# How can I show that sample mean has the smallest variance?

Let the population distribution is $N(\mu,1)$.
Sample mean: $\bar{X_n}=\frac{\sum_{i=1}^{n} X_i}{n}$
Then $E(\bar{X_n})=\mu$ and $V(\bar{X_n})=\frac{1}{n}$
It is an unbiased estimator, and as $n \rightarrow \infty$ it converges to $\mu$.
But I want to show that it is the minimun variance estimator, for example,
take another unbiased estimator $Y_n$ and show $V(\bar{X_n}) \le V(Y_n)$.

• You can show that the sample mean is a complete, sufficient statistic for the estimation of $\mu$. The Lehmann-Scheffe theorem than states that it is the MVUE estimator of $\mu$. If you need further assistance, try pointing out what have you already tried.
– SBM
Commented Oct 12, 2013 at 16:05
• @SBM I use MSE convergence and show that smaple mean is a consistent estimator. Is it not sufficient for this proof?
– dont
Commented Oct 12, 2013 at 16:10
• Well, consistency is an asymptotic property, which has nothing to do with MVUE (Infact, being an MVU estimator doesn't imply consistency). Have you heard of the cramer-rao bound or Lehmann Scheffe theorem?
– SBM
Commented Oct 12, 2013 at 16:14
• @SBM Yes, I studied that. Actually, the question is to show that sample mean is a "good estimator of $\mu$" and I wonder what "good estimator" means... So I just tried to show that sample mean has the minimum variance.
– dont
Commented Oct 12, 2013 at 16:15
• If you learned about the CRB, than I guess a good estimator is an efficient one (Though I'm guessing), This means that this estimator is unbiased, MVU and attains the cramer rao bound for both the small and large sample sizes.
– SBM
Commented Oct 12, 2013 at 16:19

If you wanted to show only that the sample mean has a smaller variance than every other weighted average of the observations, then this would be an exercise in Lagrange multipliers. But if you want to include all unbiased estimators of $\mu$ based on $X_1,\ldots,X_n$ (for example, the sample median is one such estimator, and is not a weighted average of the observations), then this becomes equivalent to the one-to-one nature of the two-sided Laplace transform.
Observe that the conditional distribution of $(X_1,\ldots,X_n)$ given $\bar X = (X_1+\cdots+X_n)/n$ does not depend on $\mu$. (I could add the details of how to find the conditional distribution if necessary.) In other words, the sample mean $\bar X$ is a sufficient statistic for $\mu$. Therefore, the Rao–Blackwell theorem tells us that any minimum-variance estimator is to be found only among functions of $\bar X$.
Therefore it is enough to show that the only function $g(\bar X)$ of $\bar X$ (where of course, which function $g$ is, is not allowed to depend on $\mu$; i.e. $g(\bar X)$ is actually a statistic) that is an unbiased estimator of $\mu$ is $\bar X$ itself.
The density function of $\bar X$ is $$x\mapsto \text{constant}\cdot \exp\left(\frac{-1}{2}\cdot\left(\frac{x-\mu}{1/\sqrt{n}}\right)^2\right).$$ In order that the function $g(\bar X)$ be an unbiased estimator of $\mu$, we must $g(\bar X)-\bar X$ being an unbiased estimator of $0$. Let $h(x) = g(x)-x$; then we must have $$\int_{-\infty}^\infty (\text{same constant})\cdot h(x) \exp\left(\frac{-1}{2}\cdot\left(\frac{x-\mu}{1/\sqrt{n}}\right)^2\right) \, dx = 0$$ for all values of $\mu$. Hence $$\text{same constant}\cdot \exp\left(\frac{-n\mu^2}{2}\right) \cdot \int_{-\infty}^\infty \left(h(x) \exp\left(\frac{-n}{2} x^2\right)\right) \exp\left(nx\mu\right) \, dx = 0$$ regardless of the value of $\mu$. Thus the two-sided Laplace transform of the function $$x\mapsto h(x)\exp\left( \frac{-nx^2}{2} \right)$$ is $0$ for all values of $\mu$.