How do you find the expected value of the sum?

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.

You 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:

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