Expected Values

Example 1| Example 2 | Example 3 | Useful Web Resources| Solutions

 

Definition 2.3-1: Expected Value

Let the random variable X have probability mass function p(x). The expected value of X , denoted as E(X) is defined by
    E(X) = S xp(x)
    • An expected value can be thought of as an average value of a random variable
    • Example: If the random variable is midterm marks, the expected value would be the midterm mark we would see on average in a class.
    • The expected value is also called the mean of the random variable and is denoted by u.
    • E(X) and u only defined if this quantity is finite.

     

Example 1:

    Suppose the random variable X has the following probability distribution:

    x -3 -2 -1 0 1 2 3
    P(X) 0.01 0.05 0.20 0.04 0.3 0.3 0.1

    A: What is E(X) ?

    (Round to 2 decimal places)

    Solution

Sometimes we may be interested in finding the expected value of a function of a random variable. These functions are usually in the form Y=X^2 and Y=a+bX. The value of E(Y) can be computed in two ways. We can find the probability mass function of Y and apply Definition 2.3-1, or we can use the probability mass function of X and apply Theorem 2.3-1 found below:

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Theorem 2.3-1:

If Y is a function of random variable X, where, say Y=Q(x), then

E(Y) = E(Q(X)) = S Q(x)p(x)

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    Example 2:

Theorem 2.3-2:

If Y = a + bX, then E(Y) = a + bE(X)

Theorem 2.3-3:

If Y=g1(X) +g2(X) + ... + gn(X) then

E(Y) = E(g1(X))+ E(g2(X))+ ... +E(gn(X))

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Example 3:

Let the random variable Y have the probabiliy mass function from Example 2.

y 0 1 4 9
p(y) 0.04 0.50 0.35 0.11

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A: If Y= 45+3E(X), what is E(Y)?

(Round to 2 decimal places)

B: If Y= 4(X^2)-9X +1, what is E(Y)?

(Round to 2 decimal places)

 

Solution

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Useful Web Resources

Expected Value (Expectation or Mean)

Expected Value of an Estimator- www.warnercnr.colostate.edu/ class_info/fw663/ExpectedValue.PDF

Expected Value by PlanetMath

Expected Value by Wikipedia

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Solutions

Solution to Example 1

E(X) = 0.01(-3) +0.05(-2)+0.2(-1)+0.04(0)+0.3(1)+0.3(2)+0.1(3)

= -0.03 - 0.1-0.2+0+0.3+0.6+0.3

= 0.87

Solution to Example 2

1: The possible values of y are 9,4,1 and 0 because (-3)^2=(3^2)= 9 ,

(-2)^2=(2)^2=4 , (-1)^2=(1)^2=1, (0)^2=0

The probability mass function for Y=X^2 looks like this initially:

X^2
(-3)^2=9
(-2)^2=4
(-1)^2=1
(0)^2=0
(1)^2=1
(-2)^2=4
(-3)^2=9
p(x)
0.01
0.05
0.2
0.04
0.3
0.3
0.1

This becomes:

Y= X^2 9 4 1 0
p(y)
0.01+0.1 =0.11
0.05+0.3=0.35
0.2+0.3=0.5
0.04

4: Since P(Y=-3) does not exist, the answer is 0.

6: E(Y)= 9(0.11)+4(0.35)+1(0.5)+0(0.04) = 2.89

7: E(X^2)=E(Y)=2.89

8: [E(X)]^2=(0.87)^2=0.7569

9: [E(X)]^2=0.7569 is not equal to E(Y)=2.89

Solution to Example 3

A: E(Y)= 45 +3E(X)=45+3(0.87)=47.61

B: E(Y)=4E(X62)-9E(X)+1 =4(2.89)- 9(0.87)+1 = 4.73

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