10Cmiddle


 * Expected Value of Discrete Random Distributions **


 * Expected Value ** is another phrase for ** Mean **

The __**expected value**__ gives the average result that could be expected to occur over a large number of trials

Notation:

math . \qquad \text{Mean = } \; E(X) \; \{ \textit{ Expected Value } \} \\. \\ . \qquad \text{Mean = } \; \mu \qquad \{ \textit { mu - pronounced "m-you" } \} \qquad. math

In some places, (including the Classpad Calculator), the symbol for mean is:

math . \qquad \text{Mean } = \bar{x} \qquad. math

The Expected Value of X is the sum of x times p(x)

math \\ . \qquad E(X) = \sum \, x \, p(x) \\. \\ . \qquad E(X) = x_1\,p(x_1) + x_2\,p(x_2) + x_3 \, p(x_3) + \dots \qquad. math


 * Example 1 **

Let X be the number of tails achieved after tossing 3 coins. Find the Expected Value for X.

math \\ . \qquad E(X) = 0 \times \dfrac{1}{8} + 1 \times \dfrac{3}{8} + 2 \times\dfrac{3}{8} + 3 \times \dfrac{1}{8} \qquad. \\ . \\ . \qquad E(X) = \dfrac{0}{8}+\dfrac{3}{8} + \dfrac{6}{8} + \dfrac{3}{8} \\. \\ . \qquad E(X) = 1\dfrac{1}{2} math

NOTE: The Expected Value does __**not**__ have to be one of the possible outcomes

NOTE: The Expected Value gives the average result that is expected over a large number of trials.


 * Expectation Theorems **

For any given value of E(X), the following is true

... .... E(aX) = aE(X), where a is a constant

... ... E(aX + b) = aE(X) + b, where a, b are constants

... ... E(X + Y) = E(X) + E(Y), where X, Y are random variables

... ... E(X 2 ) is not equal to E(X) 2.

NOTE: math \\ . \qquad E(X^2) = \sum \, x^2 \, p(x) \\. \\ . \qquad E(X^2) = (x_1)^2\,p(x_1) + (x_2)^2\,p(x_2) + (x_3)^2 \, p(x_3) + \dots \qquad. math


 * Example 2 **

It is known that for the variable, X, the Expected Value E(X) = 1.5

a) find E(3X) {ie multiply the number of tails by 3} ... ... E(3X) ... ... = 3E(X) ... ... = 3 x 1.5 ... ... = 4.5

b) find E(2X + 4) {ie multiply the number of tails by 2 then add 4} ... ... E(2X + 4) ... ... = 2E(X) + 4 ... ... = 2 x 1.5 + 4 ... ... = 7


 * Example 3 **

Find the expected value the following distribution: math . \qquad p(x) = \dfrac{1}{42} \Big( 5x+3 \Big) \;\; \textit{ where } \; x \in \{ 0,1,2,3 \} \qquad. math

__**Solution**__

math . \qquad p(0) = \dfrac{3}{42}, \;\; p(1) = \dfrac{8}{42}, \;\; p(2)=\dfrac{13}{42}, \;\; p(3) = \dfrac{18}{42} \qquad. math


 * Expected Value **

math \\ . \qquad E(X) = \sum \, x \, p(x) \\. \\ . \qquad E(X) = 0 \times \dfrac{3}{42} + 1 \times \dfrac{8}{42} + 2 \times \dfrac{13}{42} + 3 \times \dfrac{18}{42} \qquad .\\. \\ . \qquad E(X) = \dfrac{0}{42} + \dfrac{8}{42} + \dfrac{26}{42} + \dfrac{54}{42} \\. \\ . \qquad E(X) = \dfrac{88}{42} = \dfrac{44}{21} \\. \\ . \qquad E(X) = 2\dfrac{2}{21} math


 * Median and Mode **


 * NO LONGER IN COURSE **

The ** median ** is the middle value of the distribution.

... ... Pr(x < median) = 0.5

... ... Pr(x > median) = 0.5


 * To find the median**, work from the left, adding up the probabilities until you get to 0.5

If you get to exactly 0.5, the median is the average of that value of x and the next. EG: In the above example, the median is 1.5

The ** mode ** is the most common value of the distribution


 * To find the mode**, choose x such that p(x) is the largest.

Some distributions are bi-modal (2 values have the equal largest probability) Some distributions have no mode ( many or all probabilities are equal)


 * Example 4 **

Find the median and mode for the following distribution: math . \qquad p(x) = \dfrac{1}{42} \Big( 5x+3 \Big) \;\; \textit{ where } \; x \in \{ 0,1,2,3 \} \qquad. math

__**Solution**__

math . \qquad p(0) = \dfrac{3}{42}, \;\; p(1) = \dfrac{8}{42}, \;\; p(2)=\dfrac{13}{42}, \;\; p(3) = \dfrac{18}{42} \qquad. math


 * Median **

math \\ . \qquad p(X \leqslant 0) = \dfrac{3}{42} \qquad. \\ . \\ . \qquad p(X \leqslant 1) = \dfrac{11}{42} \\. \\ . \qquad p(X \leqslant 2) = \dfrac{24}{42} math

Hence the median = 2


 * Mode **

... ... p(3) is largest so

... ... Mode = 3

.