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Independent Random variables$=X_1,X_2$

$Var\left(X_1\right)={蟽}_1^2$

$Var\left(X_2\right)={蟽}_2^2$

Value of $渭=?$

Variance of$位X_1+(1-位)X_2=?$

Suppose that $X_1$and $X_2$are independent random variables having a common mean $渭$. Let $位X_1+(1-位)X_2$will be used as an estimate of $渭$for some appropriate value of $位$. It is known that $Var\left(X_1\right)={蟽}_1^2$and $Var\left(X_2\right)={蟽}_2^2$.

Find the variance of $位X_1+(1-位)X_2$.

$\mathrm{Var}\left(位X_1+(1鈭�位)X_2\right)=\mathrm{Var}\left(位X_1\right)+\mathrm{Var}\left[(1鈭�位)X_2\right]$

localid="1647409774914" $={位}^2\mathrm{Var}\left(X_1\right)+(1鈭�位{)}^2\mathrm{Var}\left[X_2\right]$

$={位}^2{蟽}_1^2+(1鈭�位{)}^2{蟽}_2^2$

Find the value of $位$yields the estimate having the lowest possible variance? Differentiate with respect to $位$and then equating to $0.$

role="math" $\frac{d}{d位}\left[{位}^2{蟽}_1^2+(1鈭�位{)}^2{蟽}_2^2\right]=0$

$2位{蟽}_1^2鈭�2(1鈭�位){蟽}_2^2=0$

$位=\frac{{蟽}_2^2}{{蟽}_1^2+{蟽}_2^2}$

Therefore,

Explain it is desirable to use this value of $位.$

As$Var\left[位X_1+(1-位)X_2\right]=E\left[{\left(位X_1+(1-位)X_2\right)}^2\right]$, then $位$is to be small.