Suppose that $X_1, . . . , X_n$, $n\geq 2$, is a sample from a $N(\mu,\sigma^2)$ distribution. Suppose $\mu$ and $\sigma^2$ are both known to be nonnegative but otherwise unspecified. Now, I want to find the MLE of $\mu$ and $\sigma^2$. I have drawn the MLE for non-restrictive parameters but I am stuck on this one.

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    $\begingroup$ My instinct on this would be to carry out maximum likelihood estimation, but using Lagrange multipliers/KKT conditions to enforce inequality constraints on $\mu$ and $\sigma^2$. $\endgroup$
    – microhaus
    Commented Mar 21, 2021 at 23:10
  • $\begingroup$ Tbh, I have no idea on how to implement either of those. @microhaus $\endgroup$
    – statwoman
    Commented Mar 21, 2021 at 23:32
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    $\begingroup$ Now I've had more time to think, I'm now not even sure whether it would make sense to enforce a condition that $\sigma^2 \geq 0$, because the variance is by definition, non-negative. $\endgroup$
    – microhaus
    Commented Mar 22, 2021 at 0:29

2 Answers 2



Let $\bar{x}$ denote the sample mean:

$$\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$$

The constrained maximum likelihood mean $\hat{\mu}$ and variance $\hat{\sigma}^2$ are:

$$\hat{\mu} = \left\{ \begin{array}{cl} \bar{x} & \bar{x} \ge 0 \\ 0 & \text{Otherwise} \\ \end{array} \right.$$

$$\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \hat{\mu})^2$$

That is, we simply take the sample mean and clip it to zero if it's negative. Then, plug it into the usual expression for the (uncorrected) sample variance. I obtained these expressions by setting up the constrained optimization problem, then solving for the parameters that satisfy the KKT conditions, as described below.


Objective function

Maximizing the likelihood is equivalent to minimizing the negative log likelihood $L(\mu, \sigma^2)$, which will be more convenient to work with:

$$L(\mu, \sigma^2) = -\sum_{i=1}^n \log \mathcal{N}(x_i \mid \mu, \sigma^2)$$

$$= \frac{n}{2} \log(2 \pi) + \frac{n}{2} \log(\sigma^2) + \frac{1}{2 \sigma^2} \sum_{i=1}^n (x_i-\mu)^2$$

We'll also need its partial derivatives w.r.t. $\mu$ and $\sigma^2$:

$$\frac{\partial}{\partial \mu} L(\mu, \sigma^2) = \frac{n \mu}{\sigma^2} - \frac{1}{\sigma^2} \sum_{i=1}^n x_i$$

$$\frac{\partial}{\partial \sigma^2} L(\mu, \sigma^2) = \frac{n}{2 \sigma^2} - \frac{1}{2 \sigma^4} \sum_{i=1}^n (x_i-\mu)^2$$

Optimization problem

The goal is to find the parameters $\hat{\mu}$ and $\hat{\sigma}^2$ that minimize the negative log likelihood, subject to a non-negativity constraint on the mean. The variance is non-negative by definition and the solution below turns out to automatically respect this constraint, so we don't need to impose it explicitly. The optimization problem can be written as:

$$\hat{\mu}, \hat{\sigma}^2 = \arg \min_{\mu, \sigma^2} \ L(\mu, \sigma^2) \quad \text{s.t. } g(\mu, \sigma^2) \le 0$$

$$\text{where } \ g(\mu, \sigma^2) = -\mu$$

I've written the constraint this way to follow convention, which should hopefully make it easier to match this up with other discussions about constrained optimization. In our problem, this just amounts to the constraint $\mu \ge 0$.

KKT conditions

If $(\hat{\mu}, \hat{\sigma}^2)$ is an optimal solution, there must exist a constant $\lambda$ such that the KKT conditions hold: 1) stationarity, 2) primal feasibility, 3) dual feasibility, and 4) complementary slackness. Furthermore, we have a convex loss function with a convex, continuously differentiable constraint. This implies that the KKT conditions are sufficient for optimality, so we can find the solution by solving for the parameters that satisfy these conditions.


$$\frac{\partial}{\partial \mu} L(\hat{\mu}, \hat{\sigma}^2) + \lambda \frac{\partial}{\partial \mu} g(\hat{\mu}, \hat{\sigma}^2) = 0$$

$$\frac{\partial}{\partial \sigma^2} L(\hat{\mu}, \hat{\sigma}^2) + \lambda \frac{\partial}{\partial \sigma^2} g(\hat{\mu}, \hat{\sigma}^2) = 0$$

Plug in expressions for the derivatives and solve for the parameters:

$$\hat{\mu} = \frac{1}{n} \hat{\sigma}^2 \lambda + \frac{1}{n} \sum_{i=1}^n x_i \tag{1}$$

$$\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n (x_i-\hat{\mu})^2 \tag{2}$$

Primal feasibility:

$$g(\hat{\mu}, \hat{\sigma}^2) \le 0 \implies \hat{\mu} \ge 0$$

This just says the parameters must respect the constraints

Dual feasibility:

$$\lambda \ge 0$$

Complementary slackness:

$$\lambda g(\hat{\mu}, \hat{\sigma}^2) = 0 \implies \lambda \hat{\mu} = 0$$

This says that either $\lambda$ or $\hat{\mu}$ (or both) must be zero.


Note that the RHS of equation $(1)$ is a multiple of $\lambda$ plus the sample mean $\frac{1}{n} \sum_{i=1}^n x_i$. If the sample mean is non-negative, set $\lambda$ to zero (satisfying the dual feasibility and complementary slackness conditions). It then follows from equation $(1)$ (the stationarity condition) that $\hat{\mu}$ is equal to the sample mean. This also satisfies the primal feasibility condition, since it's non-negative.

Otherwise, if the sample mean is negative, set $\hat{\mu}$ to zero (satisfying the primal feasibility and complementary slackness conditions). To satisfy equation $(1)$ (the stationarity condition), set $\lambda = -\hat{\sigma}^{-2} \sum_{i=1}^n x_i$. Since the sample mean is negative and the variance is positive, $\lambda$ takes a positive value, satisfying the dual feasibility conditionn.

In both cases, we can plug $\hat{\mu}$ into equation $(2)$ to obtain $\hat{\sigma}^2$.

  • $\begingroup$ I could not understand the equation 1) above. Shouldnt it be $\hat{\mu} = -\frac{1}{n} \hat{\sigma}^2 \lambda + \frac{1}{n} \sum_{i=1}^n x_i \tag{1}$? $\endgroup$ Commented Feb 1, 2023 at 16:19

(The answer by @user20160 is the principled one, but the asker specified that she does not know about KKT conditions or Lagrange multipliers, so I wanted to avoid those concepts.)

Since there are only two parameters a direct attack will work. The log-likelihood function based on an iid sample of size $n$ is $$ \ell(\mu,\sigma^2)= -\frac12 n \log( 2\pi) -\frac12 n\log(\sigma^2) - \frac12\sum_i \left( \frac{x_i-\mu}{\sigma} \right)^2 $$ First, consider $\mu$ as known and find the value of $\sigma^2$ that maximizes the likelihood with that value of $\mu$. You will find $$ \hat{\sigma^2}_\mu = \frac{\sum_i (x_i-\mu)^2}{n} $$ Then substitute that value into the likelihood function, and the result is the profile (log) likelihood function for $\mu$, $$ \ell_p(\mu)= \ell(\mu, \hat{\sigma^2}_\mu) $$ which now only depends on the one parameter $\mu$, and you can maximize it over the interval $(0, \infty)$. Plot it or use a numerical optimization routine. I will leave symbolic solution for you.

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    $\begingroup$ Thank you kjetil, I didn't know about this approach! $\endgroup$
    – statwoman
    Commented Mar 22, 2021 at 14:26
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    $\begingroup$ +1 It's always nice to see multiple approaches, tailored to different backgrounds. Since profile likelihood and KKT conditions are both generally useful (beyond the original problem), I'm glad we have both of these methods demonstrated in this thread. $\endgroup$
    – user20160
    Commented Mar 22, 2021 at 18:14

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