enters the score equation, why the variance changes the curvature of the log-likelihood but not its maximizing value, and how to check that a score-equation solution really is a maximum. This treatment is aligned with the SOA VEE Mathematical Statistics syllabus.

Model and assumptions: Let \(X_1,\ldots,X_n\) be independent observations from the same normal distribution, with \(X_i\sim N(\mu,\sigma^2)\). The variance \(\sigma^2>0\) is known and fixed, while the mean \(\mu\) is unknown and may be any real number. The observed data are \(x_1,\ldots,x_n\). The result below depends on the observations having this common normal model and on the variance being known; it is not a joint estimation of the mean and variance.

Likelihood and Log-Likelihood

For a single observation, the normal density at \(x_i\), viewed as a function of \(\mu\), is \(f(x_i\mid\mu)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left[-\frac{(x_i-\mu)^2}{2\sigma^2}\right]\). Independence means the joint density is the product of the individual densities. Once the data are observed, this joint density, viewed as a function of \(\mu\), is the likelihood.

L(\mu)=\prod_{i=1}^{n}\frac{1}{\sqrt{2\pi\sigma^2}} \exp\left[-\frac{(x_i-\mu)^2}{2\sigma^2}\right] =(2\pi\sigma^2)^{-n/2} \exp\left[-\frac{1}{2\sigma^2}\sum_{i=1}^{n}(x_i-\mu)^2\right].

The logarithm is convenient because it turns the product into a sum. Since the logarithm is strictly increasing, maximizing the likelihood and maximizing its log give the same value of \(\mu\). The variance is known, so terms involving only \(\sigma^2\), \(n\), and constants do not change as \(\mu\) changes.

\ell(\mu)=\log L(\mu) =-\frac{n}{2}\log(2\pi\sigma^2) -\frac{1}{2\sigma^2}\sum_{i=1}^{n}(x_i-\mu)^2.

Only the final term depends on \(\mu\). Differentiating the log-likelihood with respect to \(\mu\) gives the score. The score equation sets this derivative equal to zero.

U(\mu)=\frac{d\ell(\mu)}{d\mu} =\frac{1}{\sigma^2}\sum_{i=1}^{n}(x_i-\mu) =\frac{\sum_{i=1}^{n}x_i-n\mu}{\sigma^2}.

Because \(\sigma^2>0\), the score equation \(U(\mu)=0\) is equivalent to \(\sum_{i=1}^{n}x_i-n\mu=0\). Solving for \(\mu\) gives the estimate.

Theorem: For an independent sample from \(N(\mu,\sigma^2)\) with known \(\sigma^2>0\) and unrestricted \(\mu\), the maximum likelihood estimate is

\widehat{\mu}_{\mathrm{MLE}} =\frac{1}{n}\sum_{i=1}^{n}x_i =\bar{x}.

Why the Score Solution Is the Maximum

A score equation gives a candidate, but a stationary point must be checked. Differentiate the score once more. The second derivative is strictly negative for every possible \(\mu\), because \(n>0\) and \(\sigma^2>0\). Thus the log-likelihood is strictly concave: it bends downward throughout the real line, so its stationary point is its unique global maximum.

\frac{d^2\ell(\mu)}{d\mu^2} =-\frac{n}{\sigma^2}<0.

There is also a direct way to see the same result. Write each deviation from a candidate mean as \(x_i-\mu=(x_i-\bar{x})+(\bar{x}-\mu)\). On squaring and summing, the cross-term vanishes because \(\sum_{i=1}^n(x_i-\bar{x})=0\). Therefore,

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

The first term on the right does not depend on \(\mu\), and the second is nonnegative, reaching its smallest value only at \(\mu=\bar{x}\). The log-likelihood is a constant minus this sum divided by \(2\sigma^2\). It is therefore largest uniquely when \(\mu=\bar{x}\). This identity also explains the result geometrically: among all possible means, the sample average minimizes the sum of squared deviations.

A Reliable Calculation Procedure

In a numerical problem, keep the unknown parameter visible until the score equation has been solved. Then check the candidate by the second derivative or by the form of the log-likelihood. A useful sequence is:

Record the model.
Identify \(n\), the observed values, and the given variance. If the problem supplies a standard deviation, square it to obtain the variance used in the density.

Write the log-likelihood.
Use independence to add the individual log-densities. Keep the terms involving \(\mu\); parameter-free terms may be retained or omitted when maximizing.

Differentiate and solve.
Set the score equal to zero and solve for \(\mu\). Check that the equation contains \(n\mu\), since the parameter appears in all \(n\) observations.

Verify the maximum and report it.
Confirm the second derivative is negative, then give the sample average in the units of the data.

Worked Applications

Worked Example: Sensor Readings With a Known Variance

A hypothetical sensor is tested four times. Its readings are \(14.6,\ 15.1,\ 14.9,\) and \(15.4\) units. Assume the readings are independent and normally distributed with unknown mean \(\mu\) and known variance \(\sigma^2=0.36\) units squared. Find the MLE of \(\mu\), write the score, and verify the result.

Method and model. Use the normal-sample log-likelihood and solve its score equation. The supplied variance is \(0.36\), corresponding to a standard deviation of \(\sqrt{0.36}=0.6\) units.

First calculate the sample size and sum: \(n=4\), and \(14.6+15.1+14.9+15.4=60.0\). Thus \(\bar{x}=60.0/4=15.0\) units.

For these observations the score is \(U(\mu)=[60.0-4\mu]/0.36\). Setting it equal to zero gives \(60.0-4\mu=0\), so \(\widehat{\mu}_{\mathrm{MLE}}=15.0\) units.

Check. At \(\mu=15.0\), the residuals are \(-0.4,\ 0.1,\ -0.1,\) and \(0.4\), whose sum is \(0\). The score is therefore \(0/0.36=0\). The second derivative is \(-4/0.36<0\), confirming a unique maximum.

Common mistake avoided. Do not divide the sample sum by the variance when solving the score equation. The variance is a positive factor multiplying the score, so it cancels when the score is set to zero.

Worked Example: Converting a Known Standard Deviation

A hypothetical laboratory records five independent readings of a concentration, in milligrams per litre: \(31,\ 28,\ 35,\ 30,\) and \(26\). The normal model has an unknown mean and a known standard deviation of \(2\) milligrams per litre. Find the MLE of the mean and show the score equation using the correct variance.

Method and model. The standard deviation is not the variance. Convert it first: \(\sigma=2\), so \(\sigma^2=2^2=4\) in squared concentration units. The sample size is \(5\), and the sum of the readings is \(31+28+35+30+26=150\).

The score as a function of the candidate mean is \(U(\mu)=(150-5\mu)/4\). Setting \(U(\mu)=0\) gives \(150-5\mu=0\), hence \(\widehat{\mu}_{\mathrm{MLE}}=150/5=30\) milligrams per litre.

Check. The residuals at \(30\) are \(1,\ -2,\ 5,\ 0,\) and \(-4\), which sum to \(1-2+5+0-4=0\). The score is zero there, and the second derivative is \(-5/4<0\). Thus the estimate is the unique maximum.

Common mistake avoided. Inserting \(2\) instead of \(4\) in the denominator would give an incorrect log-likelihood and score scale. In this particular model the score equation still has the same root if that positive scale is used consistently, but it is not the likelihood specified by the problem.

Worked Example: Comparing the Maximum With a Nearby Candidate

A hypothetical process produces five independent normal measurements: \(19,\ 22,\ 20,\ 25,\) and \(24\). The variance is known to be \(4\) squared units. Find the MLE of the mean and compare the log-likelihood at that estimate with the log-likelihood at \(\mu=21\).

Method and model. Sum the observations: \(19+22+20+25+24=110\), so \(n=5\) and \(\bar{x}=110/5=22\). The score is \(U(\mu)=(110-5\mu)/4\), which is zero at \(\mu=22\). The MLE is therefore \(22\) units.

To compare the log-likelihoods, use \(\ell(\mu)=-\frac{5}{2}\log(2\pi\cdot4)-\frac{1}{8}\sum_{i=1}^{5}(x_i-\mu)^2\). At \(\mu=22\), the squared residuals sum to \((19-22)^2+(22-22)^2+(20-22)^2+(25-22)^2+(24-22)^2=9+0+4+9+4=26\).

At \(\mu=21\), the squared residuals sum to \((19-21)^2+(22-21)^2+(20-21)^2+(25-21)^2+(24-21)^2=4+1+1+16+9=31\). The parameter-free log-likelihood term is the same at both candidates, so \(\ell(22)-\ell(21)=(31-26)/8=5/8=0.625\). The log-likelihood is higher at \(22\).

Check. The second derivative is \(-5/4<0\), so the score solution at \(22\) is the unique global maximum, not merely better than the single comparison value \(21\).

Common mistake avoided. When comparing log-likelihoods, do not reverse the sign of the squared-residual difference. A smaller residual sum produces a larger log-likelihood because the sum is subtracted.

Common Exam Traps and Interpretation

The known variance appears in the score as a divisor, but because it is positive and fixed, it does not affect the score equation's root. It does affect the shape of the log-likelihood: for fixed \(n\), a smaller variance produces greater curvature in magnitude. That fact does not change the estimated mean in this model.

Dropping the sample-size factor. The derivative of \(\sum_{i=1}^n(x_i-\mu)^2\) is \(-2\sum_{i=1}^n(x_i-\mu)\), which includes \(n\) copies of \(\mu\). The score numerator is \(\sum x_i-n\mu\), not \(\sum x_i-\mu\).
Mixing up variance and standard deviation. The normal density and log-likelihood use \(\sigma^2\). If a problem gives \(\sigma\), square it before writing those expressions.
Stopping at the score equation. A zero score identifies a stationary point. Here the negative second derivative, or the squared-deviation identity, verifies that it is the unique maximum.
Confusing the estimate with a known parameter. The observations determine the estimate \(\bar{x}\); \(\mu\) remains the unknown population mean in the model. The estimate is reported in the measurement units, not squared units.

The sample average is the MLE under the stated normal model, but that conclusion should not be detached from its assumptions. The derivation here treats \(\sigma^2\) as known. It does not estimate the variance or maximize over mean and variance simultaneously. It also assumes an unrestricted real-valued mean and independent observations with the same normal distribution.

Key takeaway: For independent observations from \(N(\mu,\sigma^2)\) with known positive variance, the log-likelihood score is \((\sum x_i-n\mu)/\sigma^2\). Its unique zero is \(\widehat{\mu}_{\mathrm{MLE}}=\bar{x}\), and the negative second derivative confirms the maximum.

Check Your Understanding

For each problem, use the stated normal model and known variance. Keep the units consistent and check that the score is zero at the estimate.

A hypothetical monitor gives readings \(6.8,\ 7.0,\ 6.5,\ 7.4,\) and \(7.3\). They are independent normal observations with known variance \(0.25\) squared units. Find the MLE of the mean. (Answer: \(7.0\) units.)
Six hypothetical claim-processing times, in hours, are \(3.1,\ 2.7,\ 3.4,\ 2.8,\ 3.0,\) and \(3.6\). Assume independent normal observations with known standard deviation \(0.4\) hours. Find the variance used in the likelihood and the MLE. (Answer: variance \(0.16\) hours squared; MLE \(3.1\) hours.)
Four hypothetical measurements are \(102,\ 98,\ 105,\) and \(95\), with known variance \(25\) squared units. Find the MLE and the score at the candidate \(\mu=99\). (Answer: MLE \(100\); score at \(99\) is \(4/25=0.16\).)
Five hypothetical readings are \(12,\ 15,\ 13,\ 16,\) and \(14\), with known variance \(9\). Find the MLE and the log-likelihood difference \(\ell(14)-\ell(13)\). (Answer: MLE \(14\); difference \(5/18\).)
Eight hypothetical observations have sum \(52.8\), and are independent from a normal distribution with unknown mean and known standard deviation \(1.5\). Find the variance in the model, the MLE, and the score at \(\mu=6.5\). (Answer: variance \(2.25\); MLE \(6.6\); score at \(6.5\) is \(0.8/2.25=0.3556\), to four decimal places.)

What You'll Learn