Question-23

Consider a random variable \(X\) with mean \(\mu\) and variance \(\sigma^2\). Let \(\bar{X_1}\) and \(\bar{X_2}\) be the sample means of two i.i.d. random samples of \(X\) with \(E[\bar{X_1}]=E[\bar{X_2}] = \mu\). Let \(\hat{\mu} = a\bar{X_1} +(1-a)\bar{X_2}\) be another estimator for \(\mu\), where \(0<a<1\). Find the value of \(a\) for which the variance of \(\hat{\mu}\) is minimum, assuming \(\bar{X_1}\) and \(\bar{X_2}\) are independent.

• \(a = \dfrac{\text{Var}(\bar{X_1})}{\text{Var}(\bar{X_1})+\text{Var}(\bar{X_2})}\)

• \(a = \dfrac{\text{Var}(\bar{X_1}+\bar{X_2})}{\text{Var}(\bar{X_2})}\)

• \(a = \dfrac{\text{Var}(\bar{X_2})}{\text{Var}(\bar{X_1})+\text{Var}(\bar{X_2})}\)

• \(a= \dfrac{1}{2}\)

Hint

Use the concept of Estimators.

Answer

• \(a = \dfrac{\text{Var}(\bar{X_1})}{\text{Var}(\bar{X_1})+\text{Var}(\bar{X_2})}\)

• \(a = \dfrac{\text{Var}(\bar{X_1}+\bar{X_2})}{\text{Var}(\bar{X_2})}\)

• \(a = \dfrac{\text{Var}(\bar{X_2})}{\text{Var}(\bar{X_1})+\text{Var}(\bar{X_2})}\)

• \(a= \dfrac{1}{2}\)

Solution