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 erro.
(a) The statistical properties of $\hat{\mu}_1=\overline{X}$ are well-known. For $\hat{\mu}_2$ check whether the esimator is:
i) unbiased (if not, determine the bias)
ii) asymptotically unbiased
My attempt:
i) We have that the Bias is equal to $E(\hat{\mu_2})-\mu$
So, $E(\hat{\mu_2})=E\left(\frac{1}{n+c}\sum^n_{i=1}X_i\right)=\frac{1}{n+c}\sum^n_{i=1}E(X_i)=\frac{n\mu}{n+c}$
The bias is:
$\frac{n \mu}{n+c}-\mu=\frac{n \mu -(n+c)\mu}{n+c}=\frac{\mu(n-n-c)}{n+c}=\frac{-c\mu}{n+c}<0$ Hence the bias is negative.
(ii) We have that $\hat{\mu_2}$ is asymptotically biased if Bias $\to 0$ as $n \to \infty$. In our case this is true, hence it's asymptotically biased.
Is this correct?
