# Mean and Variance of Sample Mean

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We'll finally accomplish what we set out to do in this lesson, namely to determine the theoretical mean and variance of the continuous random variable $\bar{X}$. In doing so, we'll discover the major implications of the theorem that we learned on the previous page.

Let X1, X2, ... , Xn  be a random sample of size n from a distribution (population) with mean μ and variance σ2. What is the mean, that is, the expected value, of the sample mean $\bar{X}$?

Solution. Starting with the definition of the sample mean, we have:

$E(\bar{X})=E\left(\dfrac{X_1+X_2+\cdots+X_n}{n}\right)$

Then, using the linear operator property of expectation, we get:

$E(\bar{X})=\dfrac{1}{n} [E(X_1)+E(X_2)+\cdots+E(X_n)]$

Now, the Xi are identically distributed, which means they have the same mean μ. Therefore, replacing E(Xi) with the alternative notation μ, we get:

$E(\bar{X})=\dfrac{1}{n}[\mu+\mu+\cdots+\mu]$

Now, because there are n μ's in the above formula, we can rewrite the expected value as:

$E(\bar{X})=\dfrac{1}{n}[n \mu]=\mu$

We have shown that the mean (or expected value, if you prefer) of the sample mean $\bar{X}$ is μ. That is, we have shown that the mean of $\bar{X}$ is the same as the mean of the individual Xi.

Let X1X2, ... , Xn  be a random sample of size from a distribution (population) with mean μ and variance σ2. What is the variance of $\bar{X}$?

Solution. Starting with the definition of the sample mean, we have:

$Var(\bar{X})=Var\left(\dfrac{X_1+X_2+\cdots+X_n}{n}\right)$

Rewriting the term on the right so that it is clear that we have a linear combination of Xi's, we get:

$Var(\bar{X})=Var\left(\dfrac{1}{n}X_1+\dfrac{1}{n}X_2+\cdots+\dfrac{1}{n}X_n\right)$

Then, applying the theorem on the last page, we get:

$Var(\bar{X})=\dfrac{1}{n^2}Var(X_1)+\dfrac{1}{n^2}Var(X_2)+\cdots+\dfrac{1}{n^2}Var(X_n)$

Now, the Xi are identically distributed, which means they have the same variance σ2. Therefore, replacing Var(Xi) with the alternative notation σ2, we get:

$Var(\bar{X})=\dfrac{1}{n^2}[\sigma^2+\sigma^2+\cdots+\sigma^2]$

Now, because there are n σ2's in the above formula, we can rewrite the expected value as:

$Var(\bar{X})=\dfrac{1}{n^2}[n\sigma^2]=\dfrac{\sigma^2}{n}$

Our result indicates that as the sample size increases, the variance of the sample mean decreases. That suggests that on the previous page, if the instructor had taken larger samples of students, she would have seen less variability in the sample means that she was obtaining. This is a good thing, but of course, in general, the costs of research studies no doubt increase as the sample size n increases.  There is always a trade-off!