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Also suppose the coin is tossed 80 times: i.e., the sample might be something like <span class="texhtml">''x''<sub>1</sub></span> = H, <br><span class="texhtml">''x''<sub>2</sub></span>&nbsp;= T, …, <span class="texhtml">''x''<sub>8</sub></span>&nbsp;= T, and the count of number of HEADS, "H" is observed.  
 
Also suppose the coin is tossed 80 times: i.e., the sample might be something like <span class="texhtml">''x''<sub>1</sub></span> = H, <br><span class="texhtml">''x''<sub>2</sub></span>&nbsp;= T, …, <span class="texhtml">''x''<sub>8</sub></span>&nbsp;= T, and the count of number of HEADS, "H" is observed.  
  
The probability of tossing TAILS is <span class="texhtml">1 − ''p''</span>. Suppose the outcome is 49 HEADS and 31 TAILS, <br>and suppose the coin was taken from a box containing three coins: one which gives HEADS<br>with probability <span class="texhtml">''p'' = 1 / 3</span>, one which gives HEADS with probability <span class="texhtml">''p'' = 1 / 2</span>&nbsp;and another which<br>gives HEADS with probability <span class="texhtml">''p'' = 2 / 3</span>. The coins have lost their labels, so which one it was is<br>unknown. Clearly the probability mass function for this experiment is binomial distribution with<br>sample size equal to 80, number of successes equal to 49 but different values of <span class="texhtml">''p''</span>. We have<br>the following probability mass functions for each of the above mentioned cases:  
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The probability of tossing TAILS is <span class="texhtml">1 − ''p''</span>. Suppose the outcome is 49 HEADS and 31 TAILS, <br>and suppose the coin was taken from a box containing three coins: one which gives HEADS<br>with probability <span class="texhtml">''p'' = 1 / 3</span>, one which gives HEADS with probability <span class="texhtml">''p'' = 1 / 2</span>&nbsp;and another which<br>gives HEADS with probability <span class="texhtml">''p'' = 2 / 3</span>. The coins have lost their labels, so which one it was is<br>unknown. Clearly the probability mass function for this experiment is binomial distribution with<br>sample size equal to 80, number of successes equal to 49 but different values of <span class="texhtml">''p''</span>. We have<br>the following probability mass functions for each of the above mentioned cases:'''<br>'''
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<math>Pr(H = 49 | p = {1}/{3}) = \binom{80}{49}(1/3)^{49}(1 - 1/3)^31 \approx 0.000</math><br>
 +
 
 +
<math>Pr(H = 49 | p = {1}/{2}) = \binom{80}{49}(1/2)^{49}(1 - 1/2)^31 \approx 0.012</math>
 +
 
 +
'''<math>Pr(H = 49 | p = {2}/{3}) = \binom{80}{49}(2/3)^{49}(1 - 2/3)^31 \approx 0.054</math><br>'''
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Based on the above equations, we can conclude that the coin with <span class="texhtml">''p'' = 2 / 3</span>&nbsp;was more likely<br>to be picked up for the observations which we were given to begin with. <br> <br>
  
 
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= Definition  =
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----
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The generic situation is that we observe a n-dimensional random vector X with probability<br>density (or mass) function <span class="texhtml">''f''(''x'' / θ)</span>. It is assumed that <span class="texhtml">θ</span>&nbsp;is a fixed, unknown constant<br>belonging to the set&nbsp;<math>\Theta \subset \mathbb{R}^{n}</math>

Revision as of 06:24, 5 April 2014

Tutorial on Maximum Likelihood Estimation: A Parametric Density Estimation Method



MLE Tutorial in PDF Format


Motivation


Suppose one wishes to determine just how biased an unfair coin is. Call the probability of
tossing a HEAD is p. The goal then is to determine p.

Also suppose the coin is tossed 80 times: i.e., the sample might be something like x1 = H,
x2 = T, …, x8 = T, and the count of number of HEADS, "H" is observed.

The probability of tossing TAILS is 1 − p. Suppose the outcome is 49 HEADS and 31 TAILS,
and suppose the coin was taken from a box containing three coins: one which gives HEADS
with probability p = 1 / 3, one which gives HEADS with probability p = 1 / 2 and another which
gives HEADS with probability p = 2 / 3. The coins have lost their labels, so which one it was is
unknown. Clearly the probability mass function for this experiment is binomial distribution with
sample size equal to 80, number of successes equal to 49 but different values of p. We have
the following probability mass functions for each of the above mentioned cases:

$ Pr(H = 49 | p = {1}/{3}) = \binom{80}{49}(1/3)^{49}(1 - 1/3)^31 \approx 0.000 $

$ Pr(H = 49 | p = {1}/{2}) = \binom{80}{49}(1/2)^{49}(1 - 1/2)^31 \approx 0.012 $

$ Pr(H = 49 | p = {2}/{3}) = \binom{80}{49}(2/3)^{49}(1 - 2/3)^31 \approx 0.054 $

Based on the above equations, we can conclude that the coin with p = 2 / 3 was more likely
to be picked up for the observations which we were given to begin with.



Definition


The generic situation is that we observe a n-dimensional random vector X with probability
density (or mass) function f(x / θ). It is assumed that θ is a fixed, unknown constant
belonging to the set $ \Theta \subset \mathbb{R}^{n} $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood