Find efficient estimator

Let $x_1$, $x_2$,...,$x_n$ be a simple random sample from $X\sim \mathcal N(\mu,\sigma^2)$. Which estimator is more efficient between $T_1=\frac{1}{n}\sum_{i=1}^n X_i=\bar X$ and $T_2=2\bar X-X_1$?

Both $T_1$ and $T_2$ are unbiased estimator:

$$\mathbb{E}(T_1)=\mathbb{E}(\bar X)=\mu$$

$$\mathbb{E}(T_2)=\mathbb{E}(2\bar X-X_1)=\mathbb{E}(2\bar X)-\mathbb{E}(X_1)=2\mu-\mu=\mu$$

I have to find each mean square error, so:

$$MSE(T_1)=\mathbb{E}((T_1-\mu)^2)=\mathbb{Var}(T_1)=\mathbb{Var}(\bar X)=\frac{\sigma^2}{n}$$

$$MSE(T_2)=\mathbb{E}((T_2-\mu)^2)=\mathbb{Var}(T_2)=\mathbb{Var}(2\bar X-X_1)=4\mathbb{Var}(\bar X)+\mathbb{Var}(X_1)=4\frac{\sigma^2}{n}+\sigma^2$$

In conclusion $T_1$ is more efficient than $T_2$.

Is correct?

• Do you have a question? Jun 24 '16 at 15:47
• Sorry, is that correct?
– Paul
Jun 24 '16 at 15:48

The conclusion is correct, but the derivation isn't quite correct, because you applied the variance sum rule that holds for independent variables, but $\bar X$ and $X_1$ aren't independent. One way to resolve this is to take the contribution $\frac2nX_1$ out of $2\bar X$ and combine it with $-X_1$; then you have two independent variables and can apply the sum rule.
• Right, I forgot the independent stuff. I don't get how I can take the contribution $\frac2nX_1$ out of $2\bar X$ and combine it with $-X_1$. Do you mean $T_2=2\bar X-X_1=\frac{2}{n}(X_1+X_2+...+X_n)-X_1=\frac{2}{n}(X_1+X_2+...+X_n)-\frac{2}{n}X_1+\frac{2}{n}X_1-X_1$?
• @Paul: Yes, and now cancel the first two occurrences of $X_1$; then you have $2(n-1)/n$ times the mean of $X_2$ through $X_n$ and $\frac2n-1$ times $X_1$. Jun 24 '16 at 16:06
• I'm close to solve it.The expectation is $\mathbb{E}(T_2)=\mathbb{E}(\frac{2}{n}(X_2+\ldots+X_n)+(\frac{2}{n}-1)X_1)=\mu$. I'm stuck on $MSE$ :-)
• @Paul: There was no need to recalculate the expected value. Linearity of expectation doesn't require independence. Only the variance is affected. To get the variance, you only need exactly the principles you already used in your attempt: Multiplying by a factor $\lambda$ multiplies the variance by $\lambda^2$, and the variance of the sum of independent variables is the sum of their variances. Jun 24 '16 at 16:36