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This is a very beautiful sample problem from ISI MStat PSB 2012 Problem 9, It's about restricted MLEs, how restricted MLEs are different from the unrestricted ones, if you miss delicacies you may miss the differences too . Try it! But be careful.

Suppose \(X_1\) and \(X_2\) are i.i.d. Bernoulli random variables with parameter \(p\) where it us known that \(\frac{1}{3} \le p \le \frac{2}{3} \). Find the maximum likelihood estimator \(\hat{p}\) of \(p\) based on \(X_1\) and \(X_2\).

Bernoulli trials

Restricted Maximum Likelihood Estimators

Real Analysis

This problem seems quite simple and it is simple, if and only if one observes subtle details. Lets think about the unrestricted MLE of \(p\),

Let the unrestricted MLE of \(p\) (i.e. when \(0\le p \le 1\) )based on \(X_1\) and \(X_2\) be \(p_{MLE}\), and \( p_{MLE}=\frac{X_1+X_2}{2}\) (How ??)

Now lets see the contradictions which may occur if we don't modify \(p_{MLE}\) to \(\hat{p}\) (as it is been asked).

See, that when if our sample comes such that \(X_1=X_2=0\) or \(X_1=X_2=1\), then \(p_{MLE}\) will be 0 and 1 respectively, where \(p\), the actual parameter neither takes the value 1 or 0 !! So, \(p_{MLE}\) needs serious improvement !

To, modify the \(p_{MLE}\), lets observe the log-likelihood function of Bernoulli based in two samples.

\( \log L(p|x_1,x_2)=(x_1+x_2)\log p +(2-x_1-x_2)\log (1-p) \)

Now, make two observations, when \(X_1=X_2=0\) (.i.e. \(p_{MLE}=0\)), then \(\log L(p|x_1,x_2)=2\log (1-p)\), see that \(\log L(p|x_1,x_2)\) decreases as p increase, hence under the given condition, log_likelihood will be maximum when p is least, .i.e. \(\hat{p}=\frac{1}{3}\).

Similarly, when \(p_{MLE}=1\) (i.e.when \( X_1=X_2=1\)), then for the log-likelihood function to be maximum, p has to be maximum, i.e. \(\hat{p}=\frac{2}{3}\).

So, to modify \(p_{MLE}\) to \(\hat{p}\), we have to develop a linear relationship between \(p_{MLE}\) and \(\hat{p}\). (Linear because, the relationship between \(p\) and \(p_{MLE}\) is linear. ). So, \(\hat{p}\) and \(p_{MLE}\) is on the line that is joining the points \((0,\frac{1}{3})\) ( when \(p_{MLE}= 0\) then \(\hat{p}=\frac{1}{3}\)) and \((1,\frac{2}{3})\). Hence the line is,

\(\frac{\hat{p}-\frac{1}{3}}{p_{MLE}-0}=\frac{\frac{2}{3}-\frac{1}{3}}{1-0}\)

\(\hat{p}=\frac{2-X_1-X_2}{6}\). is the required restricted MLE.

Hence the solution concludes.

Can You find out the conditions for which the Maximum Likelihood Estimators are also unbiased estimators of the parameter. For which distributions do you think this conditions holds true. Are the also Minimum Variance Unbiased Estimators !!

Can you give some examples when the MLEs are not unbiased ?Even If they are not unbiased are the Sufficient ??

Content

[hide]

This is a very beautiful sample problem from ISI MStat PSB 2012 Problem 9, It's about restricted MLEs, how restricted MLEs are different from the unrestricted ones, if you miss delicacies you may miss the differences too . Try it! But be careful.

Suppose \(X_1\) and \(X_2\) are i.i.d. Bernoulli random variables with parameter \(p\) where it us known that \(\frac{1}{3} \le p \le \frac{2}{3} \). Find the maximum likelihood estimator \(\hat{p}\) of \(p\) based on \(X_1\) and \(X_2\).

Bernoulli trials

Restricted Maximum Likelihood Estimators

Real Analysis

This problem seems quite simple and it is simple, if and only if one observes subtle details. Lets think about the unrestricted MLE of \(p\),

Let the unrestricted MLE of \(p\) (i.e. when \(0\le p \le 1\) )based on \(X_1\) and \(X_2\) be \(p_{MLE}\), and \( p_{MLE}=\frac{X_1+X_2}{2}\) (How ??)

Now lets see the contradictions which may occur if we don't modify \(p_{MLE}\) to \(\hat{p}\) (as it is been asked).

See, that when if our sample comes such that \(X_1=X_2=0\) or \(X_1=X_2=1\), then \(p_{MLE}\) will be 0 and 1 respectively, where \(p\), the actual parameter neither takes the value 1 or 0 !! So, \(p_{MLE}\) needs serious improvement !

To, modify the \(p_{MLE}\), lets observe the log-likelihood function of Bernoulli based in two samples.

\( \log L(p|x_1,x_2)=(x_1+x_2)\log p +(2-x_1-x_2)\log (1-p) \)

Now, make two observations, when \(X_1=X_2=0\) (.i.e. \(p_{MLE}=0\)), then \(\log L(p|x_1,x_2)=2\log (1-p)\), see that \(\log L(p|x_1,x_2)\) decreases as p increase, hence under the given condition, log_likelihood will be maximum when p is least, .i.e. \(\hat{p}=\frac{1}{3}\).

Similarly, when \(p_{MLE}=1\) (i.e.when \( X_1=X_2=1\)), then for the log-likelihood function to be maximum, p has to be maximum, i.e. \(\hat{p}=\frac{2}{3}\).

So, to modify \(p_{MLE}\) to \(\hat{p}\), we have to develop a linear relationship between \(p_{MLE}\) and \(\hat{p}\). (Linear because, the relationship between \(p\) and \(p_{MLE}\) is linear. ). So, \(\hat{p}\) and \(p_{MLE}\) is on the line that is joining the points \((0,\frac{1}{3})\) ( when \(p_{MLE}= 0\) then \(\hat{p}=\frac{1}{3}\)) and \((1,\frac{2}{3})\). Hence the line is,

\(\frac{\hat{p}-\frac{1}{3}}{p_{MLE}-0}=\frac{\frac{2}{3}-\frac{1}{3}}{1-0}\)

\(\hat{p}=\frac{2-X_1-X_2}{6}\). is the required restricted MLE.

Hence the solution concludes.

Can You find out the conditions for which the Maximum Likelihood Estimators are also unbiased estimators of the parameter. For which distributions do you think this conditions holds true. Are the also Minimum Variance Unbiased Estimators !!

Can you give some examples when the MLEs are not unbiased ?Even If they are not unbiased are the Sufficient ??

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