# MAP Estimation by Landis

## The Big Picture

Given observation X used to estimate an unknown parameter $\theta$ of distribution $f_x(X)$ (i.e. $f_x(X) =$ some function $g(\theta)$

Consider three expressions (distributions):

### 1. Likehood

$p(X; \theta)$ (discrete)

$f_x(X; \theta)$ (continuous)

used for MLE: $\overset{\land}\theta_{ML} = f_x(X | \theta)$

### 2. Prior

$P(\theta)$ (discrete)

$P_\theta(\theta)$ (continuous)

Indicates some prior knowledge as to what $\theta$ should be. Prior refers to before seeing observation.

### 3. Posterior

$p(\theta | x)$ (discrete)

$f_x(\theta, x)$ (continuous)

"Posterior" refers to after seeing observations. Use Posterior to define maximum a-posterior i (map) estimate:

$\overset{\land}\theta_{\mbox{MAP}} = \overset{\mbox{argmax}}\theta f_{\theta | X}(\theta | X)$

Using Bayes' Rule, we can expand the posterior $f_{\theta | X}(\theta | X)$:

$f_{\theta | X}(\theta | X) = \frac{f_{x|\theta}f_\theta(\theta)}{f_X(X)}$

$\overset{\land}\theta_{\mbox{map}} = \overset{\mbox{argmax}}\theta f_{X | \theta}(X | \theta) F_\theta(\theta)$

### So What?

So, what does this mean in a nutshell? Essentially, an ML estimator is as follows: "I know that a random variable follows some sort of pattern, and this pattern can be adjusted by changing a variable called a parameter. (I have no clue what this parameter should be.) I run some experiment to get a sample output from the random variable. Based on my experiment, I would like to find the pattern for the random variable that best matches my data. So, I will find the pattern that gives my data the most likely chance of occuring."

A MAP estimator is as follows: "I know that a random variable follows some sort of pattern, and this pattern can be adjusted by changing a variable called a parameter. (I have some idea what this parameter should be, so I will treat the parameter as a random variable. Then, the chance that the parameter will have some value will be expressed by it's PDF/PMF.) I run some experiment to get a sample output from the random variable. Based on my experiment and on my prior ideas of what the parameter should be, I would like to find the pattern that best matches both my data and my prior knowledge. So, I will find the pattern that maximizes the product of two things: how well the parameter matches the data, and how well the parameter meets my expectations of what the parameter should be."

## Example 1: Continuous

$X \sim f_x(X) = \lambda e^{-\lambda X}$

but we don't know the parameter $\lambda$. Let us assume, however, that $\lambda$ is actually itself exponentially distributed, i.e.

$\lambda \sim f_\lambda(\lambda) = \Lambda e^{-\Lambda\lambda}$

where $\Lambda$ is fixed and known.

Find $\overset{\land}\lambda_{\mbox{map}}$.

Solution:

$\overset{\land}\lambda_{\mbox{map}} = \overset{\mbox{argmax}}\lambda f_{\lambda | X}(\lambda | X)$

$\overset{\land}\lambda_{\mbox{map}} = \overset{\mbox{argmax}}\lambda f_x(\lambda)f_{x|\lambda}(x; \lambda)$

$\overset{\land}\lambda_{\mbox{map}} = \overset{\mbox{argmax}}\lambda \Lambda e^{-\lambda \Lambda}\lambda e^{-\lambda X}$

$\frac{d}{d\lambda} \lambda \Lambda e^{-\lambda(\Lambda + X)} = 0$

$\Lambda e^{-\lambda}(\Lambda + X) - \lambda \Lambda (\Lambda + X) e^{-\lambda(\lambda + X)} = 0$

$1 - \lambda(\Lambda + X) = 0$

$\overset{\land}X_{\mbox{map}} = \frac{1}{\Lambda + X}$

Recall from homework: $\overset{\land}X_{\mbox{ML}} = \frac{1}{X}$

Prior $f_{\lambda}(\lambda) = \Lambda e^{-\lambda \Lambda}$

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