Maximum Likelihood Estimates I'm having trouble following the section highlighted. The maximum likelihood estimate derived in part a) =X.


 A: You are given that
$$X_i\sim Poisson(\lambda p_i), N=\sum^{k}_{i=1}X_i$$
If we let $\xi_i=\lambda p_i$, the distribution of $x_i$ will be
$$p(x_i)=\frac{\xi_i^{x_i}e^{-\xi_i}}{x_i!}$$
Differentiating this w.r.t. $\xi_i$ we have
$$\frac{dp(x_i)}{d\xi_i}=\frac{x_i\xi_i^{x_1-1}e^{-\xi_i}-\xi_i^{x_i}e^{-\xi_i}}{x_1!}$$
To find the maximum likelihood value of $\xi_i$, which we denote by $\xi_{i,MLE}=\hat{\lambda}_{MLE}\hat{p}_{i,MLE}$, we need to set the differential to zero and solve for $\xi_i$
$$\frac{x_i\xi^{x_1-1}_{i,MLE}e^{-\xi_{i,MLE}}-\xi^{x_i}_{MLE}e^{-\xi_{i,MLE}}}{x_1!}=0\Rightarrow(x_i-\xi_{i,MLE})=0\Rightarrow\xi_{i,MLE}=x_i$$
Thus 
$$\hat{\lambda}_{MLE}\hat{p}_{i,MLE}=\xi_{i,MLE}=x_i\Rightarrow \hat{p}_{i,MLE}=\frac{x_i}{\hat{\lambda}_{MLE}}$$
Given the constraint that $\sum_{i=1}^kp_i=1$, hence $\sum_{i=1}^k\hat{p}_{i,MLE}=1$, we have
$$\sum_{i=1}^k\frac{x_i}{\hat{\lambda}_{MLE}}=1\Rightarrow \hat{\lambda}_{MLE}=\sum_{i=1}^kx_i$$
Substituting the value of $\hat{\lambda}_{MLE}$ in terms of $x_i$ back into the MLE of $p_i$ the result is
$$\hat{p}_{i,MLE}=\frac{x_i}{\sum_{i=1}^kx_i}$$ 
In vector notation, where $\hat{p}_{MLE}=[\hat{p}_{1,MLE},\hat{p}_{2,MLE},...,\hat{p}_{k,MLE}]$ and $x=[x_1,x_2,...,x_k]$ the result can be expressed as
$$\hat{p}_{MLE}=\frac{x}{\sum_{i=1}^kx_i}$$ 
