Stuck on a step in Gauss' proof of the Normal Distribution

Question

I'm reading through Gauss' proof (with small modifications) as it appears in this article. I'm stuck at this step (bottom of page 105 (text) 10 (pdf)):

\begin{equation*} 0 = \frac{\partial\Omega}{\partial\bar{M}} = \end{equation*} \begin{equation} -\phi'(M_1-\bar{M})\phi(M_2-\bar{M})\cdots\phi(M_n-\bar{M}) \label{none} \end{equation} \begin{equation*} -\phi(M_1-\bar{M})\phi'(M_2-\bar{M})\cdots\phi(M_n-\bar{M}) \end{equation*} \begin{equation*} -\cdots \end{equation*} \begin{equation*} -\phi(M_1-\bar{M})\phi(M_2-\bar{M})\cdots\phi'(M_n-\bar{M}) \end{equation*} \begin{equation*} =-\left( \frac{\phi'(M_1-\bar{M})}{\phi(M_1-\bar{M})} + \frac{\phi'(M_2-\bar{M})}{\phi(M_2-\bar{M})} + \cdots + \frac{\phi'(M_n-\bar{M})}{\phi(M_n-\bar{M})} \right)\Omega \end{equation*}

Here $\Omega$ is the joint distribution, $\phi$ is the unknown (eventually shown to be the Normal Distribution), and $\bar{M}$ the mean and therefore the term that maximizes the joint distribution. From what I can tell to get to the last step we must have:

\begin{equation*} \frac{1}{\phi(M_1-\bar{M})} =\phi(M_2-\bar{M})\phi(M_3-\bar{M})\cdots\phi(M_n-\bar{M}) \end{equation*}

But I am not sure why.

Answer 1

I think you missed to consider the factor $Ω$ at the end, so that really $$ \fracΩ{ϕ(M_1−\bar M)}=ϕ(M_2-\bar M)ϕ(M_3−\bar M)⋯ϕ(M_n−\bar M) $$ as was the construction from the beginning.

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In general, logarithmic differentiation of $f(x)=f_1(x)⋯f_n(x)$ will lead to $$ \ln|f(x)|=\ln|f_1(x)| +…+\ln|f_n(x)| \\ \implies \frac{f'(x)}{f(x)}=\frac{f_1'(x)}{f_1(x)} +…+\frac{f_n'(x)}{f_n(x)} $$