Ch. 16 Resources

Chapter 16: Random Variables

Like the previous two chapters, the computations in this chapter are fairly elementary and require no new calculator or computer techniques, so we will simply work through a few examples.

Playing the lottery

To play the Washington State Lottery's Daily Game, you pay $1 and then choose three digits (each may be 0 through 9, with repeated digits allowed). If the three digits chosen in the daily drawing match the three digits on your ticket (in the same order) you win $500; otherwise, you win nothing. (There are variations of this game, but let's stick with the basic scenario for this example.)

Let's let the random variable $\displaystyle{X}$ represent your net profit when playing this game one time. The possible values that $\displaystyle{X}$ may take on are $499 (the $500 prize if you win, minus the $1 you paid to play the game) or -$1 (in other words, the dollar you lose when you don't win the game). Let's construct a probability model for this game:

outcome	profit	probability
	$\displaystyle{x}$	$\displaystyle{P}{\left({X}={x}\right)}$
win	$499	0.001
lose	-$1	0.999

To compute the probabilities listed in this table we note that there are 10 possible digits for each of the three positions on the ticket, so the probability of matching any one digit is $\displaystyle\frac{{{1}}}{{{10}}}={0.1}$ . Since each digit is selected independently of the others, we can compute:

$\displaystyle{P}{\left(\text{win}\right)}={P}{\left(\text{match and match and match}\right)}={P}{\left(\text{match}\right)}\times{P}{\left(\text{match}\right)}\times{P}{\left(\text{match}={0.1}\times{0.1}\times{0.1}={0.001}\right.}$

and then note that

$\displaystyle{P}{\left(\text{lose}\right)}={1}-{P}{\left(\text{win}\right)}={1}-{0.001}={0.999}$

Now we compute the mean amount of money you win playing this game:

$mu = E(X) \ = \ sum x cdot P(X=x) \ = \ (499)(0.001) + (-1)(0.999) \ = \ -0.50$

In other words, each time you play the game you lose an average of $0.50. Of course, you never actually lose $0.50: you either lose $1 or gain $499 (-$0.50 is just an average).

Next we compute the standard deviation but (as we will always do) we start with the variance:

$Var(X) \ = \ sum (x-mu)^2 cdot P(X=x) = (499-(-0.5))^2 cdot 0.001 + (-1-(-0.5))^2 cdot 0.999 = 249.75$

Now take the square root of the variance to get the standard deviation:

$\displaystyle\sigma={S}{D}{\left({X}\right)}=\sqrt{{{249.75}}}\approx\${15.80}$

Double or nothing

The Daily Game offers the opportunity to wager amounts other than $1 (there is a minimum bet of $0.50) and the prizes are adjusted accordingly. If you double the amount of your original wager (to $2), then the amount of the payout is also doubled (to $1000). Now the possible profits are $998 and -$2, which are just twice the amounts from the previous problem. Thus it makes sense to call the new random variable for the profit on one play of a $2 ticket $\displaystyle{2}{X}$ .

We could note that the probabilities of winning and losing remain the same and create a new probability distribution for $\displaystyle{2}{X}$ :

outcome	profit	probability
	$\displaystyle{y}$	$\displaystyle{P}{\left({2}{X}={y}\right)}$
win	$998	0.001
lose	-$2	0.999

We could then compute the mean as before:

$\displaystyle\mu={E}{\left({2}{X}\right)}={\left({998}\right)}{\left({0.001}\right)}+{\left(-{2}\right)}{\left({0.999}\right)}=-{1}$

but notice this is the same as:

$\displaystyle{E}{\left({2}{X}\right)}={2}\times{E}{\left({X}\right)}={2}{\left(-\${0.50}\right)}=-${1.00}$

which uses the general formula $\displaystyle{E}{\left({a}{X}\right)}={a}\cdot{E}{\left({X}\right)}$ from page 415 in the text.

For the variance we could compute:

$\displaystyle{V}{a}{r}{\left({2}{X}\right)}={{\left({998}-{\left(-{1}\right)}\right)}}^{{2}}\cdot{0.001}+{{\left(-{2}-{\left(-{1}\right)}\right)}}^{{2}}\cdot{0.999}={999}$

and then take the square root to get:

$\displaystyle\sigma={S}{D}{\left({2}{X}\right)}=\sqrt{{{999}}}\approx\${31.61}$

Notice, however, that we can also use the general formulas $\displaystyle{V}{a}{r}{\left({a}{X}\right)}={{a}}^{{2}}\cdot{V}{a}{r}{\left({X}\right)}$ and $SD(aX) = |a| \cdot SD(X)$ from the text:

$\displaystyle{S}{D}{\left({2}{X}\right)}={2}{S}{D}{\left({X}\right)}={2}\times{15.803}={31.61}$

Two tickets

Now suppose that instead of doubling the bet on a single lottery ticket, you simply buy two $1 tickets. You now have two independent events: whether the first ticket is a winner has no influence on whether or not the second ticket is also a winner (assuming you let the lottery computer choose the numbers on each ticket). By contrast, when we doubled the cost and prize money our profit was either $998 or -$2; with two tickets we could gain $998, gain $498, or lose $2.

Since we now have two random variables, let's call your profit on the first ticket $\displaystyle{X}$ and your profit on the second ticket $\displaystyle{Y}$ . For any single ticket, we already know the expected value and standard deviation:

$\displaystyle{E}{\left({X}\right)}=-\${0.50}$ and $\displaystyle{S}{D}{\left({X}\right)}=\${15.80}$

while

$\displaystyle{E}{\left({Y}\right)}=-\${0.50}$ and $\displaystyle{S}{D}{\left({Y}\right)}=\${15.80}$ .

Now let's define a third random variable: $\displaystyle{T}={X}+{Y}$ , the total profit for the two tickets combined. We could constuct a probability distribution for the new random variable:

outcome	profit	probability
	$\displaystyle{t}$	$\displaystyle{P}{\left({T}={t}\right)}$
win both	$998	0.000001
win one, lose one	$498	0.001998
lose both	-$2	0.998001

To compute the probabilities in the preceding table we note that

$\displaystyle{P}{\left(\text{both win}\right)}={P}{\left(\text{first wins and second wins}\right)}={P}{\left(\text{first wins}\right)}\times{P}{\left(\text{second wins}\right)}={0.001}\times{0.001}={0.000001}$

and

$\displaystyle{P}\text{both lose}{\)}={P}{\left(\text{first loses and second loses}\right)}={P}{\left(\text{first loses}\right)}\times{P}{\left(\text{second loses}\right)}={0.999}\times{0.999}={0.998001}$

and finally deduce that

$\displaystyle{P}{\left(\text{win one, lose one}\right)}={1}-{\left[{0.000001}+{0.998001}\right]}={0.001998}$

using the "Probability Assignment Rule" Rule.

We can then compute the mean:

$\displaystyle\mu={\left({998}\right)}{\left({0.000001}\right)}+{\left({498}\right)}{\left({0.001998}\right)}+{\left(-{2}\right)}{\left({0.998001}\right)}=-{1}$

and the variance:

$\displaystyle{V}{a}{r}{\left({T}\right)}={{\left({998}-{\left(-{1}\right)}\right)}}^{{2}}{\left({0.000001}\right)}+{{\left({498}-{\left(-{1}\right)}\right)}}^{{2}}{\left({0.001998}\right)}+{{\left(-{2}-{\left(-{1}\right)}\right)}}^{{2}}{\left({0.998001}\right)}={499.50}$

and then the standard deviation:

$\displaystyle\sigma={S}{D}{\left({T}\right)}=\sqrt{{{499.5}}}\approx\${23.35}$

While the above computations do lead to the mean and the standard deviation, there is a much easier way to arrive at the same answers, using the general formulas on pages 415 and 416 of the text for the mean:

$\displaystyle\mu={E}{\left({T}\right)}={E}{\left({X}+{Y}\right)}={E}{\left({X}\right)}+{E}{\left({Y}\right)}=-\${0.50}+{\left(-${0.50}\right)}=-${1.00}$

the variance:

$\displaystyle{\sigma}^{{2}}={V}{a}{r}{\left({T}\right)}={V}{a}{r}{\left({X}+{Y}\right)}={V}{a}{r}{\left({X}\right)}+{V}{a}{r}{\left({Y}\right)}={249.75}+{249.75}={499.50}$

and the standard deviation:

$\displaystyle\sigma={S}{D}{\left({T}\right)}=\sqrt{{{V}{a}{r}{\left({T}\right)}}}=\sqrt{{{499.50}}}\approx\${23.35}$

Two players

Finally, suppose that you buy one ticket and your friend buys another ticket; we can now let the random variable $\displaystyle{X}$ represent your profit from placing a single $1 bet and $\displaystyle{Y}$ represent your friend's profit. If you and your friend have a friendly competition to see who has the greater profit, we can consider a new random variable: the difference between your profit and your friend's profit, which we can denote $\displaystyle{D}={X}-{Y}$ . As above we can create a probability distribution listing the possible outcomes:

outcome	difference	probability
	$\displaystyle{d}$	$\displaystyle{P}{\left({D}={d}\right)}$
both win	$0	0.000001
you win, she loses	$500	0.000999
you lose, she wins	-$500	0.000999
both lose	$0	0.998001

You should check that you understand where all of the numbers in table come from. It's not hard to compute the mean, variance and standard deviation using the definitions, as we have in previous cases (you can do this yourself for practice) but it's much quicker to use the appropriate formulas listed on pages 415 and 416 of the text:

$\displaystyle\mu={E}{\left({D}\right)}={E}{\left({X}-{Y}\right)}={E}{\left({X}\right)}-{E}{\left({Y}\right)}=-\${0.50}-{\left(-${0.50}\right)}=${0}$

$\displaystyle{\sigma}^{{2}}={V}{a}{r}{\left({D}\right)}={V}{a}{r}{\left({X}-{Y}\right)}={V}{a}{r}{\left({X}\right)}+{V}{a}{r}{\left({Y}\right)}={249.75}+{249.75}={499.50}$

$\displaystyle\sigma={S}{D}{\left({D}\right)}=\sqrt{{{V}{a}{r}{\left({D}\right)}}}=\sqrt{{{499.50}}}\approx\${23.35}$

Note that we always add variances, even when we're computing the variance of the difference of two independent random variables. It's also important to check, before we use these formulas, that we have two independent random variables.

Homework

Work the following exercises in Chapter 16: 1, 5, 11, 13, 23–29 odd, 37, 39, 41 and 45.

Errata

On page 415, the second equation in the "Double the love" For Example should read:

$\displaystyle{V}{a}{r}{\left({2}{X}\right)}={{2}}^{{2}}{V}{a}{r}{\left({X}\right)}={{2}}^{{2}}\cdot{{8.62}}^{{2}}={297.2176}\rightarrow{S}{D}{\left({2}{X}\right)}=\sqrt{{{292.2176}}}=\${17.24}$

On page 416, the line below the second displayed equation should read "four times as big" (not "twice as big").

The Just Checking problem on page 418 should be #2 (not #1).

The last line of the first paragraph on page 419 should read "50 tyiyn" (not "tylyn").

The TI-83/84 instructions on page 427 should read 1-VarStats L1,L2.

ActivStats

Work the activities on pages 16-1 and 16-2 in the ActivStats lesson book, as time permits.

Additional Resources

Random Variables: Episode 16 from Against All Odds features a discussion of random variables.
Carnegie Mellon: Introduction to Statistics: Module 8 of Carnegie Mellon's open source Introduction to Statistics course covers many of the same ideas.

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