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If the random variables X1,X2,...,Xn <math>\epsilon</math> R are Independent Identically Distributed (I.I.D.) then,
 
If the random variables X1,X2,...,Xn <math>\epsilon</math> R are Independent Identically Distributed (I.I.D.) then,
  
L(θ) = <math>\prod_{i=1}^{n}</math> f(xi | θ) .  
+
L(θ) = <sub>i=1</sub> ∏ <sup>n</sup>f(xi | θ) .  
 
We need to find the value <math> \hat{\theta} \epsilon </math> θ that maximizes this function.
 
We need to find the value <math> \hat{\theta} \epsilon </math> θ that maximizes this function.
  

Revision as of 20:47, 26 April 2014


Maximum Likelihood Estimation (MLE) Analysis for various Probability Distributions
A slecture by Hariharan Seshadri

(partially based on Prof. Mireille Boutin's ECE 662 lecture)


What would be the learning outcome from this slecture?

  • Basic Theory behind Maximum Likelihood Estimation (MLE)
  • Derivations for Maximum Likelihood Estimates for parameters of Exponential Distribution, Geometric Distribution, Binomial Distribution, Poisson Distribution, and Uniform Distribution

Introduction

The maximum likelihood estimate (MLE) is the value $ \hat{\theta} $ which maximizes the function L(θ) given by L(θ) = f (X1,X2,...,Xn | θ) where 'f' is the probability density function in case of continuous random variables and probability mass function in case of discrete random variables and 'θ' is the parameter being estimated.

In other words,$ \hat{\theta} $ = arg maxθ L(θ), where $ \hat{\theta} $ is the best estimate of the parameter 'θ' . Thus, we are trying to maximize the probability density (in case of continuous random variables) or the probability of the probability mass (in case of discrete random variables)

If the random variables X1,X2,...,Xn $ \epsilon $ R are Independent Identically Distributed (I.I.D.) then,

L(θ) = i=1nf(xi | θ) . We need to find the value $ \hat{\theta} \epsilon $ θ that maximizes this function.

In the course, Purdue ECE 662, Pattern Recognition and Decision Taking Processes, we have already looked at the MLE of the Normal Distribution and found that to be:

$ \widehat{\mu} = \frac{\sum_{i=1}^{n}x_{i}}{n} $

$ \hat{\sigma{}^{2}} = \frac{1}{n-1}\sum_{i=1}^{n}(x_{i}-\widehat{\mu})^{2} $


where, Xi's are the Normal (Gaussian) Random Variables $ \epsilon $ R , 'n' is the number of samples, and $ \widehat{\mu} $ and $ \hat{\sigma{}^{2}} $ are the estimated Mean and estimated Standard Deviation.


Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009