# Unbiased estimator for mu

I have to find an unbiased estimator for $$\mu$$ where $$\bar{X} = \frac{1}{n}(X_1+X_2+...+X_n)$$ is the mean for a $$\mathcal{N}(\mu,1)$$-distribution.

I know that unbiased requires $$E[\bar{X}] = \mu$$ in this case but I am not sure whether or not my calculations are correct:

$$E[\bar{X}] = E[\frac{1}{n}(X_1+X_2+...+X_n)] = \frac{1}{n} E[X_1+X_2+...+X_n] = \frac{1}{n} E[X_i] = \frac{1}{n} \mu = \mu.$$ Thus $$E[\bar{X}] = \mu$$ and therefore $$\bar{X}$$ must be an unbiased estimator for $$\mu$$. Is this correct?

Furthermore I need to calculate the MSE. Thus I need to find $$E[(\bar{X}-\mu)^2] = Var(\bar{X}-\mu) + (E[\bar{X} - \mu ])^2 = Var(\bar{X}) = Var(\frac{1}{n} ( x_1+x_2+...+x_n) = \frac{1}{n^2} \cdot Var(x_i) = \frac{1}{n} \cdot Var(x) \cdot n = \frac{1}{n} \cdot Var(x) = \frac{1}{n}.$$ Is this also correct?

I do not have answers to the above questions and I am currently studying for my exam in probability and statistics.

• How is $\frac1n\mu=\mu$? – Shubham Johri Dec 17 '19 at 14:26
You have dropped the sum in both calculations. You should have: $$\frac{1}{n}\sum_{i=1}^n \mathbb{E}[X_i] = \frac{1}{n}\sum_{i=1}^n \mu = \mu$$ and $$\frac{1}{n^2}\sum_{i=1}^n Var(X_i) =\frac{1}{n^2}\sum_{i=1}^n 1 = \frac{1}{n}.$$
• Hi thanks for the help. However can you explain why $\frac{1}{n} \sum_{i=1}^n \mu = \mu$? Is the sum just equal to n? Same for the variance. Thanks. – Mathias Dec 17 '19 at 14:22
• @MathiasNissen Since $\mu$ does not depend on the sum index, you get $\sum_{i=1}^n \mu = \mu \sum_{i=1}^n 1 = n\mu$ – gt6989b Dec 17 '19 at 14:26