# Calculate $MSE(\hat\mu_2)$

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?

$$\text{MSE}=\text{BIAS}^2+\text{VAR}$$
$$\text{MSE}_2=\frac{\mu^2c^2}{(n+c)^2}+\frac{n\sigma^2}{(n+c)^2}$$
Thus, in terms of MSE, there are particular values of $$\mu$$ and $$c$$ where the alternative estimator is preferred to the sample mean