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In a population \({\theta _i}\) denote the proportion of individuals of type i, for i = 1,...,k. Here, 0 鈮� \({\theta _i}\) 鈮� 1 and\({\theta _1} + ... + {\theta _k} = 1\) as well as \({n_1} + ... + {n_k} = n\). W e need to calculate the M.L.E. of \({\theta _1},...,{\theta _n}\)

Let \(f\left( x \right) = \theta _1^{{n_1}}...\theta _n^{{n_k}}\)and \(L\left( {{\theta _1}...{\theta _K}} \right) = \log f\left( x \right)\,\,\,and\,\,{\theta _K} = 1 - \sum\limits_{i = 1}^n {{\theta _i}} \)

By maximizing the log of the likelihood function we will get the M.L.E.

Thus,

\(\begin{array}{c}L\left( \theta \right) = \log f\left( {x;\theta } \right)\\ = \sum\limits_{i = 1}^n {{n_i}\log {\theta _i}} \end{array}\)

Now, let us take the partial derivative

\(\frac{{\partial L\left( \theta \right)}}{{\partial \theta }} = \frac{{{n_i}}}{{{\theta _i}}} - \frac{{{n_k}}}{{{\theta _k}}}\)

From the equation of partial derivatives to zero the following system of equations is obtained

\(\frac{{{\theta _1}}}{{{n_1}}} = \frac{\theta }{{{n_2}}} = ... = \frac{{{\theta _k}}}{{{n_k}}} = c\)

Now, from the fact \(\sum\limits_{i = 1}^n {{\theta _i}} = 1\) and \({\theta _i} = c{n_i}\) we get

\(1 = \sum\limits_{i = 1}^k {{\theta _i}} = cn\). So the maximum is obtained when \(c = \frac{1}{n}\)

Hence the M.L.E. is \({\hat \theta _i} = \frac{{{n_i}}}{n}\)