In my exercise we assume that $Y_i|X_i$ has distribution with density $f_i(y_i,\eta_i) $ for $i=1,...,n$ where $\eta_i=X_i^T$ is the linear predictor. The generalized linear model with an exponential dispersion distribution is a case with: $$f_i(y_i,\eta_i)=e^{\frac{\theta(\eta_i)y_i-c(\eta_i)}{\Psi}}$$ It is assumed that assumed that $f_i(y_i, \eta_i)>0$ and we define that: $$U_i(\eta_i)=\frac{\partial}{\partial \eta_i}log f_i(Y_i, \eta_i)$$ and $$w_i=-E(\frac{\partial^2}{\partial^2 \eta_i}log f_i(Y_i, \eta_i))$$
We introduce now the score vector $U(\eta)=(U_1(\eta_1),..., U_n(\eta_n))^T$ and the diagonal matrix W with $W_{ii}=w_i$
I have to prove that for the m'th Fisher Scoring step for solving the score equation amounts to solving the linear equation: $$XU(\mu_m)-(X^TW_mX)(\beta-\beta_m)=0$$ Where here $W_m$ is defined in terms of $\eta=X\beta_m$.
I think that there by Fisher Scoring step means in the Fisher Scoring Algoritm? And I have found in my book that for the Fisher Scoring Algoritm: $$\beta_{m+1}=\beta_m+(X^TW_mX)^{-1}X^TU(\eta_m)$$ But how can I show it's same solving the two equations?
See this previous question