The Big Idea
The expected value (or mean) of a random variable $X$ is the long term average of observed values of $X$.
Example: Roll a fair die. What is the long-term average?
Long-Term Average of Rolling a Fair Die
| Empirical Average | |||
| Number of Rolls |
Calculating the Mean $\mu$ of a Random Variable $X$
$$\mu=\sum_{x \in X} x P(X=x)$$
Example
Calculate the mean value $\mu$ of our fair die.
For tossing a fair die, $\displaystyle P(X=x)=\frac{1}{6}$ for any $x=1,2,3,4,5,6.$
$$
\begin{array}{l}
\mu= \displaystyle \sum\limits_{x \in \{1,2,3,4,5,6\}} x P(X=x)\\
=1 \cdot \frac{1}{6}+2 \cdot \frac{1}{6}+3 \cdot \frac{1}{6}+4 \cdot \frac{1}{6}+ 5 \cdot \frac{1}{6}+ 6 \cdot \frac{1}{6}\\
=3.5
\end{array}
$$
Calculating the Mean $\mu$ of a Random Variable $X$
Compare the how we calculated the mean of $X$ to its PDF table: $$ \mu=1 \cdot \frac{1}{6}+2 \cdot \frac{1}{6}+3 \cdot \frac{1}{6}+4 \cdot \frac{1}{6}+ 5 \cdot \frac{1}{6}+ 6 \cdot \frac{1}{6}=3.5 $$ $$ \begin{array}{|l|l|} \hline x & P(X=x)\\ \hline 1 & \frac{1}{6}\\ \hline 2 & \frac{1}{6} \\ \hline 3 & \frac{1}{6} \\ \hline 4 & \frac{1}{6} \\ \hline 5 & \frac{1}{6} \\ \hline 6 & \frac{1}{6} \\ \hline \end{array} $$
Example: A crooked die has the following distribution. Find $\mu.$ $$ \begin{array}{|l|l|} \hline x & P(X=x)\\ \hline 1 & \frac{1}{12}\\ \hline 2 & \frac{1}{12} \\ \hline 3 & \frac{1}{12} \\ \hline 4 & \frac{1}{6} \\ \hline 5 & \frac{1}{4} \\ \hline 6 & \frac{1}{3} \\ \hline \end{array} $$
Example: A crooked die has the following distribution. Find $\mu.$ $$ \begin{array}{|l|l|} \hline x & P(X=x)\\ \hline 1 & \frac{1}{12}\\ \hline 2 & \frac{1}{12} \\ \hline 3 & \frac{1}{12} \\ \hline 4 & \frac{1}{6} \\ \hline 5 & \frac{1}{4} \\ \hline 6 & \frac{1}{3} \\ \hline \end{array} $$ $$ \mu=1 \cdot \frac{1}{12}+2 \cdot \frac{1}{12}+3 \cdot \frac{1}{12}+4 \cdot \frac{1}{6}+ 5 \cdot \frac{1}{4}+ 6 \cdot \frac{1}{3}=\frac{53}{12}\approx 4.42 $$
Long-Term Average of Our Crooked Die
| Empirical Average | |||
| Number of Rolls |
Example
Sleazy P. Martini offers Billy Bob to play a game which involves selecting a card from a regular 52-card deck and tossing a coin. The coin is a crooked coin with $P(\mbox{Heads})=0.45$ and $P(\mbox{Tails})=0.55.$
• If the card is a face card, and the coin lands on Heads, Billy Bob wins $\$7.$
• If the card is a face card, and the coin lands on Tails, Billy Bob wins $\$2.$
• If the card is not a face card, Billy Bob loses $\$3,$ no matter what the coin shows.
If Billy Bob were to play this game repeatedly, what is the long-term average of Billy Bob's winnings? Round your answer to the nearest cent.
The PDF Table of Sleazy P's Game
$$ \begin{array}{|l|l|} \hline x & P(X=x)\\ \hline +7 & 0.45 \cdot \frac{3}{13}\\ \hline +2 & 0.55 \cdot \frac{3}{13} \\ \hline -3 & \frac{10}{13} \\ \hline \end{array} $$ $$\mu=7 \cdot 0.45 \cdot \frac{3}{13}+ 2 \cdot 0.55 \cdot \frac{3}{13}+(-3) \cdot \frac{10}{13}=-\frac{69}{52}\approx -1.33$$
$$ \begin{array}{|l|l|} \hline x & P(X=x)\\ \hline +7 & 0.45 \cdot \frac{3}{13}\\ \hline +2 & 0.55 \cdot \frac{3}{13} \\ \hline -3 & \frac{10}{13} \\ \hline \end{array} $$ $$\mu=7 \cdot 0.45 \cdot \frac{3}{13}+ 2 \cdot 0.55 \cdot \frac{3}{13}+(-3) \cdot \frac{10}{13}=-\frac{69}{52}\approx -1.33$$
Calculating the Standard Deviation $\sigma$ of a Random Variable $X$
$$\sigma=\sqrt{\sum_{x \in X} (x-\mu)^2 P(X=x)}$$
Example: For tossing a fair die, $P(X=x)=\frac{1}{6}$ for any $x=1,2,3,4,5,6.$ $$ \begin{array}{l} \sigma= \sqrt{\displaystyle \sum\limits_{x \in \{1,2,3,4,5,6\}} (x-3.5)^2 P(X=x)}\\ =\sqrt{(1-3.5)^2\cdot \frac{1}{6}+(2-3.5)^2 \cdot \frac{1}{6}+(3-3.5)^2 \cdot \frac{1}{6}+(4-3.5)^2 \cdot \frac{1}{6}+ (5-3.5)^2 \cdot \frac{1}{6}+ (6-3.5)^2 \cdot \frac{1}{6}}\\ \approx 1.708 \end{array} $$
Compare the how we calculated the mean of $X$ to its PDF table which we modified to include the squared deviations from the mean: $$ \begin{array}{|l|l|l|} \hline x & (x-3.5)^2 &P(X=x)\\ \hline 1 & (1-3.5)^2 &\frac{1}{6}\\ \hline 2 & (2-3.5)^2 &\frac{1}{6} \\ \hline 3 & (3-3.5)^2 &\frac{1}{6} \\ \hline 4 & (4-3.5)^2 &\frac{1}{6} \\ \hline 5 & (5-3.5)^2 &\frac{1}{6} \\ \hline 6 & (6-3.5)^2&\frac{1}{6} \\ \hline \end{array} $$ $$\sigma=\sqrt{(1-3.5)^2\cdot \frac{1}{6}+(2-3.5)^2 \cdot \frac{1}{6}+(3-3.5)^2 \cdot \frac{1}{6}+(4-3.5)^2 \cdot \frac{1}{6}+ (5-3.5)^2 \cdot \frac{1}{6}+ (6-3.5)^2 \cdot \frac{1}{6}}$$
Example: A crooked die has the following distribution. Recall $\mu=\frac{53}{12}$. Find $\sigma.$ $$ \begin{array}{|l|l|} \hline x & P(X=x)\\ \hline 1 & \frac{1}{12}\\ \hline 2 & \frac{1}{12} \\ \hline 3 & \frac{1}{12} \\ \hline 4 & \frac{1}{6} \\ \hline 5 & \frac{1}{4} \\ \hline 6 & \frac{1}{3} \\ \hline \end{array} $$
Recall $\mu=\frac{53}{12}$. Find $\sigma.$
$$
\begin{array}{l}
\sigma= \sqrt{\sum\limits_{x \in \{1,2,3,4,5,6\}} (x-\frac{53}{12})^2 P(X=x)}\\
=\sqrt{(1-\frac{53}{12})^2\cdot \frac{1}{12}+(2-\frac{53}{12})^2 \cdot \frac{1}{12}+(3-\frac{53}{12})^2 \cdot \frac{1}{12}+(4-\frac{53}{12})^2 \cdot \frac{1}{6}+ (5-\frac{53}{12})^2 \cdot \frac{1}{4}+ (6-\frac{53}{12})^2 \cdot \frac{1}{3}}\\
\approx 1.61
\end{array}
$$
Example: The following is our modified PDF table which includes the squared deviations from the mean: $$ \begin{array}{|l|l|l|} \hline x & (x-\frac{53}{12})^2 &P(X=x)\\ \hline 1 & (1-\frac{53}{12})^2 &\frac{1}{12}\\ \hline 2 & (2-\frac{53}{12})^2 &\frac{1}{12} \\ \hline 3 & (3-\frac{53}{12})^2 &\frac{1}{12} \\ \hline 4 & (4-\frac{53}{12})^2 &\frac{1}{6} \\ \hline 5 & (5-\frac{53}{12})^2 &\frac{1}{4} \\ \hline 6 & (6-\frac{53}{12})^2 &\frac{1}{3} \\ \hline \end{array} $$ $$\sigma=\sqrt{\left(1-\frac{53}{12}\right)^2\cdot \frac{1}{12}+\left(2-\frac{53}{12}\right)^2 \cdot \frac{1}{12}+\left(3-\frac{53}{12}\right)^2 \cdot \frac{1}{12}+\left(4-\frac{53}{12}\right)^2 \cdot \frac{1}{6}+ \left(5-\frac{53}{12}\right)^2 \cdot \frac{1}{4}+ \left(6-\frac{53}{12}\right)^2 \cdot \frac{1}{3}}$$
Example
Sleazy P. Martini offers Billy Bob to play a game which involves selecting a card from a regular 52-card deck and tossing a coin. The coin is a crooked coin with $P(\mbox{Heads})=0.45$ and $P(\mbox{Tails})=0.55.$
• If the card is a face card, and the coin lands on Heads, Billy Bob wins $\$7.$
• If the card is a face card, and the coin lands on Tails, Billy Bob wins $\$2.$
• If the card is not a face card, Billy Bob loses $\$3,$ no matter what the coin shows.
If Billy Bob were to play this game repeatedly, what is the long-term standard deviation of Billy Bob's winnings? Round your answer to the nearest cent.
Recall $\displaystyle \mu=-\frac{69}{52}\approx -1.33.$
$$
\begin{array}{|l|l|l|}
\hline
x & (x-(-\frac{69}{52}))^2 & P(X=x)\\ \hline
+7 & (7-(-\frac{69}{52}))^2 & 0.45 \cdot \frac{3}{13}\\ \hline
+2 & (2-(-\frac{69}{52}))^2 & 0.55 \cdot \frac{3}{13} \\ \hline
-3 & (-3-(-\frac{69}{52}))^2 & \frac{10}{13} \\ \hline
\end{array}
$$
$$
\begin{array}{l}
\sigma=\sqrt{(7-(-\frac{69}{52}))^2 \cdot 0.45 \cdot \frac{3}{13}+ (2-(-\frac{69}{52}))^2 \cdot 0.55 \cdot \frac{3}{13}+(-3-(-\frac{69}{52}))^2 \cdot \frac{10}{13}}\\
\approx 3.28
\end{array}
$$
Long-Term Average and Standard Deviation of Billy Bob's "Winnings"
| Empirical Average | |||
| Number of Times Played |
Example: Roulette
A roulette wheel has $38$ numbers: $1$ through $36,$ $0,$ and $00.$ A ball is rolled and it falls into one of the $38$ slots giving a winning number. The payout for betting on even (that is, even numbers $2$ through $36$) is $\$1$ plus the dollar that was bet. Find the expected value of betting on even.
$$
\begin{array}{|l|l|}
\hline
x & P(X=x)\\ \hline
+1 & \frac{18}{38} \\ \hline
-1 & \frac{20}{38} \\ \hline
\end{array}
$$
$$
\mu=1 \cdot \frac{18}{38}+(-1) \cdot \frac{20}{38} \approx -0.053
$$
Long-Term Average Winnings of Betting on Even in Roulette
| Empirical Average | ||||
| Number of Times Played |