## 4.5 Means and Variances of Random Variables

The probability histograms and density curves that picture the probability distributions of random variables resemble our earlier pictures of distributions of data. In describing data, we moved from graphs to numerical measures such as means and standard deviations. Now we make the same move to expand our descriptions of the distributions of random variables. We can speak of the mean winnings in a game of chance or the standard deviation of the randomly varying number of calls a travel agency receives in an hour. In this section, we learn more about how to compute these descriptive measures and about the laws they obey.

The mean of a random variable

In Chapter 1 (page 24), we learned that the mean is the average of the observations in a sample. Recall that a random variable is a numerical outcome of a random process. Think about repeating the random process many times and recording the resulting values of the random variable. In general, you can think of the mean of a random variable as the average of a very large sample. In the case of discrete random variables, the relative frequencies of the values in the very large sample are the same as their probabilities.

Here is an example for a discrete random variable.

EXAMPLE 4.28 The Tri-State Pick 3 Lottery

Most states and Canadian provinces have government-sponsored lotteries. Here is a simple lottery wager from the Tri-State Pick 3 game that New Hampshire shares with Maine and Vermont. You choose a three-digit number, 000 to 999. The state chooses a three-digit winning number at random and pays you \$500 if your number is chosen.

Because there are 1000 three-digit numbers, you have probability 1/1000 of winning. Taking to be the amount your ticket pays you, the probability distribution of is

 Payoff \$0 \$500 Probability 0.999 0.001

The random process consists of drawing a three-digit number. The population consists of the numbers 000 to 999. Each of these possible outcomes is equally likely in this example. In the setting of sampling in Chapter 3 (page 132), we can view the random process as selecting an SRS of size 1 from the population. The random variable is 500 if the selected number is equal to the one that you chose and is 0 if it is not.

What is your average payoff from many tickets? The ordinary average of the two possible outcomes \$0 and \$500 is \$250, but that makes no sense as the average because \$500 is much less likely than \$0. In the long run, you receive \$500 once in every 1000 tickets and \$0 on the remaining 999 of 1000 tickets. The long-run average payoff is

or 50 cents. That number is the mean of the random variable . (Tickets cost \$1, so in the long run, the state keeps half the money you wager.)

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If you play Tri-State Pick 3 several times, we would, as usual, call the mean of the actual amounts you win . The mean in Example 4.28 is a different quantity—it is the long-run average winnings you expect if you play a very large number of times.

### Question

4.109 Find the mean of the probability distribution.

You toss a fair coin. If the outcome is heads, you win \$5.00; if the outcome is tails, you win nothing. Let be the amount that you win in a single toss of a coin. Find the probability distribution of this random variable and its mean.

Just as probabilities are an idealized description of long-run proportions, the mean of a probability distribution describes the long-run average outcome. We can't call this mean , so we need a different symbol. The common symbol for the mean of a probability distribution is , the Greek letter mu. We used in Chapter 1 for the mean of a Normal distribution, so this is not a new notation. We will often be interested in several random variables, each having a different probability distribution with a different mean.

mean

To remind ourselves that we are talking about the mean of , we often write rather than simply . In Example 4.28, . Notice that, as often happens, the mean is not a possible value of . You will often find the mean of a random variable called the expected value of . This term can be misleading because we don't necessarily expect an observation on X to equal its expected value.

expected value

The mean of any discrete random variable is found just as in Example 4.28. It is not simply an average of the possible outcomes, but a weighted average in which each outcome is weighted by its probability. Because the probabilities add to 1, we have total weight 1 to distribute among the outcomes. An outcome that occurs half the time has probability one-half and gets one-half the weight in calculating the mean. Here is the general definition.

Mean of a Discrete Random Variable

Suppose that is a discrete random variable whose distribution is

 Value of … Probability …

To find the mean of , multiply each possible value by its probability, then add all the products:

EXAMPLE 4.29 The Mean of Equally Likely First digits

If first digits in a set of data all have the same probability, the probability distribution of the first digit is then

 First digit 1 2 3 4 5 6 7 8 9 Probability 1/9 1/9 1/9 1/9 1/9 1/9 1/9 1/9 1/9

The mean of this distribution is

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Suppose that the random digits in Example 4.29 had a different probability distribution. In Case 4.1 (pages 184–185), we described Benford's law as a probability distribution that describes first digits of numbers in many real situations. Let's calculate the mean for Benford's law.

EXAMPLE 4.30 The Mean of First digits That Follow Benford's Law

CASE 4.1

Here is the distribution of the first digit for data that follow Benford's law. We use the letter for this random variable to distinguish it from the one that we studied in Example 4.29. The distribution of is

 First digit 1 2 3 4 5 6 7 8 9 Probability 0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046

The mean of is

The mean reflects the greater probability of smaller first digits under Benford's law than when first digits 1 to 9 are equally likely.

Figure 4.16 locates the means of and on the two probability histograms. Because the discrete uniform distribution of Figure 4.16(a) is symmetric, the mean lies at the center of symmetry. We can't locate the mean of the right-skewed distribution of Figure 4.16(b) by eye—calculation is needed.

Figure 4.16: FIGURE 4.16 Locating the mean of a discrete random variable on the probability histogram: (a) digits between 1 and 9 chosen at random; and (b) digits between 1 and 9 chosen from records that obey Benford's law.

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What about continuous random variables? The probability distribution of a continuous random variable is described by a density curve. Chapter 1 showed how to find the mean of the distribution: it is the point at which the area under the density curve would balance if it were made out of solid material. The mean lies at the center of symmetric density curves such as the Normal curves. Exact calculation of the mean of a distribution with a skewed density curve requires advanced mathematics.24 The idea that the mean is the balance point of the distribution applies to discrete random variables as well, but in the discrete case, we have a formula that gives us this point.

Reminder

mean as balance point, p. 41

Mean and the law of large numbers

With probabilities in hand, we have shown that, for discrete random variables, the mean of the distribution can be determined by computing a weighted average in which each possible value of the random variable is weighted by its probability. For example, in Example 4.30, we found the mean of the first digit of numbers obeying Benford's law is 3.441.

Suppose, however, we are unaware of the probabilities of Benford's law but we still want to determine the mean of the distribution. To do so, we choose an SRS of financial statements and record the first digits of entries known to follow Benford's law. We then calculate the sample mean to estimate the unknown population mean . In the vocabulary of statistics, is referred to as a parameter and is called a statistic. These terms and their definitions are more formally described in Section 5.3 when we introduce the ideas of statistical inference.

It seems reasonable to use to estimate . An SRS should fairly represent the population, so the mean of the sample should be somewhere near the mean of the population. Of course, we don't expect to be exactly equal to , and we realize that if we choose another SRS, the luck of the draw will probably produce a different . How can we control the variability of the sample means? The answer is to increase the sample size. If we keep on adding observations to our random sample, the statistic is guaranteed to get as close as we wish to the parameter and then stay that close. We have the comfort of knowing that if we gather up more financial statements and keep recording more first digits, eventually we will estimate the mean value of the first digit very accurately. This remarkable fact is called the law of large numbers. It is remarkable because it holds for any population, not just for some special class such as Normal distributions.

Law of Large numbers

Draw independent observations at random from any population with finite mean . As the number of observations drawn increases, the mean of the observed values becomes progressively closer to the population mean .

The behavior of is similar to the idea of probability. In the long run, the proportion of outcomes taking any value gets close to the probability of that value, and the average outcome gets close to the distribution mean. Figure 4.1 (page 174) shows how proportions approach probability in one example. Here is an example of how sample means approach the distribution mean.

EXAMPLE 4.31 Applying the Law of Large Numbers

CASE 4.1

With a clipboard, we begin our sampling. The first randomly drawn financial statement entry has an 8 as its first digit. Thus, the initial sample mean is 8. We proceed to select a second financial statement entry, and find the first digit to be 3, so for the mean is now

As this stage, we might be tempted to think that digits are equally likely because we have observed a large and a small digit. The flaw in this thinking is obvious. We are believing that short-run results accurately reflect long-run behavior. With clear mind, we proceed to collect more observations and continue to update the sample mean. Figure 4.17 shows that the sample mean changes as we increase the sample size. Notice that the first point is 8 and the second point is the previously calculated mean of 5.5. More importantly, notice that the mean of the observations gets close to the distribution mean and settles down to that value. The law of large numbers says that this always happens.

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Figure 4.17: FIGURE 4.17 The law of large numbers in action. As we take more observations, the sample mean always approaches the mean (μ) of the population.

### Question

4.110 Use the Law of Large Numbers applet.

The Law of Large Numbers applet animates a graph like Figure 4.17 for rolling dice. Use it to better understand the law of large numbers by making a similar graph.

The mean of a random variable is the average value of the variable in two senses. By its definition, is the average of the possible values, weighted by their probability of occurring. The law of large numbers says that is also the long-run average of many independent observations on the variable. The law of large numbers can be proved mathematically starting from the basic laws of probability.

Thinking about the law of large numbers

The law of large numbers says broadly that the average results of many independent observations are stable and predictable. The gamblers in a casino may win or lose, but the casino will win in the long run because the law of large numbers says what the average outcome of many thousands of bets will be. An insurance company deciding how much to charge for life insurance and a fast-food restaurant deciding how many beef patties to prepare also rely on the fact that averaging over many individuals produces a stable result. It is worth the effort to think a bit more closely about so important a fact.

The “law of small numbers”

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Both the rules of probability and the law of large numbers describe the regular behavior of chance phenomena in the long run. Psychologists have discovered that our intuitive understanding of randomness is quite different from the true laws of chance.25 For example, most people believe in an incorrect “law of small numbers.” That is, we expect even short sequences of random events to show the kind of average behavior that, in fact, appears only in the long run.

Some teachers of statistics begin a course by asking students to toss a coin 50 times and bring the sequence of heads and tails to the next class. The teacher then announces which students just wrote down a random-looking sequence rather than actually tossing a coin. The faked tosses don't have enough “runs” of consecutive heads or consecutive tails. Runs of the same outcome don't look random to us but are, in fact, common. For example, the probability of a run of three or more consecutive heads or tails in just 10 tosses is greater than 0.8.26 The runs of consecutive heads or consecutive tails that appear in real coin tossing (and that are predicted by the mathematics of probability) seem surprising to us. Because we don't expect to see long runs, we may conclude that the coin tosses are not independent or that some influence is disturbing the random behavior of the coin.

EXAMPLE 4.32 The “Hot Hand” in Basketball

Belief in the law of small numbers influences behavior. If a basketball player makes several consecutive shots, both the fans and her teammates believe that she has a “hot hand” and is more likely to make the next shot. This is doubtful.

Careful study suggests that runs of baskets made or missed are no more frequent in basketball than would be expected if each shot were independent of the player's previous shots. Baskets made or missed are just like heads and tails in tossing a coin. (Of course, some players make 30% of their shots in the long run and others make 50%, so a coin-toss model for basketball must allow coins with different probabilities of a head.) Our perception of hot or cold streaks simply shows that we don't perceive random behavior very well.27

Our intuition doesn't do a good job of distinguishing random behavior from systematic influences. This is also true when we look at data. We need statistical inference to supplement exploratory analysis of data because probability calculations can help verify that what we see in the data is more than a random pattern.

How large is a large number?

The law of large numbers says that the actual mean outcome of many trials gets close to the distribution mean as more trials are made. It doesn't say how many trials are needed to guarantee a mean outcome close to . That depends on the variability of the random outcomes. The more variable the outcomes, the more trials are needed to ensure that the mean outcome is close to the distribution mean . Casinos understand this: the outcomes of games of chance are variable enough to hold the interest of gamblers. Only the casino plays often enough to rely on the law of large numbers. Gamblers get entertainment; the casino has a business.

Rules for means

Imagine yourself as a financial adviser who must provide advice to clients regarding how to distribute their assets among different investments such as individual stocks, mutual funds, bonds, and real estate. With data available on all these financial instruments, you are able to gather a variety of insights, such as the proportion of the time a particular stock outperformed the market index, the average performance of the different investments, the consistency or inconsistency of performance of the different investments, and relationships among the investments. In other words, you are seeking measures of probability, mean, standard deviation, and correlation. In general, the discipline of finance relies heavily on a solid understanding of probability and statistics. In the next case, we explore how the concepts of this chapter play a fundamental role in constructing an investment portfolio.

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CASE 4.3

Portfolio Analysis One of the fundamental measures of performance of an investment is its rate of return. For a stock, rate of return of an investment over a time period is basically the percent change in the share price during the time period. However, corporate actions such as dividend payments and stock splits can complicate the calculation. A stock's closing price can be amended to include any distributions and corporate actions to give us an adjusted closing price. The percent change of adjusted closing prices can then serve as a reasonable calculation of return.

For example, the closing adjusted price of the well-known S&P 500 market index was \$1,923.57 for April 2014 and was \$1,960.96 for May 2014. So, the index's monthly rate of return for that time period was

Investors want high positive returns, but they also want safety. Since 2000 to mid-2014, the S&P 500's monthly returns have swung to as low as and to as high as . The variability of returns, called volatility in finance, is a measure of the risk of an investment. A highly volatile stock, which may often go either up or down, is more risky than a Treasury bill, whose return is very predictable.

A portfolio is a collection of investments held by an individual or an institution. Portfolio analysis begins by studying how the risk and return of a portfolio are determined by the risk and return of the individual investments it contains. That's where statistics comes in: the return on an investment over some period of time is a random variable. We are interested in the mean return, and we measure volatility by the standard deviation of returns. Indeed, investment firms will report online the historical mean and standard deviation of returns of individual stocks or funds.28

Suppose that we are interested in building a simple portfolio based on allocating funds into one of two investments. Let's take one of the investments to be the commonly chosen S&P 500 index. The key now is to pick another investment that does not have a high positive correlation with the market index. Investing in two investments that have very high positive correlation with each other is tantamount to investing in just one.

Possible choices against the S&P 500 index are different asset classes like real estate, gold, energy, and utilities. For example, suppose we build a portfolio with 70% of funds invested in the S&P 500 index and 30% in a well-known utilities sector fund (XLU). If is the monthly return on the S&P 500 index and the monthly return on the utilities fund, the portfolio rate of return is

How can we find the mean and standard deviation of the portfolio return starting from information about and ? We must now develop the machinery to do this.

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Think first not about investments but about making refrigerators. You are studying flaws in the painted finish of refrigerators made by your firm. Dimples and paint sags are two kinds of surface flaw. Not all refrigerators have the same number of dimples: many have none, some have one, some two, and so on. You ask for the average number of imperfections on a refrigerator. The inspectors report finding an average of 0.7 dimple and 1.4 sags per refrigerator. How many total imperfections of both kinds (on the average) are there on a refrigerator? That's easy: if the average number of dimples is 0.7 and the average number of sags is 1.4, then counting both gives an average of flaws.

In more formal language, the number of dimples on a refrigerator is a random variable that varies as we inspect one refrigerator after another. We know only that the mean number of dimples is . The number of paint sags is a second random variable having mean . (As usual, the subscripts keep straight which variable we are talking about.) The total number of both dimples and sags is another random variable, the sum . Its mean is the average number of dimples and sags together. It is just the sum of the individual means and . That's an important rule for how means of random variables behave.

Here's another rule. A large lot of plastic coffee-can lids has a mean diameter of 4.2 inches. What is the mean in centimeters? There are 2.54 centimeters in an inch, so the diameter in centimeters of any lid is 2.54 times its diameter in inches. If we multiply every observation by 2.54, we also multiply their average by 2.54. The mean in centimeters must be , or about 10.7 centimeters. More formally, the diameter in inches of a lid chosen at random from the lot is a random variable with mean . The diameter in centimeters is , and this new random variable has mean .

The point of these examples is that means behave like averages. Here are the rules we need.

Rules for Means

• Rule 1. If is a random variable and and are fixed numbers, then
• Rule 2. If and are random variables, then
• Rule 3. If and are random variables, then

EXAMPLE 4.33 Aggregating demand in a Supply Chain

To remain competitive, companies worldwide are increasingly recognizing the need to effectively manage their supply chains. Let us consider a simple but realistic supply chain scenario. ElectroWorks is a company that manufactures and distributes electronic parts to various regions in the United States. To serve the Chicago– Milwaukee region, the company has a warehouse in Milwaukee and another in Chicago. Because the company produces thousands of parts, it is considering an alternative strategy of locating a single, centralized warehouse between the two markets—say, in Kenosha, Wisconsin—that will serve all customer orders. Delivery time, referred to as lead time, from manufacturing to warehouse(s) and ultimately to customers is unaffected by the new strategy.

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To illustrate the implications of the centralized warehouse, let us focus on one specific part: SurgeArrester. The lead time for this part from manufacturing to warehouses is one week. Based on historical data, the lead time demands for the part in each of the markets are Normally distributed with

If the company were to centralize, what would be the mean of the total aggregated lead time demand ? Using Rule 2, we can easily find the mean overall lead time demand is

At this stage, we only have part of the picture on the aggregated demand random variable—namely, its mean value. In Example 4.39 (pages 232–233), we continue our study of aggregated demand to include the variability dimension that, in turn, will reveal operational benefits from the proposed strategy of centralizing. Let's now consider the portfolio scenario of Case 4.3 (page 225) to demonstrate the use of a combination of the mean rules.

EXAMPLE 4.34 Portfolio Analysis

CASE 4.3

The past behavior of the two securities in the portfolio is pictured in Figure 4.18, which plots the monthly returns for S&P 500 market index against the utility sector index from January 2000 to May 2014. We can see that the returns on the two indices have a moderate level of positive correlation. This fact will be used later for gaining a complete assessment of the expected performance of the portfolio. For now, we can calculate mean returns from the 173 data points shown on the plot:29

Figure 4.18: FIGURE 4.18 Monthly returns on S&P 500 index versus returns on Utilities Sector index (January 2000 to May 2014), Example 4.34.

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By combining Rules 1 and 2, we can find the mean return on the portfolio based on a 70/30 mix of S&P index shares and utility shares:

This calculation uses historical data on returns. Next month may, of course, be very different. It is usual in finance to use the term expected return in place of mean return.

### Question

4.111 Find .

The random variable has mean . If , what is ?

### Question

4.112 Find .

The random variable has mean , and the random variable has mean . If , find .

### Question

4.113 Managing a new-product development process.

Managers often have to oversee a series of related activities directed to a desired goal or output. As a new-product development manager, you are responsible for two sequential steps of the product development process—namely, the development of product specifications followed by the design of the manufacturing process. Let be the number of weeks required to complete the development of product specifications, and let be the number of weeks required to complete the design of the manufacturing process. Based on experience, you estimate the following probability distribution for the first step:

 Weeks 1 2 3 Probability 0.3 0.5 0.2

For the second step, your estimated distribution is

 Weeks 1 2 3 4 5 Probability 0.1 0.15 0.4 0.3 0.05
1. Calculate and .
2. The cost per week for the activity of developing product specifications is \$8000, while the cost per week for the activity of designing the manufacturing process is \$30,000. Calculate the mean cost for each step.
3. Calculate the mean completion time and mean cost for the two steps combined.

CASE 4.3

### Question

4.114 Mean return on portfolio.

The addition rule for means extends to sums of any number of random variables. Let's look at a portfolio containing three mutual funds from three different industrial sectors: biotechnology, information services, and defense. The monthly returns on Fidelity Select Biotechnology Fund (FBIOX), Fidelity National Information Services Fund (FIX), and Fidelity Select Defense and Aerospace Fund (FSDAX) for the 60 months ending in July 2014 had approximately these means:30

What is the mean monthly return for a portfolio consisting of 50% biotechnology, 30% information services, and 20% defense and aerospace?

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The variance of a random variable

The mean is a measure of the center of a distribution. Another important characteristic of a distribution is its spread. The variance and the standard deviation are the standard measures of spread that accompany the choice of the mean to measure center. Just as for the mean, we need a distinct symbol to distinguish the variance of a random variable from the variance of a data set. We write the variance of a random variable as . Once again, the subscript reminds us which variable we have in mind. The definition of the variance of a random variable is similar to the definition of the sample variance given in Chapter 1. That is, the variance is an average value of the squared deviation of the variable from its mean .

As for the mean of a discrete random variable, we use a weighted average of these squared deviations based on the probability of each outcome. Calculating this weighted average is straightforward for discrete random variables but requires advanced mathematics in the continuous case. Here is the definition.

Variance of a discrete Random Variable

Suppose that is a discrete random variable whose distribution is

 Value of … Probability …

and that is the mean of . The variance of is

The standard deviation of is the square root of the variance.

EXAMPLE 4.35 Find the Mean and the Variance

CASE 4.2

In Case 4.2 (pages 210–211), we saw that the distribution of the daily demand of transfusion blood bags is

 Bags used 0 1 2 3 4 5 6 Probability 0.202 0.159 0.201 0.125 0.088 0.087 0.056 Bags used 7 8 9 10 11 12 Probability 0.025 0.022 0.018 0.008 0.006 0.003

We can find the mean and variance of by arranging the calculation in the form of a table. Both and are sums of columns in this table.

0 0.202 0.00
1 0.159 0.159
2 0.201 0.402
3 0.125 0.375
4 0.088 0.352
5 0.087 0.435
6 0.056 0.336
7 0.025 0.175
8 0.022 0.176
9 0.018 0.162
10 0.008 0.080
11 0.006 0.066
12 0.003 0.036

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We see that . The standard deviation of is . The standard deviation is a measure of the variability of the daily demand of blood bags. As in the case of distributions for data, the connection of standard deviation to probability is easiest to understand for Normal distributions (for example, 68–95–99.7 rule). For general distributions, we are content to understand that the standard deviation provides us with a basic measure of variability.

Reminder

68-95-99.7 rule, p. 43

### Question

4.115 Managing new-product development process.

Exercise 4.113 (page 228) gives the distribution of time to complete two steps in the new-product development process.

1. Calculate the variance and the standard deviation of the number of weeks to complete the development of product specifications.
2. Calculate and for the design of the manufacturing-process step.

Rules for variances and standard deviations

What are the facts for variances that parallel Rules 1, 2, and 3 for means? The mean of a sum of random variables is always the sum of their means, but this addition rule is true for variances only in special situations. To understand why, take to be the percent of a family's after-tax income that is spent, and take to be the percent that is saved. When increases, decreases by the same amount. Though and may vary widely from year to year, their sum is always 100% and does not vary at all. It is the association between the variables and that prevents their variances from adding.

If random variables are independent, this kind of association between their values is ruled out and their variances do add. As defined earlier for general events and (page 205), two random variables and are independent if knowing that any event involving alone did or did not occur tells us nothing about the occurrence of any event involving alone.

Probability models often assume independence when the random variable outcomes appear unrelated to each other. You should ask in each instance whether the assumption of independence seems reasonable.

correlation

When random variables are not independent, the variance of their sum depends on the correlation between them as well as on their individual variances. In Chapter 2, we met the correlation between two observed variables measured on the same individuals. We defined the correlation (page 75) as an average of the products of the standardized and observations. The correlation between two random variables is defined in the same way, once again using a weighted average with probabilities as weights in the case of discrete random variables. We won't give the details—it is enough to know that the correlation between two random variables has the same basic properties as the correlation calculated from data. We use , the Greek letter rho, for the correlation between two random variables. The correlation is a number between −1 and 1 that measures the direction and strength of the linear relationship between two variables. The correlation between two independent random variables is zero.

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Returning to family finances, if is the percent of a family's after-tax income that is spent and is the percent that is saved, then . This is a perfect linear relationship with a negative slope, so the correlation between and is . With the correlation at hand, we can state the rules for manipulating variances.

Rules for Variances and Standard deviations of Linear Transformations, Sums, and differences

• Rule 1. If is a random variable and and are fixed numbers, then
• Rule 2. If and are independent random variables, then

This is the addition rule for variances of independent random variables.

• Rule 3. If and have correlation , then

This is the general addition rule for variances of random variables.

To find the standard deviation, take the square root of the variance.

Because a variance is the average of squared deviations from the mean, multiplying by constant multiplies by the square of the constant. Adding a constant to a random variable changes its mean but does not change its variability. The variance of is, therefore, the same as the variance of . Because the square of 21 is 1, the addition rule says that the variance of a difference between independent random variables is the sum of the variances. For independent random variables, the difference is more variable than either or alone because variations in both and contribute to variation in their difference.

As with data, we prefer the standard deviation to the variance as a measure of the variability of a random variable. Rule 2 for variances implies that standard deviations of independent random variables do not add. To work with standard deviations, use the rules for variances rather than trying to remember separate rules for standard deviations. For example, the standard deviations of and are both equal to because this is the square root of the variance .

EXAMPLE 4.36 Payoff in the Tri-State Pick 3 Lottery

The payoff of a \$1 ticket in the Tri-State Pick 3 game is \$500 with probability 1/1000 and 0 the rest of the time. Here is the combined calculation of mean and variance:

 0 0.999 0 500 0.001 0.5

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The mean payoff is 50 cents. The standard deviation is . It is usual for games of chance to have large standard deviations because large variability makes gambling exciting.

If you buy a Pick 3 ticket, your winnings are because the dollar you paid for the ticket must be subtracted from the payoff. Let's find the mean and variance for this random variable.

EXAMPLE 4.37 Winnings in the Tri-State Pick 3 Lottery

By the rules for means, the mean amount you win is

That is, you lose an average of 50 cents on a ticket. The rules for variances remind us that the variance and standard deviation of the winnings are the same as those of . Subtracting a fixed number changes the mean but not the variance.

Suppose now that you buy a \$1 ticket on each of two different days. The payoffs and on the two tickets are independent because separate drawings are held each day. Your total payoff is . Let's find the mean and standard deviation for this payoff.

EXAMPLE 4.38 Two Tickets

The mean for the payoff for the two tickets is

Because and are independent, the variance of is

The standard deviation of the total payoff is

This is not the same as the sum of the individual standard deviations, which is . Variances of independent random variables add; standard deviations generally do not.

When we add random variables that are correlated, we need to use the correlation for the calculation of the variance, but not for the calculation of the mean. Here are two examples.

EXAMPLE 4.39 Aggregating demand in a Supply Chain

In Example 4.33, we learned that the lead time demands for SurgeArresters in two markets are Normally distributed with

Based on the given means, we found that the mean aggregated demand is 3104. The variance and standard deviation of the aggregated cannot be computed from the information given so far. Not surprisingly, demands in the two markets are not independent because of the proximity of the regions. Therefore, Rule 2 for variances does not apply. We need to know , the correlation between and , to apply Rule 3. Historically, the correlation between Milwaukee demand and Chicago demand is about . To find the variance of the overall demand, we use Rule 3:

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The variance of the sum is greater than the sum of the variances because of the positive correlation between the two markets. We find the standard deviation from the variance,

Notice that even though the variance of the sum is greater than the sum of the variances, the standard deviation of the sum is less than the sum of the standard deviations. Here lies the potential benefit of a centralized warehouse. To protect against stockouts, ElectroWorks maintains safety stock for a given product at each warehouse. Safety stock is extra stock in hand over and above the mean demand. For example, if ElectroWorks has a policy of holding two standard deviations of safety stock, then the amount of safety stock (rounded to the nearest integer) at warehouses would be

Location Safety Stock
Milwaukee warehouse
Chicago warehouse
Centralized warehouse

The combined safety stock for the Milwaukee and Chicago warehouses is 640 units, which is 40 more units required than if distribution was operated out of a centralized warehouse. Now imagine the implication for safety stock when you take into consideration not just one part but thousands of parts that need to be stored.

This example illustrates the important supply chain concept known as risk pooling. Many companies such as Walmart and e-commerce retailer Amazon take advantage of the benefits of risk pooling as illustrated by this example.

risk pooling

EXAMPLE 4.40 Portfolio Analysis

CASE 4.3 Now we can complete our initial analysis of the portfolio constructed on a 70/30 mix of S&P 500 index shares and utility sector shares. Based on monthly returns between 2000 and 2014, we have

In Example 4.34 (pages 227–228), we found that the mean return is 0.411%. To find the variance of the portfolio return, combine Rules 1 and 3:

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We see that portfolio has a smaller mean return than investing all in the utility index. However, what is gained is that the portfolio has less variability (or volatility) than investing all in one or the other index.

Example 4.40 illustrates the first step in modern finance, using the mean and standard deviation to describe the behavior of a portfolio. We illustrated a particular mix (70/30), but what is needed is an exploration of different combinations to seek the best construction of the portfolio.

EXAMPLE 4.41 Portfolio Analysis

CASE 4.3 By doing the mean computations of Example 4.34 (pages 227–228) and the standard deviation computations of Example 4.40 for different mixes, we find the following values.

S&P 500 proportion
0.0 0.675 4.403
0.1 0.637 4.201
0.2 0.600 4.038
0.3 0.562 3.919
0.4 0.524 3.848
0.5 0.487 3.828
0.6 0.449 3.860
0.7 0.411 3.942
0.8 0.373 4.071
0.9 0.336 4.243
1.0 0.298 4.453

From Figure 4.19, we see that the plot of the portfolio mean returns against the corresponding standard deviations forms a parabola. The point on the parabola where the portfolio standard deviation is lowest is the minimum variance portfolio (MVP). From the preceding table, we see that the MVP is somewhere near a 50/50 allocation between the two investments. The solid curve of the parabola provides the preferable options in that the expected return is, for a given level of risk, higher than the dashed line option.

minimum variance portfolio

Figure 4.19: FIGURE 4.19 Mean return of portfolio versus standard deviation of portfolio, Example 4.41.

235

### Question

4.116 Comparing sales.

Tamara and Derek are sales associates in a large electronics and appliance store. Their store tracks each associate's daily sales in dollars. Tamara's sales total varies from day to day with mean and standard deviation

Derek's sales total also varies, with

Because the store is large and Tamara and Derek work in different departments, we might assume that their daily sales totals vary independently of each other. What are the mean and standard deviation of the difference between Tamara's daily sales and Derek's daily sales? Tamara sells more on the average. Do you think she sells more every day? Why?

### Question

4.117 Comparing sales.

It is unlikely that the daily sales of Tamara and Derek in the previous problem are uncorrelated. They will both sell more during the weekends, for example. Suppose that the correlation between their sales is . Now what are the mean and standard deviation of the difference ? Can you explain conceptually why positive correlation between two variables reduces the variability of the difference between them?

### Question

4.118 Managing new-product development process.

Exercise 4.113 (page 228) gives the distributions of , the number of weeks to complete the development of product specifications, and , the number of weeks to complete the design of the manufacturing process. You did some useful variance calculations in Exercise 4.115 (page 230). The cost per week for developing product specifications is \$8000, while the cost per week for designing the manufacturing process is \$30,000.

1. Calculate the standard deviation of the cost for each of the two activities using Rule 1 for variances (page 231).
2. Assuming the activity times are independent, calculate the standard deviation for the total cost of both activities combined.
3. Assuming , calculate the standard deviation for the total cost of both activities combined.
4. Assuming , calculate the standard deviation for the total cost of both activities combined. How does this compare with your result in part (b)? In part (c)?
5. Assuming , calculate the standard deviation for the total cost of both activities combined. How does this compare with your result in part (b)? In part (c)? In part (d)?