The mean or expected value of a random variable can be thought of as the “long-term average”, meaning the average of the outcomes of an ever increasing number of trials of the experiment.
Denoted \(E(X)\) or \(\mu_X\).
If the variable \(X\) takes the values \(x_1, x_2, \ldots x_n\) with probabilities \(p_1, p_2, \ldots, p_n\) respectively, then the mean is defined as: \[E(X) = \mu_X = p_1x_1 + p_2x_2 + \cdots + p_nx_n\]
You can think of this as a weighted average of the values, with their probabilities as weights. This makes sense: We want to take all values into account, but those values that have a higher probability are meant to appear more often, and so should contribute more. Each value contributes an amount proportionate to its relative frequency.
As a simple example, consider the example from the last section, with probability table:
| X | 0 | 1 | 2 |
| P(X) | \(1/2\) | \(1/4\) | \(1/4\) |
Then for the mean we would have:
\[E(X) = \frac{1}{2}\cdot 0 + \frac{1}{4} \cdot 1 + \frac{1}{4} \cdot 2 = \frac{3}{4} = 0.75\]
We can think of this as saying that if you were to play that game repeatedly, you would be gaining on average \(\$0.75\) per game. You can also think of it as the “fair price to pay to play the game”.
We examined a number of games in the previous section. Compute the mean of the random variables in each of those games.
The standard deviation follows a similar formula:
\[\sigma_X^2 = p_1 (x_1 - \mu_X)^2 + p_2 (x_2 - \mu_X)^2 + \cdots p_n (x_n - \mu_X)^2\]
So we look at how far each value is from the mean, square to remove the signs, average while accounting for the different probabilities, and finally take a square root.
This square of the standard deviation, typically called the Variance \(\textrm{Var}(X)\), you will often see written as \(E\left((X-\mu_X)^2\right)\).
Compute the standard deviation for each of the examples discussed so far.