Does MLE require i.i.d. data? Or just independent parameters?

Question

Estimating parameters using maximum likelihood estimation (MLE) involves evaluating the likelihood function, which maps the probability of the sample (X) occurring to values (x) on the parameter space (θ) given a distribution family (P(X=x|θ) over possible values of θ (note: am I right on this?). All examples I've seen involve calculating P(X=x|θ) by taking the product of F(X) where F is the distribution with the local value for θ and X is the sample (a vector).

Since we're just multiplying the data, does it follow that the data be independent? E.g. could we not use MLE to fit time-series data? Or do the parameters just have to be independent?

Answer 1

The likelihood function is defined as the probability of an event $E$ (data set ${\bf x}$) as a function of the model parameters $\theta$

$${\mathcal L}(\theta;{\bf x})\propto {\mathbb P}(\text{Event }E;\theta)= {\mathbb P}(\text{observing } {\bf x};\theta).$$

Therefore, there is no assumption of independence of the observations. In the classical approach there is no definition for independence of parameters since they are not random variables; some related concepts could be identifiability, parameter orthogonality, and independence of the Maximum Likelihood Estimators (which are random variables).

Some examples,

(1). Discrete case. ${\bf x}=(x_1,...,x_n)$ is a sample of (independent) discrete observations with ${\mathbb P}(\text{observing } x_j ; \theta)>0$, then

$${\mathcal L}(\theta;{\bf x})\propto \prod_{j=1}^n{\mathbb P}(\text{observing } x_j ; \theta).$$

Particularly, if $x_j\sim \text{Binomial}(N,\theta)$, with $N$ known, we have that

$${\mathcal L}(\theta;{\bf x})\propto \prod_{j=1}^n \theta^{x_j}(1-\theta)^{N-x_j}.$$

(2). Continuous approximation. Let ${\bf x}=(x_1,...,x_n)$ be a sample from a continuous random variable $X$, with distribution $F$ and density $f$, with measurement error $\epsilon$, this is, you observe the sets $(x_j-\epsilon,x_j+\epsilon)$. Then

\begin{eqnarray*} {\mathcal L}(\theta;{\bf x})\propto \prod_{j=1}^n {\mathbb P}[\text{observing } (x_j-\epsilon,x_j+\epsilon);\theta] = \prod_{j=1}^n[F(x_j+\epsilon;\theta)-F(x_j-\epsilon;\theta)] \end{eqnarray*}

When $\epsilon$ is small, this can be approximated (using the Mean Value Theorem) by

\begin{eqnarray*} {\mathcal L}(\theta;{\bf x})\propto \prod_{j=1}^n f(x_j;\theta) \end{eqnarray*}

For an example with the normal case, take a look at this.

(3). Dependent and Markov model. Suppose that ${\bf x}=(x_1,...,x_n)$ is a set of observations possibly dependent and let $f$ be the joint density of ${\bf x}$, then

\begin{eqnarray*} {\mathcal L}(\theta;{\bf x})\propto f({\bf x}; \theta). \end{eqnarray*}

If additionally the Markov property is satisfied, then

\begin{eqnarray*} {\mathcal L}(\theta;{\bf x})\propto f({\bf x}; \theta) = f(x_1;\theta)\prod_{j=1}^{n-1} f(x_{j+1} \vert x_j ;\theta). \end{eqnarray*}

Take also a look at this.

Answer 2

(+1) Very good question.

Minor thing, MLE stands for maximum likelihood estimate (not multiple), which means that you just maximize the likelihood. This does not specify that the likelihood has to be produced by IID sampling.

If the dependence of the sampling can be written in the statistical model, you just write the likelihood accordingly and maximize it as usual.

The one case worth mentioning when you do not assume dependence is that of the multivariate Gaussian sampling (in time series analysis for example). The dependence between two Gaussian variables can be modelled by their covariance term, which you incoroporate in the likelihood.

To give a simplistic example, assume that you draw a sample of size $2$ from correlated Gaussian variables with same mean and variance. You would write the likelihood as

$$\frac{1}{2\pi\sigma^2\sqrt{1-\rho^2}}\exp\left(-\frac{z}{2\sigma^2(1-\rho^2)}\right),$$

where $z$ is

$$z = (x_1-\mu)^2-2\rho(x_1-\mu)(x_2-\mu)+(x_2-\mu)^2.$$

This is not the product of the individual likelihoods. Still, you would maximize this with parameters $(\mu, \sigma, \rho)$ to get their MLE.

Answer 3

Of course, Gaussian ARMA models possess a likelihood, as their covariance function can be derived explicitly. This is basically an extension of gui11ame's answer to more than 2 observations. Minimal googling produces papers like this one where the likelihood is given in the general form.

Another, to an extent, more intriguing, class of examples is given by multilevel random effect models. If you have data of the form $$y_{ij} = x_{ij}'\beta + u_i + \epsilon_{ij},$$ where indices $j$ are nested in $i$ (think of students $j$ in classrooms $i$, say, for a classic application of multilevel models), then, assuming $\epsilon_{ij} \perp u_i$, the likelihood is $$ \ln L \sim \sum_i \ln \int \prod_j f(y_{ij}|\beta,u_i) {\rm d}F(u_i) $$ and is a sum over the likelihood contributions defined at the level of clusters, not individual observations. (Of course, in the Gaussian case, you can push the integrals around to produce an analytic ANOVA-like solution. However, if you have say a logit model for your response $y_{ij}$, then there is no way out of numerical integration.)