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%$\alpha =\frac{\sigma _{2}^{2}}{\sigma _{1}^{2}+\sigma _{2}^{2}}$
%}


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\begin{document}


Suppose $X_{1}$ and $X_{2}$ are RV's both with mean $\mu $

Then $\alpha X_{1}+(1-\alpha )X_{2}$ estimates $\mu $

Combining two statistics estimating the same quantity.

Suppose $X_{1}$ and $X_{2}$ are two rv's for which

\qquad $E(X_{1})=E(X_{2})=\mu $

Consider the linear combination $Y=\alpha X_{1}+(1-\alpha )X_{2}$

$Var(\alpha X_{1}+(1-\alpha )X_{2})=\alpha ^{2}\sigma _{1}^{2}+(1-\alpha
)^{2}\sigma _{2}^{2}$

$\frac{d}{d\alpha }(\alpha ^{2}\sigma _{1}^{2}+(1-\alpha )^{2}\sigma
_{2}^{2})=\allowbreak 2\alpha \sigma _{1}^{2}-2\sigma _{2}^{2}+2\alpha
\sigma _{2}^{2}$

$2\alpha \sigma _{1}^{2}-2\sigma _{2}^{2}+2\alpha \sigma _{2}^{2}=0$,
Solution is:$\alpha =\frac{\sigma _{2}^{2}}{\sigma _{1}^{2}+\sigma _{2}^{2}}$

Thus minimized variance is

$\alpha ^{2}\sigma _{1}^{2}+(1-\alpha )^{2}\sigma _{2}^{2}=\allowbreak
\sigma _{1}^{2}\frac{\sigma _{2}^{4}}{\sigma _{1}^{4}+\sigma
_{2}^{4}+2\sigma _{1}^{2}\sigma _{2}^{2}}+\sigma _{1}^{4}\frac{\sigma
_{2}^{2}}{\sigma _{1}^{4}+\sigma _{2}^{4}+2\sigma _{1}^{2}\sigma _{2}^{2}}%
=\allowbreak \left( \sigma _{1}^{2}+\sigma _{2}^{2}\right) ^{-1}\sigma
_{2}^{2}\sigma _{1}^{2}$

\end{document}
