5/27/2015

## Overview

In Chapters 12–13, we saw:

• How to decribe random phenomena as trials
• How to use probability rules to describe the outcomes of these trials

In this chapter, we will:

• Describe the long-term behavior of random phenomena using probability distributions
• See how many phenomena can be modeled using several common named distributions

## Background: Insurance

Say a particular insurance company offers a "death and disability" policy with the following payouts:

• $10,000 for death •$5,000 for disability

They charge a $50 premium per year. How profitable do they think they'll be? • We need to know the probability that a given client will be killed or disabled • With this information, we can find the expected value of their product • We can also find the standard deviation, which can tell us about the uncertainty they'll face ## Probability Models Recall that a mathematical model is just a formula that is used to represent the real world • $$Area = Base \times Height$$ • $$Speed = \frac{distance}{time}$$ A statistical model is a mathematical model which accounts for the uncertainty of random events. • $$\hat{y} = b_0 + b_1x$$ A probability distribution is a specific type of statistical model which describes the probability of certain events happening. • $$X \sim N(\mu,\; \sigma)$$ ## Definitions A Random Variable: • A variable whose value is random. • Each time we observe the variable, it is a random trial. • A particular coin flip is a random trial, but flipping a coin is a random variable • We typically denote random variables with capital letters, like $$X$$ and its value using lower-case letters • "$$X=x$$" means "the variable $$X$$ taking the value $$x$$" • $$P(X = x)$$ is the probability that $$X$$ takes the value $$x$$ • We usually use the shorthand $$P(x)$$ ## Insurance: The Variable For our insurance example, we'll denote the amount paid out $$X$$. Suppose that the following is true: • One in one thousand people will be killed in a given year, on average. • Two in one thousand people will be disabled, on average. How do we represent this? Usually with a table • Outcome Payout $$(x)$$ Probability $$P(x)$$ Death$10,000 $$\frac{1}{1000}$$
Disability $5,000 $$\frac{2}{1000}$$ Neither$0 $$\frac{997}{1000}$$

## Discrete vs. Continuous

A discrete random variable:

• A variable whose outcomes we can list
• The payout is discrete, because we can list all three outcomes

A continuous random variable:

• There are too many possibilities to list
• We usually deal with ranges
• Usually measurements are continuous
• Something can weigh 1 kg, 1.1 kg, 1.11 kg, 1.111 kg, etc.
• Technically, there are an infinite number of possible outcomes

## Valid Distributions

The outcomes in a probability distribution make up the sample space, so we need to follow the same rules as in Chapter 12

In the discrete case:

• All of the probabilities need to add up to exactly 1
• Outcomes cannot overlap
• Every probability needs to $$\ge 0$$

In the continuous case:

• The same basic rules apply, but we need calculus to verify them
• For this class, just trust that they're valid

## 14.1 The Expected Value

The expected value is the long-term average outcome or population mean of a random variable.

• If we repeatedly observe it, what's the average?
• If we haven't observed a particular trial yet, what do we expect to happen?

Notation

• $$E(X)$$ or $$\mu$$
• We use $$E(X)$$ if we're just describing the variable
• We use $$\mu$$ if it's the parameter of a model, e.g. in the Normal Distribution

Calculation

• $$E(X) = \sum xP(x)$$

## Insurance: The Expected Value

Outcome Payout $$(x)$$ Probability $$P(x)$$
Death $10,000 $$\frac{1}{1000}$$ Disability$5,000 $$\frac{2}{1000}$$
Neither $0 $$\frac{997}{1000}$$ Finding $$E(X)$$ • $$E(X) = \sum xP(x)$$ • $$E(X) = \10000\left(\frac{1}{1000}\right) + \5000\left(\frac{2}{1000}\right) + \0\left(\frac{997}{1000}\right)$$ • $$E(X) = \frac{\10000}{1000} + \frac{\10000}{1000} = \10 + \10 = \20$$ ## Insurace: Interpreting $$E(X)$$ We found $$E(X) = \20$$. What does this tell us? • For a given customer, the company expects to spend$20
• Remember that they charge $50 for the policy • For each customer, we have an expected profit of$30

Note

• They only ever pay out $10,000,$5,000 or $0 • They'll either lose$9,950 or $4950, or they can keep all$50
• Since most people will not get injured or killed, the larger number of customers who give them pure profit balance out those who cost them thousands
• This is the basis for all insurance/warranty plans

## 14.2 The Standard Deviation

For the insurance example, we had a wide range of outcomes. This means that there's a large amount of uncertainty or variability from customer to customer.

• Just like with the spread of samples, we describe the spread of probability distributions with the standard deviation.
• In samples, we found the variance as the average squared distance from observations to the mean
• The standard deviation was the square root of the variance
• For distributions, we use the expected squared distance from each outcome to the expected value

## The Standard Deviation

Notation:

• We use $$VAR(X)$$ or $$\sigma^2$$ to denote the distribution's variance
• We use $$SD(X)$$ or $$\sigma$$ to denote the distribution's standard devation

Calculation:

• $$VAR(X) = \sigma^2 = \sum (x - \mu)^2P(x) = \sum x^2P(x) - \mu^2$$
• $$SD(X) = \sqrt{VAR(X)}$$ or $$\sigma = \sqrt{\sigma^2}$$

## Insurance: The Standard Deviation

For our insurance company, $$X = \{\10000,\; \5000,\; \0\}$$ and $$E(X) = \20$$. Finding the standard deviation:

• $$\sigma^2 = \sum (x - \mu)^2P(X)$$
• $$\sigma^2 = (10000 - 20)^2\left(\frac{1}{1000}\right) + (5000 - 20)^2\left(\frac{2}{1000}\right) + (0 - 20)^2\left(\frac{997}{1000}\right)$$
• $$\sigma^2 = (9980)^2\left(\frac{1}{1000}\right) + (4980)^2\left(\frac{2}{1000}\right) + (-20)^2\left(\frac{997}{1000}\right)$$
• $$\sigma^2 = (9960040)\left(\frac{1}{1000}\right) + (24800400)\left(\frac{2}{1000}\right) + (400)\left(\frac{997}{1000}\right)$$
• $$\sigma^2 = 99600.4 + 49600.8 + 398.8$$
• $$\sigma^2 = 149600$$
• $$\sigma = \sqrt{149600} = 386.7816$$

## Insurance: The Standard Deviation

What does it tell us that $$\sigma = \386.78$$?

• There's a big difference between paying out thousands or pocketing $50 • While we expect to make$30 per person in the long term, there is a lot of uncertainty about individual customers
• They'll probably make a lot of profit if they insure thousands, but insuring a small number of people is a lot of risk

## In StatCrunch

This isn't an algebra class, so we can use StatCrunch to do the heavy lifting.

• Open a blank data set
• Enter the values of $$X$$ as on column
• Enter the probabilities as another column
• Stat $$\to$$ Calculators $$\to$$ Custom
• Select the columns
• Hit Compute

## Example: Refurbished Computers

Say a company sells custom computers to businesses. A particular client orders two new computers, but they refuse to take refurbished (repaired) computers.

• Someone made an error in stocking, and of the 15 computers in the stock room 4 were refurbished
• Since the company filled the order by randomly picking up computers from the stock room, there was a chance they shipped 0, 1, or 2 refurbished computers
• If the client just gets one refurbished computer, they'll ship it back at the computer company's expense, costing them $100 • If they get two refurbished machines, they'll cancel the entire order, and the computer comany will lose$1000.

## Refurbished Computers: Probabilities

Getting Two New Computers

• There are 11 new machines out of 15
• $$P(\text{first new}) = \frac{11}{15}$$
• $$P(\text{second new } | \text{ first new}) = \frac{10}{14}$$
• $$P(\text{both new}) = P(\text{first new} \text{ AND } \text{second new } | \text{ first new})$$
• $$P(\text{both new}) = P(\text{first new}) \times P(\text{second new } | \text{ first new})$$
• $$P(\text{both new}) = \frac{11}{15} \times \frac{10}{14} = \frac{110}{210} \approx 0.524$$
• There's a 52.4% chance the computer company doesn't lose money

## Refurbished Computers: Probabilities

Getting Two Refurbished Computers

• There are 4 refurbished machines out of 15
• $$P(\text{first refurb.}) = \frac{4}{15}$$
• $$P(\text{second refurb. } | \text{ first refurb.}) = \frac{3}{14}$$
• $$P(\text{both refurb.}) = P(\text{first refurb.} \text{ AND } \text{second refurb. } | \text{ first refurb.})$$
• $$P(\text{both refurb.}) = P(\text{first refub.}) \times P(\text{second refub. } | \text{ first refurb.})$$
• $$P(\text{both refurb.}) = \frac{4}{15}\times\frac{3}{14} = \frac{12}{210} \approx 0.057$$

## Refurbished Computers: The Probability Distribution

Outcome Money Lost $$(X)$$ Probability $$P(x)$$
Both Refurbished $1000 0.057 One New$100 0.419

• $$E(X) = \20$$, so the expected profit was $50 -$20 = $30 • If we lose an additional$5 from each customer, the expected profit is $45 -$20 = $25 • The standard deviation will stay the same ## Multiplying by a Constant In earlier chapters, we saw that the mean and standard deviation were both scaled when multiplying by a constant. The same holds true for random variables. • $$E(aX) = aE(X)$$ • $$VAR(aX) = a^2VAR(X)$$, because we square all the differences from the means • $$SD(aX) = |a|SD(X)$$, again because we are squaring things then taking the square root So what happens if we double all payouts for the insurance company? ## Doubling Payouts Outcome Payout $$(x)$$ Probability $$P(x)$$ Death$20,000 $$\frac{1}{1000}$$
Disability $10,000 $$\frac{2}{1000}$$ Neither$0 $$\frac{997}{1000}$$

Using StatCrunch,

• $$E(X) = \40$$
• $$SD(X) = \773.56$$

## What Happened?

$$E(X) = \40$$

• Because we doubled all payouts, the expected payout has doubled

$$SD(X) = \773.56 = 2\times \386.78$$

• By doubling all payouts, we doubled the differences between the outcomes
• Since the outcomes are further apart, we doubled the range of outcomes
• For any one customer, there is much more uncertainty in how much the company will lose

Instead of looking at a single customer, we'll look at two. Call them Mr. $$X$$. and Mrs. $$Y$$.

Isn't looking at two customers the same as multiplying one customer's payouts by two?

• Not quite.
• Mr. $$X$$ might die, but Mrs. $$Y$$ survives the year unharmed
• Mr. $$X$$ could stay safe, while Mrs. $$Y$$ gets maimed
• We have different rules for adding random variables

If $$X$$ and $$Y$$ are independent,

• $$E(X + Y) = E(X) + E(Y)$$
• $$VAR(X + Y) = VAR(X) + VAR(Y)$$
• $$SD(X + Y) = \sqrt{VAR(X + Y)}$$

So what should the insurance company expect with Mr. $$X$$ and Mrs. $$Y$$?

$$E(X + Y)$$:

• $$E(X + Y) = E(X) + E(Y) = 20 + 20 = 40$$

$$SD(X + Y)$$

• $$VAR(X + Y) = VAR(X) + VAR(Y)$$
• $$VAR(X + Y) = 149600 + 149600 = 299200$$
• $$SD(X + Y) = \sqrt{299200} = \546.99$$

What happened?

• By doubling the number of policies, we've doubled the expected payout (but also the premiums we collect)
• Notice that $$SD(X + Y) < SD(2X)$$
• By insuring multiple people, we spread the risk around between the customers
• Even though the expected payout is the same as offering one policy with twice the coverage, we've reduced the uncertainty involved
• It's the same profit with less uncertainty

## Subtracting Variables

• $$E(X \pm Y) = E(X) \pm E(Y)$$
• $$VAR(X \pm Y) = VAR(X) + VAR(Y)$$
• $$SD(X \pm Y) = \sqrt{VAR(X) + VAR(Y)}$$

Note that, even when we subtract the variables, we always add the variances

## $$X + X \ne 2X$$

Like we saw with the insurance example, adding two random variables with the same distribution is not the same as multiplying one of them by two

• For the insurance, $$2X = \{\0, \10000, \20000\}$$
• For both customers, the sample space includes all possible combinations of $$X$$ added together

We need to be careful with notation

• For a small number of variables, we might use $$X$$, $$Y$$, and $$Z$$
• For more variables, we often number them $$X_1, X_2, \ldots, X_n$$ where $$n$$ is our number of variables

## Multiple Observations

When we observe the same variable multiple times, like having two insurance customers, we can simplify things. For each observation, the mean and variances are the same, so:

• $$E(X_1 + X_2) = E(X_1) + E(X_2) = 2\times E(X)$$
• $$E(X_1 + X_2 + \ldots + X_n) = n \times E(X)$$
• $$VAR(X_1 + X_2) = VAR(X_1) + VAR(X_2) = 2\times VAR(X)$$
• $$VAR(X_1 + X_2 + \ldots + X_n) = n \times VAR(X)$$
• $$SD(X_1 + X_2 + \ldots + X_n) = \sqrt{VAR(X_1 + X_2 + \ldots + X_n)}$$

## Example: Week and Weekend Shifts

Say you're a waiter at a restauarant and a large portion of your income is in tips. During a typical 5-day work week, you make an average of $1200 with a standard deviation of$150. On the weekends, you average $400 in tips with a standard deviation of$70. Let $$X$$ represent the 5-day work week and $$Y$$ represent the weekend.

What do you expect to make for an entire 7-day week?

• $$E(X + Y) = E(X) + E(Y) = \1200 + \400 = \1600$$

What is the standard deviation for the entire week?

• $$VAR(X + Y) = VAR(X) + VAR(Y) = 150^2 + 70^2$$
• $$VAR(X + Y) = 22500 + 4900 = 27400$$
• $$SD(X + Y) = \sqrt{X + Y} = \sqrt{27400} \approx \165.53$$

## Example: Week and Weekend Shifts

Say that you typically make within one standard deviation of the mean. What's a typical weekly salary for you?

• $$E(X + Y) - SD(X + Y) = 1600 - 165.53 = \1434.47$$
• $$E(X + Y) + SD(X + Y) = 1600 + 165.53 = \1765.53$$
• You typically make between $1434.47 and 1765.53 Only on really good weeks do you make more than two standard deviations above the mean. How much do you make in a really good week? • $$E(X + Y) + 2\times SD(X + Y) = 1600 + 2\times 165.53 = 1600 + 331.06$$ • $$E(X + Y) + 2\times SD(X + Y) = \1931.06$$ • Having a really good week means earning at least$1931.06

## Example: Monthly Income

Using the same information from before, let's call weekly income $$W$$. $$E(W) = \1600$$, $$SD(W) = \165.53$$

How much would you expect to make in a month?

• $$E(M) = E(W_1 + W_2 + W_3 + W_4) = 4 \times E(W)$$
• $$E(M) = 4 \times 1600 = \6400$$

What is the standard deviation for a month?

• $$VAR(M) = VAR(W_1 + W_2 + W_3 + W_4) = 4\times VAR(W)$$
• $$VAR(M) = 4 \times 165.53^2 = 4 \times 27400.18 = 109600.7$$
• $$SD(M) = \sqrt{VAR(M)} = \sqrt{109600.7} = 331.06$$

## Summary

• A random variable is a variable with a random outcome
• We describe the behavior of a random variable with a probability distribution
• The expected value or mean of a random variable describe the long-term average
• The variance and standard deviation describe the uncertainty or spread of a distribution
• When we add random variables, we can add the expected values and variances (but not the standard deviation)