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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

Score statistic and Fisher information

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  • $\begingroup$ can you Taylor expand? $\endgroup$ Commented Nov 2, 2021 at 19:52
  • $\begingroup$ Yes have learned about Taylor expand but on simple examples. Can I use Taylor Expand to show it? $\endgroup$
    – Lifeni
    Commented Nov 2, 2021 at 19:54
  • $\begingroup$ Can you help me? $\endgroup$
    – Lifeni
    Commented Nov 2, 2021 at 20:48

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I actually do not know how to do this exactly, but you can write $U(\beta)\approx U(\beta_0)+(\beta-\beta_0)U'(\beta_0)$ (Taylor expansion). Note the similarity to your second to last display. Now take this and solve for $(\beta-\beta_0)$ and pull $\beta_0$ onto the other side, and note the similarity to your last display. You must show that the solution to those two displays is the same, and this seems to be a way to connect them.

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  • $\begingroup$ Now it makes a bit sense with the Taylor Expand. But I'm not totally sure. You use that $\eta_m=X\beta_m$ and Taylor expaned and get $𝑈(\beta)≈𝑈(\beta_0)+(\beta−\beta_0)𝑈′(\beta_0)$, it makes sense, and when isolating $\beta$ I get that $\beta=\beta_0+\frac{U(\beta)-U(\beta_0)}{U'(\beta_0)}$. It seems a bit similar to my Fisher Scoring Algoritm equation, but it's not totally clear for my. Can you help showing that is the same? $\endgroup$
    – Lifeni
    Commented Nov 3, 2021 at 10:11
  • $\begingroup$ One thing is that if $\beta_n$ minimizes $U,$ $U(\beta_n)=0.$ Hence replace $\beta$ with $\beta_n$ in the Taylor expansion, and you will get rid of the term $U(\beta_n).$ $\endgroup$ Commented Nov 3, 2021 at 13:35
  • $\begingroup$ Note also that something like $X^TX/n$ goes in probability usually to the expectation of the second derivative of $U$ $\endgroup$ Commented Nov 3, 2021 at 13:36
  • $\begingroup$ I think you are getting closer. My strategy would be to do this expansion, etc for a simpler function like a univariate $f(x)=x^2$ and then see how maybe the Taylor expansion works with Fisher scoring. Then add randomness to $f(x).$ $\endgroup$ Commented Nov 3, 2021 at 13:41

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