5.10 Maximum Likelihood

Premise: find values of the parameters that maximize the probability of observing the data In other words, we try to maximize the value of theta in the likelihood function $L(\theta)=\prod_{i=1}^{n}f(y_i|\theta)$ $$f(y|\theta)$$ is the probability density of observing a single value of Y given some value of $$\theta$$ $$f(y|\theta)$$ can be specify as various type of distributions. You can review back section Distributions. For example If y is a dichotomous variable, then $L(\theta)=\prod_{i=1}^{n}\theta^{y_i}(1-\theta)^{1-y_i}$

$$\hat{\theta}$$ is the Maximum Likelihood estimate if $$L(\hat{\theta}) > L(\theta_0)$$ for all values of $$\theta_0$$ in the parameter space.

5.10.1 Motivation for MLE

Suppose we know the conditional distribution of y given x:

$f_{Y|X}(y,x;\theta)$ where $$\theta$$ is the unknown parameter of distribution. Sometimes we are only concerned with the unconditional distribution $$f_{Y}(y;\theta)$$

Then given a sample of iid data, we can calculate the joint distribution of the entire sample,

$f_{Y_1,...,Y_n|X_1,...,X_n(y_1,...y_n,x_1,...,x_n;\theta)}= \prod_{i=1}^{n}f_{Y|X}(y_i,x_i;\theta)$ The joint distribution evaluated at the sample is the likelihood (probability) that we observed this particular sample (depends on $$\theta$$)

Idea for MLE: Given a sample, we choose our estimates of the parameters that gives the highest likelihood (probability) of observing our particular sample

$max_{\theta} \prod_{i=1}^{n}f_{Y|X}(y_i,x_i; \theta)$

Equivalently,

$max_{\theta} \prod_{i=1}^{n} ln(f_{Y|X}(y_i,x_i; \theta))$ Solving for the Maximum Likelihood Estimator

1. Solve First Order Condition

$\frac{\partial}{\partial \theta}\sum_{i=1}^{n} ln(f_{Y|X}(y_i,x_i;\hat{\theta}_{MLE})) = 0$ where $$\hat{\theta}_{MLE}$$ is defined.

1. Evaluate Second Order Condition

$\frac{\partial^2}{\partial \theta^2} \sum_{i=1}^{n} ln(f_{Y|X}(y_i,x_i;\hat{\theta}_{MLE})) < 0$

where the above condition ensures we can solve for a maximum

Examples:
Unconditional Poisson Distribution: Number of products ordered on Amazon within an hour, number of website visits a day for a political campaign.

Exponential Distribution: Length of time until an earthquake occurs, length of time a car battery lasts.

$f_{Y|X}(y,x;\theta) = exp(-y/x\theta)/x\theta \\ f_{Y_1,..Y_n|X_1,...,X_n(y_1,...,y_n,x_1,...,x_n;\theta)} = \prod_{i=1}^{n}exp(-y_i/x_i \theta)/x_i \theta$

5.10.2 Assumption

• High Level Regulatory Assumptions is the sufficient condition used to show large sample properties
• Hence, for each MLE, we will need to either assume or verify if the regulatory assumptiosn holds.
• observations are independent and have the same density function.
• Under multivariate normal assumption, ML yields consistent estimates of the means and the covariance matrix for multivariate distribution with finite fourth moments

To find the MLE, we usually differentiate the log-likelihood function and set it equal to 0.

$\frac{d}{d\theta}l(\theta) = 0$ This is the score equation

Our confidence in the MLE is quantified by the “pointedness” of the log-likelihood $I_O(\theta)= \frac{d^2}{d\theta^2}l(\theta) = 0$ called the observed information

while $I(\theta)=E[I_O(\theta;Y)]$ is the expected information. (also known as Fisher Information). which we base our variance of the estimator.

$V(\hat{\Theta}) \approx I(\theta)^{-1}$

Consistency of MLE
Suppose that $$y_i$$ and $$x_i$$ are iid drawn from the true conditional pdf $$f_{Y|X}(y_i,x_i;\theta_0)$$. If the following regulatory assumptions hold,

R1: If $$\theta \neq \theta_0$$ then $$f_{Y|X}(y_i,x_i;\theta) \neq f_{Y|X}(y_i,x_i;\theta_0)$$
R2: The set $$\Theta$$ that contains the true parameters $$\theta_0$$ is compact
R3: The log-likelihood $$ln(f_{Y|X}(y_i,x_i;\theta_0))$$ is continuous at each $$\theta$$ with probability 1
R4: $$E(sup_{\theta \in \Theta}|ln(f_{Y|X}(y_i,x_i;\theta_0))|)$$

then the MLE estimator is consistent,

$\hat{\theta}_{MLE} \to^p \theta_0$

Asymptotic Normality of MLE

Suppose that $$y_1$$ and $$x_i$$ are iid drawn from the true conditional pdf $$f_{Y|X}(y_i,x_i;\theta)$$. If R1-R4 and the following hold

R5: $$\theta_0$$ is in the interior of the set $$\Theta$$
R6: $$f_{Y|X}(y_i,x_i;\theta)$$ is twice continuously differentiable in $$\theta$$ and $$f_{Y|X}(y_i,x_i;\theta) >0$$ for a neighborhood $$N \in \Theta$$ around $$\theta_0$$
R7: $$\int sup_{\theta \in N}||\partial f_{Y|X}(y_i,x_i;\theta)\partial\theta||d(y,x) <\infty$$, $$\int sup_{\theta \in N} || \partial^2 f_{Y|X}(y_i,x_i;\theta)/\partial \theta \partial \theta' || d(y,x) < \infty$$ and $$E(sup_{\theta \in N} || \partial^2ln(f_{Y|X}(y_i,x_i;\theta)) / \partial \theta \partial \theta' ||) < \infty$$
R8: The information matrix $$I(\theta_0) = Var(\partial f_{Y|X}(y,x_i; \theta_0)/\partial \theta)$$ exists and is non-singular

then the MLE estimator is asymptotically normal,

$\sqrt{n}(\hat{\theta}_{MLE} - \theta_0) \to^d N(0,I(\theta_0)^{-1})$

5.10.3 Properties

1. Consistent: estimates are approximately unbiased in large samples
2. Asymptotically efficient: approximately smaller standard errors compared to other estimator
3. Asymptotically normal: with repeated sampling, the estimates will have an approximately normal distribution. Suppose that $$\hat{\theta}_n$$ is the MLE for $$\theta$$ based on n independent observations. then $$\hat{\theta}_n \sim N(\theta,H^{-1})$$.
+ where H is called the Fisher information matrix. It contains the expected values of the second partial derivatives of the log-likelihood function. The (i.j)th element of H is $$-E(\frac{\partial^2l(\theta)}{\partial \theta_i \partial \theta_j})$$
+ We can estimate H by finding the form determined above, and evaluating it at $$\theta = \hat{\theta}_n$$
4. Invariance: MLE for $$g(\theta) = g(\hat{\theta})$$ for any function g(.)

$\hat{\Theta} \approx^d (\theta,I(\hat{\theta)^{-1}}))$

Explicit vs Implicit MLE

• If we solve the score equation to get an expression of MLE, then it’s called explicit
• If there is no closed form for MLE, and we need some algorithms to derive its expression, it’s called implicit

Large Sample Property of MLE

Implicit in these theorems is the assumption that we know what the conditional distribution,

$f_{Y|X}(y_i,x_i;\theta_0)$ but just do now know the exact parameter value.

• Any Distributional mis-specification will result in inconsistent parameter estimates.
• Quasi-MLE: Particular settings/ assumption that allow for certain types of distributional mis-specification (Ex: as long as the distribution is part of particular class or satisfies a particular assumption, then estimating with a wrong distribution will not lead to inconsistent parameter estimates).
• non-parametric/ Semi-parametric estimation: no or very little distributional assumption are made. (hard to implement, derive properties, and interpret)

The asymptotic variance of the MLE achieves the Cramer-Rao Lower Bound

• The Cramer-Rao Lower Bound is a lower brand for the asymptotic variance of a consistent and asymptotically normally distributed estimator.
• If an estimator achieves the lower bound then it is the most efficient estimator.

The maximum Likelihood estimator (assuming the distribution is correctly specified and R1-R8 hold) is the most efficient consistent and asymptotically normal estimator.
* most efficient among ALL consistent estimators (not limited to unbiased or linear estimators).

Note

• ML is better choice for binary, strictly positive, count, or inherent heteroskedasticity than linear model.
• ML will assume that we know the conditional distribution of the outcome, and derive an estimator using that information.
• Adds an assumption that we know the distribution (which is similar to A6 Normal Distribution in linear model)
• will produce a more efficient estimator.

5.10.4 Compare to OLS

MLE is not a cure for most of OLS problems:

• To do joint inference in MLE, we typically use log-likelihood calculation, instead of F-score
• Functional form affects estimation of MLE and OLS.
• Perfect Collinearity/Multicollinearity: highly correlated are likely to yield large standard errors.
• Endogeneity (Omitted variables bias, Simultaneous equations): Like OLS, MLE is also biased against this problem

5.10.5 Application

Other applications of MLE

• Corner Solution
• Ex: hours worked, donations to charity
• Estimate with Tobit
• Non-negative count
• Ex: Numbers of arrest, Number of cigarettes smoked a day
• Estimate with Poisson regression
• Multinomial Choice
• Ex: Demand for cars, votes for primary election
• Estimate with mutinomial probit or logit
• Ordinal Choice
• Ex: Levels of Happiness, Levels of Income
• Ordered Probit

Model for binary Response
A binary variable will have a Bernoulli distribution:

$f_Y(y_i;p) = p^{y_i}(1-p)^{(1-y_i)}$ where p is the probability of success. The conditional distribution is:

$f_{Y|X}(y_i,x_i;p(.)) = p(x_i)^{y_i} (1-p(x_i))^{(1-y_i)}$

So choose $$p(x_i)$$ to be a reasonable function of $$x_i$$ and unknown parameters $$\theta$$

We can use latent variable model as probability functions

$y_i = 1\{y_i^* > 0 \} \\ y_i^* = x_i \beta-\epsilon_i$

• $$y_i^*$$ is a latent variable (unobserved) that is not well-defined in terms of units/magnitudes
• $$\epsilon_i$$ is a mean 0 unobserved random variable.

We can rewrite the mdoel without the latent variable,

$y_i = 1\{x_i beta > \epsilon_i \}$

Then the probability function,

$p(x_i) = P(y_i = 1|x_i) \\ = P(x_i \beta > \epsilon_i | x_i) \\ = F_{\epsilon|X}(x_i \beta | x_i)$

then we need to choose a conditional distribution for $$\epsilon_i$$. Hence, we can make additional strong independence assumption

$$\epsilon_i$$ is independent of $$x_i$$

Then the probability function is simply,

$p(x_i) = F_\epsilon(x_i \beta)$ The probability function is also the conditional expectation function,

$E(y_i | x_i) = P(y_i = 1|x_i) = F_\epsilon (x_i \beta)$

so we allow the conditional expectation function to be non-linear.

Common distributional assumption

1. Probit: Assume $$\epsilon_i$$ is standard normally distributed, then $$F_\epsilon(.) = \Phi(.)$$ is the standard normal CDF.
2. Logit: Assume $$\epsilon_i$$ is standard logistically distributed, then $$F_\epsilon(.) = \Lambda(.)$$ is the standard normal CDF.

Step to derive

1. Choose a distribution (normal or logistic) and plug into the following log likelihood,

$ln(f_{Y|X} (y_i , x_i; \beta)) = y_i ln(F_\epsilon(x_i \beta)) + (1-y_i)ln(1-F_\epsilon(x_i \beta))$

1. Solve the MLE by finding the Maximum of

$\hat{\beta}_{MLE} = argmax \sum_{i=1}^{n}ln(f_{Y|X}(y_i,x_i; \beta))$

Properties of the Probit and Logit Estimators

• Probit or Logit is consistent and asymptotically normal if
• [A2][] holds: $$E(x_i' x_i)$$ exists and is non-singular
• [A5][] (or A5a) holds: {y_i,x_i} are iid (or stationary and weakly dependent).
• Distributional assumptions on $$\epsilon_i$$ hold: Normal/Logistic and independent of $$x_i$$
• Under the same assumptions, Probit or Logit is also asymptotically efficient with asymptotic variance,

$I(\beta_0)^{-1} = [E(\frac{(f_\epsilon(x_i \beta_0))^2}{F_\epsilon(x_i\beta_0)(1-F_\epsilon(x_i\beta_0))}x_i' x_i)]^{-1}$

where $$F_\epsilon(x_i\beta_0)$$ is the probability density function (derivative of the CDF)

5.10.5.1 Interpretation

$$\beta$$ is the average response in the latent variable associated with a change in $$x_i$$

• Magnitudes do not have meaning
• Direction does have meaning

The partial effect for a Non-linear binary response model

$E(y_i |x_i) = F_\epsilon (x_i \beta) \\ PE(x_{ij}) = \frac{\partial E(y_i |x_i)}{\partial x_{ij}} = f_\epsilon (x_i \beta)\beta_j$

• The partial effect is the coefficient parameter $$\beta_j$$ multiplied by a scaling factor $$f_\epsilon (x_i \beta)$$
• The scaling factor depends on $$x_i$$ so the partial effect changes depending on what $$x_i$$ is

Single value for the partial effect

• Partial Effect at the Average (PEA) is the partial effect for an average individual

$f_{\epsilon}(\bar{x}\hat{\beta})\hat{\beta}_j$

• Average Partial Effect (APE) is the average of all partial effect for each individual.

$\frac{1}{n}\sum_{i=1}^{n}f_\epsilon(x_i \hat{\beta})\hat{\beta}_j$

In the linear model, APE = PEA.
In a non-linear model (e.g., binary response), APE $$\neq$$ PEA