On a certain track team, the runners all take between 4 and 7 minutes to finish a mile.
The probability density function for the length of time it takes a runner to finish a
mile is $f(x)=\frac{4}{21}(x-4.5)^2, 4\leq x \leq 7$. The expected time for a runner on this team
to complete the mile is therefore 6.36 minutes. Suppose 4 runners are randomly selected from the team
to run a relay in which each leg is 1 mile long. How long would we expect it to take them to complete
the relay?
It is often useful to combine random variables to create new
random variables. Consider a multiple choice quiz with 5 prompts, each with 1 correct and 3 incorrect response. If
a student randomly chooses a response for each prompt, their expected score on the quiz is 1.25.
What is the expected mean scored for 10 students who all take the quiz the same way?
What is the expected total score if a student follows the same procedure on two quizzes?
How much is the score likely to vary if the process is repeated?
The answers to these questions are the means and variances of the new random variables.
If the mean and variance of the original random variables are known,
the mean and variance of the new random variables can be found from them using simple properties of random variables referred to here as the
'linearity properties'.
A linear combination of random variables is formed by adding them and multiplying by constants.
Let \(X_1, X_2, \ldots X_n\) be random variables and let \(a_1, a_2, \ldots a_n\) and $b$ be constants. Then
\(b+\sum_{i=1}^na_iX_i\) is a linear combination of \(X_1, X_2, \ldots X_n\).
The Linearity Properties: Expected Values and Variances of Combinations of Random Variables
The "linearity properties" indicate how to find the expected value and variance of a linear combination of random variables.
For random variables \(X_1, X_2, \ldots X_n\) and constants \(a_1, a_2, \ldots a_n\) and $b$:
\[E\left(b+\sum_{i=1}^na_iX_i\right) = b + \sum_{i=1}^na_iE(X_i)\]
Furthermore, if $X_1, X_2, \ldots X_n$ are independent:
\[Var\left(b+\sum_{i=1}^na_iX_i\right) = \sum_{i=1}^na_i^2Var(X_i)\]
These properties apply equally to discrete and continuous random variables.
Example: Let X and Y be independent random variables such that E(X) = 3, E(Y) = 25, Var(X) = 12, Var(Y) = 2, E(Z) = 0, and Var(Z) = 1.
Find the expected values and variances of the following linear combinations of X, Y, and Z:
$\small{4X - 8Z = 4X + (-8)^2Z}$.
A variance is never negative.
Example: Let \(X_1, X_2, \ldots X_n\) be independent random variables such that \(E(X_i) = m\) and \(Var(X_i) = v\).
Find the expected value and variance of \(T = \sum_{i=1}^n4X_i+ 10\).
The variance of a constant is a 0.
\(\small{X_1, X_2, \ldots X_n}\) are independent.
$\small{Var(X_i)=v}$ for each $i$.
Variable names in alphabetical order.
At the HuHot Mongolgian Grill, there are 1-flame, 2-flame, 3-flame, up to 7-flame sauces to choose from (the number of flames indicates how hot the sauce is).
One customer, Reggie, asks two random diners which sauces they put on their dishes and then puts both of these on his. Let $X_1$ and $X_2$
denote the flame ratings for two diners. \(E(X_1) = E(X_2) = 3.39\) and \(Var(X_1) = Var(X_2) = 2.938\).
(See this previous
example.) If the flame rating for Reggie's dish is the mean of the flame ratings for the two sauces he used, a $R$ denotes the flame rating for Reggie's dish
then $R=\frac{X_1+X_2}{2}$.
Use the linearity properties to find the expected value and variance of $R$.
$R$ is a linear combination of $X_1$ and $X_2$, random variables with known expected value and variance.
$X_1$ and $X_2$ are independent since they are chosen at random.
$\small{\begin{array}{lcl}Var(R)&=& Var\left(\frac{X_1+X_2}{2}\right)\\
&=& Var\left(\frac{X_1}{2}+\frac{X_2}{2}\right)\\
&=& Var\left(\frac{X_1}{2}\right)+Var\left(\frac{X_2}{2}\right)\\
&=& \left(\frac 12\right)^2Var(X_1)+\left(\frac 12\right)^2Var(X_2)\\
&=& \frac 14(2.938)+\frac 14(2.938)\\
&=& 1.469\end{array}}$.
The standard deviation of $R$ is $\sigma=\sqrt{1.469}=1.212$.
Reggie's dish with a combination of two random diners' choices of sauce will have an expected flame rating
of 3.39, give or take 1.469 flames.
Vivian decides to try her hand at randomly guessing on all five quiz prompts in class today.
However, the instructor has decided to make the quiz worth 15 points,
with each question worth 3 points. If $X$ denotes the score someone gets by guessing on a
5-point multiple-choice quiz, $E(X) = 1.25$ and $Var(X) = 0.9375$.
Let $V$ denote the score Vivian gets by guessing on this
modified 15-point quiz. $V=3X$. Find $E(V)$ and $Var(V)$.
$V$ is a linear transformation of $X$. By the linearity properties:
The standard devivation is $\sigma=\sqrt{8.4375}=2.9047$. Vivian's expected score, out of 15, is 3.75, give or
take 2.9047 points.
On a certain track team, the runners all take between 4 and 7 minutes to finish a mile.
The expected value of the time for a runner on this team to complete the mile is 6.36 minutes with a variance of 0.275 minutes2.
Suppose 4 runners are randomly selected from the team
to run a relay in which each leg is 1 mile long. How long would we expect it to take them to complete the relay? What is the variance?
Let $R_i$ denote the time it takes for 1 randomly selected track team member to run a mile. $E(X_i) = 6.36$.
Let $T$ denote the time it takes for 4 randomly chosen team members to each run a mile. $T=\sum_{i=1}^4R_i$. Since the runners are randomly
chosen, $R_1, R_2, R_3$ and $R_4$ are independent.
