Chapter 16 Random Variables

math2200

Life insurance

• Insurance company: a “death and disability” policy– Pay $10,000 when the client dies– Pay $5,000 if the client is permanently

disabled– Charge $50 per year

• Why $50?– Using actuarial information, the company can

calculate the expected value of the policy.

Random variable

• The amount the company pays out on an individual policy is called a random variable.

• A random variable assumes a value based on the outcome of a random event. – Random variable is often denoted by a capital

letter, e.g., X– A particular value that it can have is often den

oted by the corresponding lower case letter, e.g., x

Random variable

• Discrete– We can list all the outcomes (finite or countabl

e)– E.g. the amount the insurance pays out is eith

er $10,000, $5,000 or $0

• Continuous– any numeric value within a range of values.

• Example: the time you spend from home to school

Probability model

• The collection of all possible values and the probabilities that they occur is called the probability model for the random variable.

Example

• Death rate in any year is 1 out of every 1000 people

• 2 out of 1000 suffer some kind of disability• Probability model

Policyholder outcome

Payment

(x)

Probability (Pr(X=x))

Death 10,000 1/1000

Disability 5,000 2/1000

Neither 0 997/1000

What does the insurance company expect?

• Suppose it insures exactly 1000 people• In a year,

– 1 customer dies– 2 are disabled– The insurance company pays $10,000 +

$5,000*2 = $20,000– Payment per customer: $20,000/1000 = $20– Earnings per customer: $50– Profit : $30 per customer!

Expected value

• $20 is the expected payment per customer

• E(X) = 20=(10000 * 1 + 5000 * 2 + 0*997) / 1000

=10000*(1/1000) + 5000*(2/1000) + 0*(997/1000)

• E(X) = Σx* P(X=x)– Center of the distribution– A parameter of the model

Expected value

• Of particular interest is the value we expect a random variable to take on, notated μ (for population mean) or E(X) for expected value.

• The expected value of a (discrete) random variable can be found by summing the products of each possible value and the probability that it occurs:

• Note: Be sure that every possible outcome is included in the sum and verify that you have a valid probability model to start with.

E X x P x

• Most of the time, the company makes $50 per customer

• But, with small probabilities, the company needs to pay a lot ($10000 or $5000)

• The variation is big

• How to measure the variation?

Spread

• For data, we calculated the standard deviation by first computing the deviation from the mean and squaring it. We do that with discrete random variables as well.

• The variance for a random variable is:

• The standard deviation for a random variable is:

22 Var X x P x

SD X Var X

Variance and standard deviation

Policyholder outcome

Payment (x) Probability Pr(X=x)

Deviation

Death 10,000 1/1000 (10000-20) = 9980

Disability 5,000 2/1000 5000-20 =4980

Neither 0 997/1000 0 -20 = -20

Variance = 99802 (1/1000)+49802 (2/1000)+(-20)2 (997/1000) = 149,600

Standard deviation = square root of variance

Var(X) = Σ[x-E(X)]2 * P(X=x)

SD(X) = $386.78

Properties of expected value and standard deviation

• Shifting– E(X+c) = E(X) + c– Var(X+c) = Var(X)– Example: Consider everyone in a company receiving

a $5000 increase in salary.

• Scaling– E(aX) = aE(X)– Var(aX) = a2 Var(X)– Example: Consider everyone in a company receiving

a 10% increase in salary.

Properties of expected value and standard deviation

• Additivity– E(X ± Y) = E(X) ± E(Y)– If X and Y are independent

• Var(X ± Y) = Var(X) + Var(Y)• SD(X+Y) is NOT SD(X)+SD(Y)

• Suppose the outcomes for two customers are independent, what is the variance for the total payment to these two customers?– Var(X+Y) = Var(X)+Var(Y) = 149600 + 149600 = 2992

00• If one customer is insured twice as much, the va

riance is– Var(2X) = 4Var(X) = 4*149600 = 598400– SD(2X) = 2SD(X)

X+Y and 2X

• Random variables do not simply add up together!– X and Y have the same probability model– But they are not the same random variables– Can NOT be written as X + X

Example

• Sell used Isuzu Trooper and purchase a new Honda motor scooter– Selling Isuzu for a mean of $6940 with a

standard deviation $250– Purchase a new scooter for a mean of $1413

with a standard deviation $11

• How much money do I expect to have after the transaction? What is the standard deviation?

Combining Random Variables (The Bad News)

• It would be nice if we could go directly from models of each random variable to a model for their sum.

• But, the probability model for the sum of two random variables is not necessarily the same as the model we started with even when the variables are independent.

• Thus, even though expected values may add, the probability model itself is different.

• When two independent continuous random variables have Normal models, so does their sum or difference.

• This fact will let us apply our knowledge of Normal probabilities to questions about the sum or difference of independent random variables.

Combining Random Variables (The Good News)

Combining normal random variables

• Example: packaging stereos– Stage 1: packing

• Normal with mean 9min and sd 1.5min

– Stage 2: boxing• Normal with mean 6min and sd 1min

• What is the probability that packing two consecutive systems take over 20 minutes?

• X1: time for packing the first system– mean=9, sd = 1.5

• X2: time for packing the second system• T=X1+X2: total time to pack two systems

– E(T) = E(X1)+E(X2) = 9+9=18– Var(T) = Var(X1)+Var(X2) = 1.52 + 1.52 (assuming ind

ependence)– T is Normal with mean 18 and sd 2.12

• z-score = (20-18)/2.12 = 0.94– P(T>20) = P(Z>0.94) = 0.1736

• What percentage of the stereo systems take longer to pack than to box?– P: time for packing a system– B: time for boxing a system– D=P-B: difference in times to pack and box a

system– The questions is P(D>0)=?– Assuming P and B are independent

• D is still Normal• E(D) = E(P-B) = E(P)-E(B) = 9-6=3• Var(D) = Var(P-B) = Var(P)+Var(B) = 1.52 + 12 =

3.25• SD(D) = 1.80• D is Normal with mean 3 and sd 1.80• P(D>0) = 0.9525• About 95% of all the stereo systems will require

more time for packing than for boxing

• If X is a random variable with expected value E(X)=µ and Y is a random variable with expected value E(Y)=ν, then the covariance of X and Y is defined as

Cov(X,Y)=E((X-µ)(Y- ν))

• The covariance measures how X and Y vary together.

Correlation and Covariance

Some properties of covariance

• Cov(X,Y)=Cov(Y,X)

• Cov(X,X)=Var(X)

• Cov(cX,dY)=c*dCov(X,Y)

• Cov(X,Y) = E(XY)- µν

• If X and Y are independent, Cov(X,Y)=0– The converse is NOT true

• Var(X ± Y) = Var(X) + Var(Y) ± 2Cov(X,Y)

• Covariance, unlike correlation, doesn’t have to be between -1 and 1.

• To fix the “problem” we can divide the covariance by each of the standard deviations to get the correlation:

Correlation and Covariance (cont.)

( , )( , )

X Y

Cov X YCorr X Y

What Can Go Wrong?

• Don’t assume everything’s Normal.– You must Think about whether the Normality

Assumption is justified.

• Watch out for variables that aren’t independent:– You can add expected values for any two

random variables, but – you can only add variances of independent

random variables.

What Can Go Wrong? (cont.)

• Don’t forget: Variances of independent random variables add. Standard deviations don’t.

• Don’t forget: Variances of independent random variables add, even when you’re looking at the difference between them.

• Don’t write independent instances of a random variable with notation that looks like they are the same variables.