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Chapter 16 Random Variables math2200
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Life insurance A life insurance policy: –Pay $10,000 when the client dies –Pay $5,000 if the client is permanently disabled –Charge $50 per year
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Random variable We call the variable X a random variable if the numeric value of X is based on the outcome of a random event. e.g. The amount the company pays out on one policy –Random variable is often denoted by a capital letter, e.g. X, Y and Z. A particular value of the variable is often denoted by the corresponding lower case letter, e.g. x, y and z
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Random variable Discrete –If we can list all the outcomes (finite or countable) e.g. the amount the insurance pays out is either $10,000, $5,000 or $0 Continuous –any numeric value within a range of values. Example: the time you spend from home to school
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Probability model The collection of all possible values and the probabilities that they occur is called the probability model for the random variable.
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Example Death rate :1 out of every 1000 people per year Disability rate: 2 out of 1000 per year Probability model Policyholder outcome Payment (x) Probability (Pr (X=x)) Death10,0001/1000 Disability5,0002/1000 Neither0997/1000
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What does the insurance company expect? 1000 people insured and in a year, –1 dies –2 disabled –pays $10,000 + $5,000*2 = $20,000 –payment per customer: $20,000/1000 = $20 –charge per customer: $50 –profit : $30 per customer!
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Expected value $20 is the expected payment per customer E(X) = 20 =10000 * 1/1000 + 5000 * 2/1000 + 0*997/1000 If X is a discrete random variable
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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.
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How about spread? 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?
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Spread The variance of a random variable is: The standard deviation for a random variable is:
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Variance and standard deviation Policyholder outcome Payment (x)Probability Pr(X=x) Deviation Death10,0001/1000(10000-20) = 9980 Disability5,0002/10005000-20 =4980 Neither0997/10000 -20 = -20 Variance = 9980 2 (1/1000)+4980 2 (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
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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. Rescaling –E(aX) = aE(X) –Var(aX) = a 2 Var(X) Example: Consider everyone in a company receiving a 10% increase in salary.
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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) Suppose the payments for two customers are independent, the variance for the total payment to these two customers Var (X+Y) = Var (X)+ Var (Y) = 149600 + 149600 = 299200 If one customer is insured twice as much, the variance is –Var(2X) = 4Var(X) = 4*149600 = 598400 –SD(2X) = 2SD(X)
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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
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Example :Combine Random Variables 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?
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Combining Random Variables Bad News: the probability model for the sum of two variables is often different from what we start with. Good news: the magical normal model the probability model for the sum of independent Normal random variables is still normal.
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Example: Combining normal random variables packaging stereos –Packing the system Normal with mean 9 min and sd 1.5min –Boxing the system Normal with mean 6 min and sd 1min What is the probability that packing two consecutive systems take over 20 minutes? What percentage of the stereo systems take longer to pack than to box ?
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X1: mean=9, sd = 1.5 X2: mean=9, sd = 1.5 T=X1+X2: total time to pack two systems –E(T) = E(X1)+E(X2) = 9+9=18 –Var(T) = Var(X1)+Var(X2) = 1.5 2 + 1.5 2 = 4.5 (assuming independence) –T is Normal with mean 18 and sd 2.12 –P(T>20) = normalcdf(20,1E99, 18, 2.12) =0.1736
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What percentage of the stereo systems take longer to pack than to box ? –P: time for packing –B: time for boxing –D=P-B: difference in times to pack and box a system –The questions is P(D>0)=? –Assuming P and B are independent
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E(D) = E(P-B) = E(P)-E(B) = 9-6=3 Var(D) = Var(P-B) = Var(P)+Var(B) = 1.5 2 + 1 2 = 3.25 SD(D) = 1.80 D is Normal with mean 3 and sd 1.80 P(D>0) =normalcdf(0,1E99,3,1.80)= 0.9525 About 95% of all the stereo systems will require more time for packing than for boxing
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If E(X)=µ and E(Y)=ν, then the covariance of the random variables 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 (OPTIONAL)
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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)
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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 coefficient: Correlation and Covariance (cont.)
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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.
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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.
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