(Maximum A-Posteriori Estimation (MAP))
(Likelihood Ratio Test)
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#as T increases Type II Error Increases
 
#as T increases Type II Error Increases
 
(<math>T = 0 \Rightarrow R = \{x|P_{X|\theta}(x|\theta_1) > 0\}</math>.  So, Type I error (<math>Pr(x\in R | H_0)</math>) is maximized as T is minimized.)
 
(<math>T = 0 \Rightarrow R = \{x|P_{X|\theta}(x|\theta_1) > 0\}</math>.  So, Type I error (<math>Pr(x\in R | H_0)</math>) is maximized as T is minimized.)
 +
 +
The threshold value T=1, corresponds to the ML rule.

Revision as of 03:51, 15 December 2008

Maximum Likelihood Estimation (ML)

$ \hat a_{ML} = \overset{max}{a} f_{X}(x_i;a) $ continuous
$ \hat a_{ML} = \overset{max}{a} Pr(x_i;a) $ discrete


Chebyshev Inequality

"Any RV is likely to be close to its mean"

$ \Pr(\left|X-E[X]\right|\geq C)\leq\frac{var(X)}{C^2}. $


Maximum A-Posteriori Estimation (MAP)

$ \hat \theta_{MAP}(x) = \text{arg }\overset{max}{\theta} P_{X|\theta}(x|\theta)P_ {\theta}(\theta) $
$ \hat \theta_{MAP}(x) = \text{arg }\overset{max}{\theta} f_{X|\theta}(x|\theta)P_ {\theta}(\theta) $

Minimum Mean-Square Estimation (MMSE)

$ \hat{y}_{\rm MMSE}(x) = \int_{-\infty}^{\infty} {y}{f}_{\rm Y|X}(y|x)\, dy={E}[Y|X=x] $

Law Of Iterated Expectation

$ E[E[X|Y]] = \begin{cases} \sum_{y} E[X|Y = y]p_Y(y),\,\,\,\,\,\,\,\,\,\,\mbox{ Y discrete,}\\ \int_{-\infty}^{+\infty} E[X|Y = y]f_Y(y)\,dy,\mbox{ Y continuous.} \end{cases} $

Using the total expectation theorem:

$ E\Big[ E[X|Y]] = E[X] $

Mean Square Error

$ MSE = E[(\Theta - \hat \theta(x))^2] $
$ MSE(E(\Theta)) = var(\Theta) \, $

Linear Minimum Mean-Square Estimation (LMMSE)

The LMMS estimator $ \hat{Y} $ of Y based on the variable X is

$ \hat{Y}_{LMMSE}(x) = E[Y]+\frac{COV(Y,X)}{Var(X)}(X-E[X]) = E[Y] + \rho \frac{\sigma_{Y}}{\sigma_{X}}(X-E[X]) $

where

$ \rho = \frac{COV(Y,X)}{\sigma_{Y}\sigma_{X}} $

Law of Iterated Expectation: E[E[X|Y]]=E[X]

COV(X,Y)=E[XY] - E[X]E[Y]

Hypothesis Testing

In hypothesis testing $ \Theta $ takes on one of m values, $ \theta_1,...,\theta_m $ where m is usually small; often m = 2, in which case it is a binary hypthothesis testing problem.

The event $ \Theta = \theta_i $ is the $ i^{th} $ hypothesis denoted by $ H_i $

ML Rule

Given a value of X, we will say H1 is true if X is in region R, else will will say H0 is true.

Type I Error: False Rejection

Say $ H_1 $ when truth is $ H_0 $. Probability of this is:

$ Pr(\mbox{Say } H_1|H_0) = Pr(x \in R|\theta_0) $

Type II Error: False Acceptance

Say $ H_0 $ when truth is $ H_1 $. Probability of this is:

$ Pr(\mbox{Say }H_0|H_1) = Pr(x \in R^C|\theta_1) $

MAP Rule

$ \mbox{Overall P(err)} = P_{\theta}(\theta_{0})Pr\Big[\mbox{Say }H_{1}|H_{0}\Big] +P_{\theta}(\theta_{1})Pr\Big[\mbox{Say }H_{0}|H_{1}\Big] $

Likelihood Ratio Test

How to find a good rule? --Khosla 16:44, 13 December 2008 (UTC)

For X is discrete

$ \ L(x) = \frac{p_{X|\theta} (x|\theta_1)}{p_{X|\theta} (x|\theta_0)} $

Choose threshold (T),

$ \mbox{Say } \begin{cases} H_{1}; \mbox{ if } L(x) > T\\ H_{0}; \mbox{ if } L(x) < T \end{cases} $

The Maximum Likelihood rule is a Likelihood Ratio Test with T = 1 The MAP rule is a Likelihood Ratio Test with $ T=\frac{P_\theta(\theta_0)}{P_\theta(\theta_1)} $

Observations:

  1. as T decreases Type I Error Increases
  2. as T decreases Type II Error Decreases
  3. as T increases Type I Error Decreases
  4. as T increases Type II Error Increases

($ T = 0 \Rightarrow R = \{x|P_{X|\theta}(x|\theta_1) > 0\} $. So, Type I error ($ Pr(x\in R | H_0) $) is maximized as T is minimized.)

The threshold value T=1, corresponds to the ML rule.

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