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 \(X_1,X_2,\ldots, X_n\) be a random sample of size \(n\) from a distribution (population) with mean \(\mu\) and variance \(\sigma^2\). 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 \(X_i\) are identically distributed, which means they have the same mean \(\mu\). Therefore, replacing \(E(X_i)\) with the alternative notation \(\mu\), we get:

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

Now, because there are \(n\) \(\mu\)'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 \(\mu\). That is, we have shown that the mean of \(\bar{X}\) is the same as the mean of the individual \(X_i\).

Let \(X_1,X_2,\ldots, X_n\) be a random sample of size \(n\) from a distribution (population) with mean \(\mu\) and variance \(\sigma^2\). 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 \(X_i\)'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 \(X_i\) are identically distributed, which means they have the same variance \(\sigma^2\). Therefore, replacing \(\text{Var}(X_i)\) with the alternative notation \(\sigma^2\), we get:

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

Now, because there are \(n\) \(\sigma^2\)'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 \(n\) 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!