4.3 Mean of \(\bar{Y}\)

SW 2.5, 3.1

Another important thing to notice about \(\bar{Y}\) is that it is a random variable (as it is the average of random variables). This is in sharp contrast to \(\mathbb{E}[Y]\) which is non-random.

One related thought experiment is the following: if we could repeatedly collect new samples of size \(n\) from the same population and each time were able to estimate \(\bar{Y}\), these estimates would be different from each other.

In fact, this means that \(\bar{Y}\) has a distribution. The distribution of a statistic, like \(\bar{Y}\), is called its sampling distribution. We’d like to know about the features of the sampling distribution. Let’s start with its mean. That is, let’s calculate

\[ \begin{aligned} \mathbb{E}[\bar{Y}] &= \mathbb{E}\left[ \frac{1}{n} \sum_{i=1}^n Y_i \right] \\ &= \frac{1}{n} \mathbb{E}\left[ \sum_{i=1}^n Y_i \right] \\ &= \frac{1}{n} \sum_{i=1}^n \mathbb{E}[Y_i] \\ &= \frac{1}{n} \sum_{i=1}^n \mathbb{E}[Y] \\ &= \frac{1}{n} n \mathbb{E}[Y] \\ &= \mathbb{E}[Y] \end{aligned} \] Let’s think carefully about each step here — the arguments rely heavily on the properties of expectations and summations that we have learned earlier. The first equality holds from the definition of \(\bar{Y}\). The second equality holds because \(1/n\) is a constant and can therefore come out of the expectation. The third equality holds because the expectation can pass through the sum. The fourth equality holds because \(Y_i\) are all from the same distribution which implies that they all of the same mean and that it is equal to \(\mathbb{E}[Y]\). The fifth equality holds because \(\mathbb{E}[Y]\) is a constant and we add it up \(n\) times. And the last equality just cancels the \(n\) in the numerator with the \(n\) in the denominator.

Before moving on, let me make an additional comment:

  • The fourth equality might be a little confusing. Certainly it is not saying that all the \(Y_i\)’s are equal to each other. Rather, they come from the same distribution. For example, if you roll a die \(n\) times, you get different outcomes on different rolls, but they are all from the same distribution so that the population expectation of each roll is always 3.5, but you get different realizations on different particular rolls. Another example is if \(Y\) is a person’s income. Again, we are not saying that everyone has the same income, but just that we are thinking of income as being a draw from some distribution — sometimes you get a draw of a person with a very high income; other times you get a draw of a person with a low income, but \(\mathbb{E}[Y]\) is a feature of the underlying distribution itself where these draws come from.

How should interpret the above result? It says that, \(\mathbb{E}[\bar{Y}] = \mathbb{E}[Y]\). This doesn’t mean that \(\bar{Y}\) itself is equal to \(\mathbb{E}[Y]\). Rather, it means that, if we could repeatedly obtain (a huge number of times) new samples of size \(n\) and compute \(\bar{Y}\) each time, the average of \(\bar{Y}\) across repeated samples would be equal to \(\mathbb{E}[Y]\).