# Maximum-likelihood asymptotics

In this post, we review two facts about maximum-likelihood estimators: 1) They are consistent, meaning that they converge to the correct values given a large number of samples, $N$, and 2) They satisfy the Cramer-Rao lower bound for unbiased parameter estimates in this same limit — that is, they have the lowest possible variance of any unbiased estimator, in the $N\gg 1$ limit.

### Introduction

We begin with a simple example maximum-likelihood inference problem: Suppose one has obtained $N$ independent samples $\{x_1, x_2, \ldots, x_N\}$ from a Gaussian distribution of unknown mean $\mu$ and variance $\sigma^2$. In order to obtain a maximum-likelihood estimate for these parameters, one asks which $\hat{\mu}$ and $\hat{\sigma}^2$ would be most likely to generate the samples observed. To find these, we first write down the probability of observing the samples, given our model. This is simply
$$P(\{x_1, x_2, \ldots, x_N\} \vert \mu, \sigma^2) = \exp\left [ \sum_{i=1}^N \left (-\frac{1}{2} \log (2 \pi \sigma^2) -\frac{1}{2 \sigma^2} (x_i – \mu)^2\right ) \right ]. \tag{1} \label{1}$$
To obtain the maximum-likelihood estimates, we maximize (\ref{1}): Setting its derivatives with respect to $\mu$ and $\sigma^2$ to zero and solving gives
\begin{align}\label{mean}
\hat{\mu} &= \frac{1}{N} \sum_i x_i \tag{2} \\
\hat{\sigma}^2 &= \frac{1}{N} \sum_i (x_i – \hat{\mu})^2. \tag{3} \label{varhat}
\end{align}
These are mean and variance values that would be most likely to generate our observation set $\{x_i\}$. Our solutions show that they are both functions of the random observation set. Because of this, $\hat{\mu}$ and $\hat{\sigma}^2$ are themselves random variables, changing with each sample set that happens to be observed. Their distributions can be characterized by their mean values, variances, etc.

The average squared error of a parameter estimator is determined entirely by its bias and variance — see eq (2) of prior post. Now, one can show that the $\hat{\mu}$ estimate of (\ref{mean}) is unbiased, but this is not the case for the variance estimator (\ref{varhat}) — one should (famously) divide by $N-1$ instead of $N$ here to obtain an unbiased estimator$^1$. This shows that maximum-likelihood estimators need not be unbiased. Why then are they so popular? One reason is that these estimators are guaranteed to be unbiased when $N$, the sample size, is large. Further, in this same limit, these estimators achieve the minimum possible variance for any unbiased parameter estimate — as set by the fundamental Cramer-Rao bound. The purpose of this post is to review simple proofs of these latter two facts about maximum-likelihood estimators$^2$.

### Consistency

Let $P(x \vert \theta^*)$ be some distribution characterized by a parameter $\theta^*$ that is unknown. We will show that the maximum-likelihood estimator converges to $\theta^*$ when $N$ is large: As in (\ref{1}), the maximum-likelihood solution is that $\theta$ that maximizes
$$\tag{4} \label{4} J \equiv \frac{1}{N}\sum_{i=1}^N \log P(x_i \vert \theta),$$
where the $\{x_i\}$ are the independent samples taken from $P(x \vert \theta^*)$. By the law of large numbers, when $N$ is large, this average over the samples converges to its population mean. In other words,
$$\tag{5} \lim_{N \to \infty}J \rightarrow \int_x P(x \vert \theta^*) \log P(x \vert \theta) dx.$$
We will show that $\theta^*$ is the $\theta$ value that maximizes the above. We can do this directly, writing
\begin{align} J(\theta) – J(\theta^*) & = \int_x P(x \vert \theta^*) \log \left ( \frac{P(x \vert \theta) }{P(x \vert \theta^*)}\right) \\ & \leq \int_x P(x \vert \theta^*) \left ( \frac{P(x \vert \theta) }{P(x \vert \theta^*)} – 1 \right) \\ & = \int_x P(x \vert \theta) – P(x \vert \theta^*) = 1 – 1 = 0. \tag{6} \label{6} \end{align}
Here, we have used $\log t \leq t-1$ in the second line. Rearranging the above shows that $J(\theta^*) \geq J(\theta)$ for all $\theta$ — when $N \gg 1$, meaning that $J$ is maximized at $\theta^*$. That is, the maximum-likelihood estimator $\hat{\theta} \to \theta^*$ in this limit$^3$.

### Optimal variance

To derive the variance of a general maximum-likelihood estimator, we will see how its average value changes upon introduction of a small Bayesian prior, $P(\theta) \sim \exp(\Lambda \theta)$. The trick will be to evaluate the change in two separate ways — this takes a few lines, but is quite straightforward. In the first approach, we do a direct maximization: The quantity to be maximized is now
$$\label{7} J = \sum_{i=1}^N \log P(x_i \vert \theta) + \Lambda \theta. \tag{7}$$
Because we take $\Lambda$ small, we can use a Taylor expansion to find the new solution, writing
$$\label{8} \hat{\theta} = \theta^* + \theta_1 \Lambda + O(\Lambda^2). \tag{8}$$
Setting the derivative of (\ref{7}) to zero, with $\theta$ given by its value in (\ref{8}), we obtain
$$\sum_{i=1}^N \partial_{\theta} \left . \log P(x_i \vert \theta) \right \vert_{\theta^*} + \sum_{i=1}^N \partial_{\theta}^2 \left . \log P(x_i \vert \theta) \right \vert_{\theta^*} \times \theta_1 \Lambda + \Lambda + O(\Lambda^2) = 0. \tag{9} \label{9}$$
The first term here goes to zero at large $N$, as above. Setting the terms at $O(\Lambda^1)$ to zero gives
$$\theta_1 = – \frac{1}{ \sum_{i=1}^N \partial_{\theta}^2 \left . \log P(x_i \vert \theta) \right \vert_{\theta^*} }. \tag{10} \label{10}$$
Plugging this back into (\ref{8}) gives the first order correction to $\hat{\theta}$ due to the perturbation. Next, as an alternative approach, we evaluate the change in $\theta$ by maximizing the $P(\theta)$ distribution, expanding about its unperturbed global maximum, $\theta^*$: We write, formally,
$$\tag{11} \label{11} P(\theta) = \exp\left [ – a_0 – a_2 (\theta – \theta^*)^2 – a_3 (\theta – \theta^*)^3 + \ldots + \Lambda \theta \right].$$
Differentiating to maximize (\ref{11}), and again assuming a solution of form (\ref{8}), we obtain
$$\label{12} \tag{12} -2 a_2 \times \theta_1 \Lambda + \Lambda + O(\Lambda^2) = 0 \ \ \to \ \ \theta_1 = \frac{1}{2 a_2}.$$
We now require consistency between our two approaches, equating (\ref{10}) and (\ref{12}). This gives an expression for $a_2$. Plugging this back into (\ref{11}) then gives (for the unperturbed distribution)
$$\tag{13} \label{13} P(\theta) = \mathcal{N} \exp \left [ N \frac{ \langle \partial_{\theta}^2 \left . \log P(x, \theta) \right \vert_{\theta^*} \rangle }{2} (\theta – \theta^*)^2 + \ldots \right].$$
Using this Gaussian approximation$^4$, we can now read off the large $N$ variance of $\hat{\theta}$ as
$$\tag{14} \label{14} var(\hat{\theta}) = – \frac{1}{N} \times \frac{1}{\langle \partial_{\theta}^2 \left . \log P(x, \theta) \right \vert_{\theta^*} \rangle }.$$
This is the lowest possible value for any unbiased estimator, as set by the Cramer-Rao bound. The proof shows that maximum-likelihood estimators always saturate this bound, in the large $N$ limit — a remarkable result. We discuss the intuitive meaning of the Cramer-Rao bound in a prior post.

### Footnotes

[1] To see that (\ref{varhat}) is biased, we just need to evaluate the average of $\sum_i (x_i – \hat{\mu})^2$. This is

$$\overline{\sum_i x_i^2 – 2 \sum_{i,j} \frac{x_i x_j}{N} + \sum_{i,j,k} \frac{x_j x_k}{N^2}} = N \overline{x^2} – (N-1) \overline{x}^2 – \overline{x^2} \\ = (N-1) \left ( \overline{x^2} – \overline{x}^2 \right) \equiv (N-1) \sigma^2.$$
Dividing through by $N$, we see that $\overline{\hat{\sigma}^2} = \left(\frac{N-1}{N}\right)\sigma^2$. The deviation from the true variance $\sigma^2$ goes to zero at large $N$, but is non-zero for any finite $N$: The estimator is biased, but the bias goes to zero at large $N$.

[2] The consistency proof is taken from lecture notes by D. Panchenko, see here. Professor Panchenko is quite famous for having proven the correctness of the Parisi ansatz in replica theory. Our variance proof is original — please let us know if you have seen it elsewhere. Note that it can also be easily extended to derive the covariance matrix of a set of maximum-likelihood estimators that are jointly distributed — we cover only the scalar case here, for simplicity.

[3] The proof here actually only shows that there is no $\theta$ that gives larger likelihood than $\theta^*$ in the large $N$ limit. However, for some problems, it is possible that more than one $\theta$ maximizes the likelihood. A trivial example is given by the case where the distribution is actually only a function of $(\theta – \theta_0)^2$. In this case, both values $\theta_0 \pm (\theta^* – \theta_0)$ will necessarily maximize the likelihood.

[4] It’s a simple matter to carry this analysis further, including the cubic and higher order terms in the expansion (\ref{11}). These lead to correction terms for (\ref{14}), smaller in magnitude than that given there. These terms become important when $N$ decreases in magnitude.