Question

Using a long rod that has length μ, you are going to lay out asquare plot in which the length of each side is μ. Thus thearea of the plot will be μ². However, you do notknow the value of μ, so you decide to make n independentmeasurements X₁, X₂, ..., X_n ofthe length. Assume that each X_i has mean μ(unbiased measurements) and varianceσ_².
a. Show that x ² is not an unbiased estimator for μ².
b. For what value k is the estimator x ² -kS² unbiased for μ²?

Answer 1

We know that \(E(\bar{X})=\mu\) \(V(X)=\frac{\sigma^{2}}{n}\)

and \(V(X)=E\left(X^{2}\right)-[E(X)]^{2}\)

\(\Rightarrow E\left(X^{2}\right)=V(X)+[E(X)]^{2}\)

\(\therefore E\left(\bar{X}^{2}\right)=V(\bar{X})+[E(\bar{X})]^{2}\)

\(\quad=\frac{\sigma^{2}}{n}+\mu^{2}\)

\(\neq \mu\)

THerefore \(\bar{X}^{2}\) is not an unbiased estimator for \(\mu^{2}\)

(b) Now, E \(\left[\bar{X}^{2}-K S^{2}\right]=\mathrm{E}\left(\bar{X}^{2}\right)-k E\left(S^{2}\right)\)

\(=\frac{\sigma^{2}}{n}+\mu^{2}-k \sigma^{2}\)

For, \(\bar{X}^{2}-K S^{2}\) to be an unbiased estimate for \(\mu^{2}\) \(E\left[\bar{X}^{2}-K S^{2}\right]=\mu^{2}\)

Thus, \(\frac{\sigma^{2}}{n}+\mu^{2}-k \sigma^{2}=\mu^{2}\)

\(\Rightarrow \frac{\sigma^{2}}{n}=k \sigma^{2}\)

\(\Rightarrow k=\frac{1}{n}\)