Question

1. Let Xi, X2,.., Xn be a random sample drawn from some population with mean μ--2λ and variance σ2-4, where λ is a parameter. Define 2n We use V, to estimate λ. (a) Show that is an unbiased estimator for λ. (b) Let ơin be the variance of V,, . Show that lin ơi,-

Answer 1

Given the estimator $V_n=\frac{\sum_{i=1}^{n}X_i}{2n}$

a) The expected value of the estimator is

$E\left (V_n \right )=E\left (\frac{\sum_{i=1}^{n}X_i}{2n} \right )\\ E\left (V_n \right )=\frac{\sum_{i=1}^{n}E\left (X_i \right )}{2n} \\ E\left (V_n \right )=\frac{\sum_{i=1}^{n}2\lambda }{2n} \\ E\left (V_n \right )=\frac{2n\lambda }{2n} \\ E\left (V_n \right )=\lambda$

Thus, $V_n=\frac{\sum_{i=1}^{n}X_i}{2n}$ is an unbiased estimator of $\lambda$

b) The variance of the estimator is

$\sigma _{V_n}^2=Var\left (V_n \right )\\ \sigma _{V_n}^2=\frac{\sum_{i=1}^{n}Var\left (X_i \right )}{\left (2n \right )^2}\\ \sigma _{V_n}^2=\frac{\sum_{i=1}^{n}4}{4n^2}\\ \sigma _{V_n}^2=\frac{4n}{4n^2}\\ \sigma _{V_n}^2=\frac{1}{n}$

Taking limits,

$\lim_{n\rightarrow \infty }\sigma _{V_n}^2=\lim_{n\rightarrow \infty }\frac{1}{n}=0$

The proof is complete.