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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