I am trying to maximize a particular log likelihood function but I am stuck on the differentiation step.
Given: $ \Theta_1 + ....... + \Theta_k = 1 $
The likelihood function is:
$f_n(x|\Theta_1,.........,\Theta_k) = \Theta^{n_1}_1........\Theta^{n_k}_k$
Let $L(\Theta_1,......,\Theta_k) = log\,\,f_n(x|\Theta_1,.........,\Theta_k)$
and let $\Theta_k = 1 - \sum_{i=1}^{k-1} \Theta_i \qquad - (i)$
Then, $$ \frac {\partial L(\Theta_1,.......,\Theta_k)}{\partial\Theta_i} = \frac{n_i}{\Theta_i} - \frac{n_k}{\Theta_k}\qquad for \,\; i=1,.....,k-1 \qquad - (ii)$$
Case 1: We may write L as $\quad\sum_{i=1}^{k-1}n_i\,ln\,\Theta_i\,+\,n_k\;ln(1\,-\,\sum_{i=1}^{k-1} \Theta_i)\quad$ if we make the substitution in (i)
Case 2: We may write L as $\quad\sum_{i=1}^{k}n_i\,ln\,\Theta_i\quad$ if we don't make the substitution in (i)
For Case 1 derivative would be: $\quad\frac{n_i}{\Theta_i} - \frac{n_k}{\Theta_k}\qquad for \,\; i=1,.....,k-1$
For Case 2 derivative would be: $\quad\frac{n_i}{\Theta_i}\qquad for \,\; i=1,.....,k$
Thus for an $i\neq k$ depending upon if we make the substitution in (i) or not, we get two different results for the same partial derivative i.e. $\frac{\partial L}{\partial\Theta_i}$
Case 1 is the solution. But by this logic derivative can be anything depending on our choice of k in the set.
Where am I going wrong in Case 2? Am I making an error by not making the substitution and simply differentiating L.
Please help.
The denominator of $k$-th term is the sum of all the other $\Theta_i$'s. Formally, you'd get $$ \frac {\partial L(\Theta_1, \dots ,\Theta_k)}{\partial\Theta_i} = \frac{n_i}{\Theta_i} - \frac{n_k}{1 - \sum_{i=1}^{k-1}\Theta_i}\qquad \text{ for all } \,\; i=1,.....,k-1.$$ Can you continue from here?