The properties of a probability distribution can be summarized with a set of numerical measures known as moments. One of these moments is called the expected value, or mean. In order to calculate an expected value, you use a summation operator.
The summation operator is used to indicate that a set of values should be added together. The formulas used to compute moments for a probability distribution are based on the summation operator. This is because each calculation must be repeated for each possible value of a random variable and the results must be summed.
As an example of the summation operator, suppose a data set contains five elements. The summation operator tells you to perform the following calculations:
Xi represents a single element in a data set; i is an index, and n is the number of elements to be summed.
The expected value of a random variable X represents the average value of X that occurs if the random experiment is repeated a large number of times. You can think of the expected value as the center of the distribution.
The expected value is a weighted average of its possible values, with weights equal to probabilities. The formula for computing expected value of X is
Here are the key terms in this formula:
E[X] = the expected value of X
n = the number of possible values of X
i = an index
Xi = one possible value of X
P[Xi] = the probability of Xi
Suppose that a biopharmaceutical firm is planning to release several new drugs during the coming year, depending on whether or not the patents are approved. You can use the random variable X to represent the number of new drugs that will be released.
The table shows the probability distribution of these results.
Probability Distribution for Release of New DrugsXP[X]00.1010.2520.5030.15You can then use the probability distribution to determine the expected [average] value of X by setting up the possible values of X and the corresponding probabilities, like so:
X1 = 0 P[X1] = 0.10
X2 = 1 P[X2] = 0.25
X3 = 2 P[X3] = 0.50
X4 = 3 P[X4] = 0.15
The corresponding histogram is shown here.
Probability distribution for the number of new drugs released.
Next, you substitute these numbers into the expected value formula:
This result shows that the expected [average] number of new drugs that will be released during the coming year is 1.7. Although it's physically impossible to release 1.7 new drugs [since 1.7 is not an integer or whole number], if this experiment is repeated many times, the average number of new drugs released will be 1.7.
The expected value [or mean] of X, where X is a discrete random variable, is a weighted average of the possible values that X can take, each value being weighted according to the probability of that event occurring. The expected value of X is usually written as E[X] or m.
E[X] = S x P[X = x]
So the expected value is the sum of: [[each of the possible outcomes] × [the probability of the outcome occurring]].
In more concrete terms, the expectation is what you would expect the outcome of an experiment to be on average.
Example
What is the expected value when we roll a fair die?
There are six possible outcomes: 1, 2, 3, 4, 5, 6. Each of these has a probability of 1/6 of occurring. Let X represent the outcome of the experiment.
Therefore P[X = 1] = 1/6 [this means that the probability that the outcome of the experiment is 1 is 1/6]
P[X = 2] = 1/6 [the probability that you throw a 2 is 1/6]
P[X = 3] = 1/6 [the probability that you throw a 3 is 1/6]
P[X = 4] = 1/6 [the probability that you throw a 4 is 1/6]
P[X = 5] = 1/6 [the probability that you throw a 5 is 1/6]
P[X = 6] = 1/6 [the probability that you throw a 6 is 1/6]
E[X] = 1×P[X = 1] + 2×P[X = 2] + 3×P[X = 3] + 4×P[X=4] + 5×P[X=5] + 6×P[X=6]
Therefore E[X] = 1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 = 7/2
So the expectation is 3.5 . If you think about it, 3.5 is halfway between the possible values the die can take and so this is what you should have expected.
Expected Value of a Function of X
To find E[ f[X] ], where f[X] is a function of X, use the following formula:
E[ f[X] ] = S f[x]P[X = x]
Example
For the above experiment [with the die], calculate E[X2]
Using our notation above, f[x] = x2
f[1] = 1, f[2] = 4, f[3] = 9, f[4] = 16, f[5] = 25, f[6] = 36
P[X = 1] = 1/6, P[X = 2] = 1/6, etc
So E[X2] = 1/6 + 4/6 + 9/6 + 16/6 + 25/6 + 36/6 = 91/6 = 15.167
The expected value of a constant is just the constant, so for example E[1] = 1. Multiplying a random variable by a constant multiplies the expected value by that constant, so E[2X] = 2E[X].
A useful formula, where a and b are constants, is:
E[aX + b] = aE[X] + b
[This says that expectation is a linear operator].
Variance
The variance of a random variable tells us something about the spread of the possible values of the variable. For a discrete random variable X, the variance of X is written as Var[X].
Var[X] = E[ [X – m]2 ] where m is the expected value E[X]
This can also be written as:
Var[X] = E[X2] – m2
The standard deviation of X is the square root of Var[X].
Note that the variance does not behave in the same way as expectation when we multiply and add constants to random variables. In fact: