Mean of a Discrete Random Variable
The mean of a discrete random variable \(X\) is denoted \(\mu_X\) or, when no confusion will arise, simply \(\mu\). The terms expected value, \(E(X)\), and expectation are commonly used in place of the term mean.
\[ E(X) = \sum_{i=1}^{N}x_iP(X=x_i) \]
In a large number of independent observations of a random variable \(X\), the \(E(X)\) of those observations - the sample - will approximate the mean, \(\mu\), of the population. The larger the number of observations, the closer \(E(X)\) is to \(\mu\) (Weiss 2010).
Recall our experiment from the previous section, where we picked 1,000 individuals and asked for their number of siblings. Let us again take a look at the table, which summarizes the experiment:
\[ \begin{array}{c|lcr} \text{Siblings} & \text{Frequency} & \text{Relative}\\ \ x & f & \text{frequency}\\ \hline 0 & 205 & 0.205 \\ 1 & 419 & 0.419 \\ 2 & 280 & 0.28 \\ 3 & 65 & 0.065 \\ 4 & 29 & 0.029 \\ 5 & 2 & 0.002 \\ \hline & 1000 & 1 \end{array} \]
We can now calculate the expected value (mean) for that experiment.
\[\begin{align} \\ & E(X) = \sum_{i=1}^{N}x_iP(X=x_i) \\ & = 0 \cdot P(X=0) + 1 \cdot P(X=1)+ 2 \cdot P(X=2) + 3 \cdot P(X=3) +4 \cdot P(X=4)+ 5 \cdot P(X \ge 5) \\ & = 0 \cdot 0.205 + 1 \cdot 0.419 + 2 \cdot 0.28+ 3 \cdot 0.065 + 4 \cdot 0.029+ 5 \cdot 0.002 \\ & = 1.3 \end{align}\]
The resulting expected value of 1.3 is close to the mean \(\mu\), which we calculate using the population’s probabilities (population probabilities are taken from the lower right figure in the previous section).
\[\mu = 1 \cdot 0.2 + 2 \cdot 0.425 + 3 \cdot 0.275 + 4 \cdot 0.07 + 5 \cdot 0.025 = 1.31\]
Exercise
Let us consider a fair six sided dice. We can easily compute the expected value \(E(X)\) using R. The term fair means that each random variable \(X=x_i,\; x \in 1,2,3,4,5,6\) is equally likely to occur. Therefore \(P(X=x_i) = \frac{1}{6}\).
\[E(X) = \sum_{i=1}^{6}x_iP(X=x_i) = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 5 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6} = 3.5\]
In R we write the following code:
# Expected value of a fair six sided dice
p_die <- 1 / 6
die <- c(1, 2, 3, 4, 5, 6)
sum(die * p_die)## [1] 3.5However, what if we are not sure, if the dice really is fair? How can we know that we are not being cheated? Or, put in other words: How often do we need to roll a dice before we can be more confident?
Let us do a computational experiment: We know from the reasoning
above, that the expected value of a 6-sided fair dice is 3.5. We conduct
an experiment by rolling a dice over and over again. We store the result
and before we roll the dice again we calculate the average of all dice
rolls so far. In order to conduct that little experiment, we write a
for loop in R.
### Simulation ###
eyes <- seq(1, 6) # possible events
probs <- rep(1 / length(eyes), length(eyes)) # probabilities
expected_value <- round(sum(eyes * probs), 2) # calculate expected value
n <- 500 # number of maximum rolls
values <- NULL # initialize an empty vector to store the value
averages <- NULL # initialize an empty vector to store the average values so far
# for-loop
for (roll in 1:n) {
values <- c(values, sample(x = eyes, size = 1, prob = probs)) # sample method, type help(sample()) for further information
averages <- c(averages, mean(values))
}
### Plot ###
par(xpd = FALSE)
plot(
x = seq(1:length(averages)),
y = averages,
type = "l",
ylim = c(min(eyes), max(eyes)),
lwd = 2,
ylab = "Expected value",
xlab = "number of trials",
col = "#3366FF"
)
abline(h = expected_value, lty = 2, col = "red")
legend("topright",
legend = paste("Expected value: ", as.character(expected_value)),
col = "red",
lty = 2
)
The graph shows that after some initial volatile behavior, the curve finally flattens and approximates the \(E(X)\) of 3.5.
Standard Deviation of a Discrete Random Variable
The standard deviation of a discrete random variable \(X\) is denoted \(\sigma_X\) or, when no confusion will arise, simply \(\sigma\). It is defined as
\[ \sigma = \sqrt{\sum_{i=1}^{N}(x_i-\mu)^2P(X=x_i)} \]
Exercise
Let us turn to R and calculate the standard deviation for the dice
roll experiment from above. During the experiment we rolled 500 times.
The outcome of these rolls is stored in the vector values.
The probability for each of these numbers in the vector
values approximates \(\frac{1}{6}= 0.167\). So, we just put those
numbers in the equation for the standard deviation from above. Remember,
that the mean is stored in the expected_value variable.
x <- seq(1, 6)
p_x <- prop.table(table(values))
p_x## values
## 1 2 3 4 5 6
## 0.160 0.184 0.170 0.140 0.180 0.166sqrt(sum((x - mean(values))^2 * p_x))## [1] 1.712882Roughly speaking, our experiment showed that after 500 rolls the value of a dice number is, on average, 1.71 away from the experimental mean of 3.494.
Citation
The E-Learning project SOGA-R was developed at the Department of Earth Sciences by Kai Hartmann, Joachim Krois and Annette Rudolph. You can reach us via mail by soga[at]zedat.fu-berlin.de.
You may use this project freely under the Creative Commons Attribution-ShareAlike 4.0 International License.
Please cite as follow: Hartmann, K., Krois, J., Rudolph, A. (2023): Statistics and Geodata Analysis using R (SOGA-R). Department of Earth Sciences, Freie Universitaet Berlin.