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.

Thanks in advance everyone for your help.

  • $\begingroup$ How is $\frac1n\mu=\mu$? $\endgroup$ – Shubham Johri Dec 17 '19 at 14:26
  • $\begingroup$ @ShubhamJohri I guess it's because there are 2 sum signs missing in the equations $\endgroup$ – Jeanba Dec 17 '19 at 19:44

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 $$\frac{1}{n^2}\sum_{i=1}^n Var(X_i) =\frac{1}{n^2}\sum_{i=1}^n 1 = \frac{1}{n}.$$

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  • $\begingroup$ 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. $\endgroup$ – Mathias Dec 17 '19 at 14:22
  • 3
    $\begingroup$ @MathiasNissen Since $\mu$ does not depend on the sum index, you get $\sum_{i=1}^n \mu = \mu \sum_{i=1}^n 1 = n\mu$ $\endgroup$ – gt6989b Dec 17 '19 at 14:26

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