I have the following problem:

Based on a random sample $\{X_1,X_2,...,X_n\}$ of size $n$, two statisticians disagree on which estimator to use to estimate the population mean $\mu$ (where $\mu>0$), of a population distribution with variance $\sigma^2$. The two proposed estimators are:

$$\hat{\mu}_1=\overline{X} \text{ and } \hat{\mu}_2=\frac{1}{n+c}\sum^n_{i=1}X_i$$

where $c$ is some integer. one of the two statistician argues that $\hat{\mu_2}$ has a smaller variance than the sample mean and, for a suitable choice of $c$, is better in terms of mean squared error.

Q: It is known that the mean squared error of $\hat{\mu_1}$ is $\frac{\sigma^2}{n}$. Derive the mean squared error of $\mu_2$ and check whether the estimator $\hat{\mu_2}$ is mean square consistent.

Since $Var\left(\frac{1}{n}\sum^n_{i=1}X_i\right)=\frac{\sigma^2}{n}\Rightarrow Var\left(\sum^n_{i=1}X_i\right)=n\sigma^2$

From this question I have calculate the bias:

So, $MSE(\hat{\mu_2})=Bias(\hat{\mu_2})+Var(\hat{\mu_2})=\frac{-c\mu}{n+c}+Var\left(\frac{1}{n+c}\sum^n_{i=1}X_i\right)=\frac{-c\mu}{n+c}+\frac{n\sigma^2}{(n+c)^2}$

Would this be correct?


1 Answer 1


Your last formula is wrong because




Thus, in terms of MSE, there are particular values of $\mu$ and $c$ where the alternative estimator is preferred to the sample mean


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