91Ó°ÊÓ

Let the random variables be IID and defined as\({{\rm{X}}_{\rm{1}}}{\rm{ \ldots }}{{\rm{X}}_{\rm{n}}}\).Every one of these random variables is assumed to be a sample from the same Bernoulli, with the same p, \({X_i} \sim Ber\left( p \right)\)

The PMF of X is:

\(f\left( x \right) = {p^x}{\left( {1 - p} \right)^{1 - x}},x = 0,1\)

In the method of moments method, the parameter and the sample estimator moments are equated.

Let the parameter moment for the population be be defined as follows:

\({\mu _j}\left( \theta \right) = E\left( {X_i^j} \right)\)

Here the j term varies.

Therefore, for j=1,

\(\begin{array}{c}{\mu _1}\left( \theta \right) = E\left( {X_i^1} \right)\\ = E\left( X \right) \ldots \left( 1 \right)\end{array}\)

This denotes that \({\mu _1}\left( \theta \right)\) is equal to the population mean, that is,\(E\left( {{X_1}} \right)\)

Let the sample moment for the population be be defined as follows:

\({m_j} = \frac{1}{n}\sum\limits_{i = 1}^n {X_i^j} \)Here the j term varies.

Therefore, for j=1,

\(\begin{array}{c}{m_1} = \frac{1}{n}\sum\limits_{i = 1}^n {X_i^1} \\ = \frac{1}{n}\sum\limits_{i = 1}^n X \ldots \left( 2 \right)\end{array}\)

This denotes that \({m_1}\) is equal to the sample mean, that is,\(\frac{1}{n}\sum\limits_{i = 1}^n X \)

Therefore, for equating the first-order moment with mean,

Equating (1) and (2)

\(\begin{array}{c}{\mu _1} = {m_1}\\ = \frac{1}{n}\sum\limits_{i = 1}^n {{X_i}} \end{array}\)

Therefore, the MOM estimate is the sample mean.

The maximum likelihood to estimate the p parameter of a Bernoulli distribution.

\(\begin{array}{c}{\bf{L}}\left( {\bf{p}} \right){\bf{ = }}{{\bf{p}}^{{{\bf{x}}_{\bf{1}}}}}{\left( {{\bf{1 - p}}} \right)^{{\bf{1 - }}{{\bf{x}}_{\bf{1}}}}}{\bf{ \times }}{{\bf{p}}^{{{\bf{x}}_{\bf{2}}}}}{\left( {{\bf{1 - p}}} \right)^{{\bf{1 - }}{{\bf{x}}_{\bf{2}}}}}{\bf{ \times \ldots \times }}{{\bf{p}}^{{{\bf{x}}_{\bf{n}}}}}{\left( {{\bf{1 - p}}} \right)^{{\bf{1 - }}{{\bf{x}}_{\bf{n}}}}}\\{\bf{ = }}\prod\limits_{{\bf{i = 1}}}^{\bf{n}} {{{\bf{p}}^{{{\bf{x}}_{\bf{i}}}}}{{\left( {{\bf{1 - p}}} \right)}^{{\bf{1 - }}{{\bf{x}}_{\bf{i}}}}}} \end{array}\)

Apply log to the likelihood function.

\({\rm{log}}L\left( p \right) = \sum\limits_{i = 1}^n {{x_i}\left( {\log \left( p \right)} \right) + \left( {n - \sum\limits_{i = 1}^n {{x_i}} } \right)} \log \left( {1 - p} \right)\)

To find the value of p that maximised the log-likehood we find its first derivative and equate it to 0.

\(\frac{\partial }{{\partial p}}{\rm{log}}L\left( p \right) = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{p} + \left( {n - \sum\limits_{i = 1}^n {{x_i}} } \right)\frac{{ - 1}}{{1 - p}}\)

Equating it with 0, we get

\(p = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\)

Therefore, the MLE estimate is the sample mean.

Hence, the method of moments estimator for the parameter of a Bernoulli distribution is the M.L.E.