10 3centre

# Expected Value of Discrete Random Distributions

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Expected Value is another phrase for Mean
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The expected value gives the average result that could be expected to occur over a large number of trials
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## Notation:

… … Mean = $E(X)$ … Expected Value

… … Mean = $\mu$ … … { mu - pronounced "m-you" }

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

… … Mean = $\bar{x}$

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The Expected Value of X is the sum of x times p(x), for each possible x

… … $E(X) = \sum \, x \times p(x)$
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… … $E(X) = x_1\,p(x_1) + x_2\,p(x_2) + x_3 \, p(x_3) + \dots$

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### Example 1

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

Solution

… … The probability distribution for X is given by the table:

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… … $E(X) = \sum \, x \times p(x)$
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… … $E(X) = 0 \times \dfrac{1}{8} + 1 \times \dfrac{3}{8} + 2 \times\dfrac{3}{8} + 3 \times \dfrac{1}{8}$
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… … $E(X) = \dfrac{0}{8}+\dfrac{3}{8} + \dfrac{6}{8} + \dfrac{3}{8}$
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… … $E(X) = \dfrac{3}{2}$

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

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

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## Expected Value Theorems

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For any given value of E(X), the following is true
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… …. $E(aX) = aE(X)$ … … where a is a constant
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… … $E(aX + b) = aE(X) + b$ … … where a, b are constants
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… … $E(X + Y) = E(X) + E(Y)$ … … where X, Y are random variables
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… … $E(X^2)$ is not equal to $\Big(E(X)\Big)^2$ .

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NOTE:

… … $E(X^2) = \sum \, x^2 \times p(x)$
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… … $E(X^2) = (x_1)^2\,p(x_1) + (x_2)^2\,p(x_2) + (x_3)^2 \, p(x_3) + \dots$

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### Example 2

Let X be the number of tails achieved in 3 coin tosses.

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}

b) .. find $E(2X + 4)$ … {ie multiply the number of tails by 2 then add 4}

Solution:

a) .. find $E(3X)$ … {ie multiply the number of tails by 3}

… … $E(3X)$

… … $= 3E(X)$

… … $= 3 \times 1.5$

… … $= 4.5$

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b) .. find $E(2X + 4)$ … {ie multiply the number of tails by 2 then add 4}

… … $E(2X + 4)$

… … $= 2E(X) + 4$

… … $= 2 \times 1.5 + 4$

… … $= 7$

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### Example 3

Find the expected value the following distribution:

… … $p(x) = \dfrac{1}{42} \Big( 5x+3 \Big) \;\; \textit{ where } \; x \in \big\{ 0,\; 1,\; 2,\; 3 \big\}$

Solution

… … $p(0) = \dfrac{3}{42}, \;\; p(1) = \dfrac{8}{42}, \;\; p(2)=\dfrac{13}{42}, \;\; p(3) = \dfrac{18}{42}$
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… … Expected Value
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… … … … $E(X) = \sum \, x \times p(x)$
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… … … … $E(X) = 0 \times \dfrac{3}{42} + 1 \times \dfrac{8}{42} + 2 \times \dfrac{13}{42} + 3 \times \dfrac{18}{42}$
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… … … … $E(X) = \dfrac{0}{42} + \dfrac{8}{42} + \dfrac{26}{42} + \dfrac{54}{42}$
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… … … … $E(X) = \dfrac{88}{42} = \dfrac{44}{21}$
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… … … … $E(X) = 2\dfrac{2}{21}$

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## Median and Mode

NO LONGER IN COURSE

The median is the middle value of the distribution.

… … $Pr(x < median) = 0.5$

… … $Pr(x > median) = 0.5$

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

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… … If you get to exactly 0.5, the median is the average of that value of x and the next.

… … For example: In the above example, the median is 1.5

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The mode is the most common value of the distribution

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

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

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### Example 4

Find the median and mode for the following distribution:

Solution:

… … Median
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… … … … $p(X \leqslant 0) = \dfrac{1}{8} \quad \Rightarrow \quad p(X \leqslant 0) < 0.5$
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… … … … $p(X \leqslant 1) = \dfrac{1}{2} \quad \Rightarrow \quad p(X \leqslant 1) = 0.5$
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… … … … $p(X \geqslant 2) = \dfrac{1}{2} \quad \Rightarrow \quad p(X \geqslant 2) = 0.5$
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… … Hence median is $\dfrac{1 + 2}{2} = \dfrac{3}{2}$

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… … Mode

… … … … $p(X = 1) = P(X = 2) = \dfrac{3}{8}$

… … Hence the distribution is bi-modal with modes of 1 and 2.

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### Example 5

Find the median and mode for the following distribution:

… … $p(x) = \dfrac{1}{42} \Big( 5x+3 \Big) \;\; \textit{ where } \; x \in \big\{ 0,\; 1,\; 2,\; 3 \big\}$

Solution

… … $p(0) = \dfrac{3}{42}, \;\; p(1) = \dfrac{8}{42}, \;\; p(2)=\dfrac{13}{42}, \;\; p(3) = \dfrac{18}{42}$
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… … Median
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… … … … $p(X \leqslant 0) = \dfrac{3}{42} \quad \Rightarrow \quad p(X \leqslant 0) < 0.5$
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… … … … $p(X \leqslant 1) = \dfrac{11}{42} \quad \Rightarrow \quad p(X \leqslant 1) < 0.5$
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… … … … $p(X \leqslant 2) = \dfrac{24}{42} \quad \Rightarrow \quad p(X \leqslant 2) > 0.5$
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… … Hence the median is 2

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… … Mode
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… … … … $p(3) = \dfrac{18}{42}$ is largest
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… … Hence the mode = 3

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