I take over the following question since I have not all understood the calculation done :
Let $s_p^2 = bs_1^2 + (1-b)s_2^2$, this can be an unbiased estimator of population variance, provided we find the correct value for $b$; in particular, $s_p^2 = \frac{(n1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}$ is an unbiased estimator
To minimize variance of $s_p^2 = bs_1^2 + (1-b)s_2^2:V(s_p^2)=b^2 > V(s_1^2) + (1-b)^2 V(s_2^2)$ we need to find $\frac{d}{db}V(s_p^2)=0$ but I cannot solve this for $s_p^2$.
and the answer :
Lets start with $s_p^2 = bs_1^2 + (1-b)s_2^2:V(s_p^2)=b^2 V(s_1^2) + > (1-b)^2 V(s_2^2)$
$\frac{d}{db} V(s_p^2)=2bV(s_1^2)-2(1-b)V(s_2^2)=0 \implies > \frac{b}{1-b} = \frac{Var(s_2^2)}{Var(s_1^2)}$
As you can see, $b,(1-b)$ is inversely proportional to the variance of $s_1^2,s_2^2$ respectively.
Now, if they both come from the same population (with assumed same variance), this implies that $Var(s_1^2)=\frac{n_2}{n_1} Var(s_2^2) > \rightarrow \frac{b}{1-b} = > \frac{Var(s_2^2)}{Var(s_1^2)}=\frac{n_1}{n_2} \implies > n_2b-n_1(1-b)=b(n_2+n_1)-n_1=0 \rightarrow b=\frac{n_1}{n_1+n_2}$
This means that the larger the sample size for sample 1 relative to sample 2, the more weight we give to sample 1.
How can I prove that $s_p^2 = b\,s_1^2 + (1-b)\,s_2^2$ can be an unbiased estimator of population variance with appropriate $b$ factor ?
Could anyone tell me please the difference between the maximum likelihood estimator initialy used in the original answer and the fix to replace $n$ by $n-1$ (located in comments) ?
Indeed, I am looking for using pooled variance method by determining the parameter "$b$" that I would like to fix in order to get a mathematical and mostly physical better results.
Sorry, it is not really a duplicate, rather a problem of understand well what is done.