1Econ 310Professor WallaceOctober 3, 2013Lecture: Specific continuous probability distributions- Standard normal distribution and standardizing transformation- Student t Distribution (Section 8.4 - skipping this)- Exponential Distribution- Chi-squared distribution- F-distribution2Linear transformations of normal random variables and standardizingIt is useful to be able to take expectations (expected values) and variances of transformations of random variablesConsider the following linear transformations of the random variable X where and are constantsY a X b a b  From our properties of expected value and variance we know that if ( )E Xand 2( )Var X, then E Y a b  2 2 and ( )Var Y aRule: If Yis a linear transformation of a normal random variable X, then Yis also normal.$ Now I can take any normal variable 2~ ( , )X N and transform it into a new variable ~ (0,1)Z NWhat are the values of aand bassociated with this transformation?- We need 0E Z E a X b a b      - We need 2 21Var Z Var a X b a     This gives is two equations in two variables (aand b)(1)2 2 221 1( ) 1 1Var Z a a a        (2)1( ) 0 0E Z b b        Thus, 1 XZ X      is the standardizing transformation.3Example 1: Male height in the US is normally distributed with a mean 69 inches and variance of 36. a. What is the probability of randomly selecting a male from the population with a height between 5’4’’ and 5’8”?5’4’’=64”5’8”=68” 64 69 68 6964 68 ( 0.83 0.17)36 36 ( 0.17) ( 0.83) (0.83) (0.17) 0.7967 0.5675 0.2292P H P Z P ZF F F F                   4b. What is the probability of randomly selecting a male from the population with a height between 5’10’’ and 6’1’’?5’10”=70”6`1”=73” 70 69 73 6970 73 (0.17 0.67)36 36 (0.67) (0.17) 0.7486 0.5675 0.1811P H P Z P ZF F               5Example 2: The natural log of income has a normal distribution with mean 10.35 and variance 2.5. What is the probability that a person selected at random has income below $10,000?We need to go from dollars to log dollarsln(10,000) 9.21 9.21 10.35( $10,000) ln( ) 9.21 0.722.5 ( 0.72) 1 (0.72) 10.7642 0.2358P Y P Y P ZF F               6Exponential Distribution Relationship between exponential and Poisson distributions– if the number of events occurring in an interval of time or space has a Poisson distribution with mean Pthen waiting time or distance between events has an exponential distribution with mean 1 1EP  Exponential PDF: ( )tf t efor 0tExponential CDF: ( ) 1tF t e Survivor function: ( ) 1 ( )tS t F t e  where P is a Poisson mean number of events an interval of time or spaceExpectation and Variance of an exponential random variableExpected value:1( )E TVariance:1( )Var T$ The exponential, binomial, and Poisson distributions are what are known as one parameter distributions because we only need on e parameter to completely characterize the distribution. In contrast the normal distribution is a two parameter distribution because we need to specify both the mean and variance to completely characterize the distribution.7Example 1: The number of job offers received by an unemployed worker has a Poisson distribution. On average he receives an average of 0.25 job offers per week. What is the chance that he goes 1 week without receiving a job offer?0.25(1)( 1) (1) 0.7788P T S e   Note that this is the same probability that I would get from evaluating the Poisson PF at 0. Example 2: Waiting time between job offers for an unemployed worker has an exponential distribution with mean 4 weeks. What is the chance that he that he waits less than 2 week for job offer?0.25(2)( 2) (2) 1 0.3935P T F e    Example 3: Waiting time between job offers for an unemployed worker has an exponential distribution with mean 4 weeks. What is the chance that he that he waits less than a month for a job offers?0.25(4)( 4) (4) 1 0.6321P T F e    Example 4: The number of job offers received by an unemployed worker has a Poisson distribution. On average he receives an average of 0.25 job offers per week. What is the chance that he goes 6 months without receiving a job offer?6 months is roughly 26 weeks0.25(26)( 26) (26) 0.0015P T S e   8Chi-Squared DistributionLet 1 2, ,...,kZ Z Zbe independently distributed standard normal random variables. Then 21kk iiQ Zhas a chi-squared distribution with kdegrees of freedom2( )kQ k . $ Incidentally – if we were just to sum up a bunch of normal random variables the resultant sum would have a normal distribution. Any sum of normal random variables is normal. Expected value and variance of the chi-squared distributionExpected value: ( )kE Q kVariance:( ) 2kVar Q kThings to note about the chi-squared distribution:1. There are infinitely many chi-squared distributions because kcan take on any positive integer value2. Chi-squared random variables only take on positive values3. Unlike the normal distribution there is no standardizing transformation that allows to go from say 2( )kto 2(1). - The chi-squared table in our book (Appendix B, Table 5) provides the values of xsuch that 1 ( )F xis equal to 0.995, 0.990, 0.975, 0.950, 0.900, 0.100, 0.050, 0.025, and 0.001 for degrees of freedom 1 through 30, 40, 50, 60, 70, 80, 90, and 100. - The notation for the critical value associated with the top 100 %Aof the distribution for a random variable that is distributed 2( )kis 2( )Ak. In other if words if 2( )kQ kthen2( )k AP Q k A .9Example 1: Let 2(95)Q. Is the value of 125 in the rightmost 5% of the distribution?Yes - it would have to in the upper 5% of the distribution because 20.05(100) 124 is the cutoff value (critical value) that separates the top 5% and bottom 95% of the distribution for a2(100)random variable. The 5% critical value for a 2(95)random variable will be lower.Example 2: Use a