4.6 The Gamma Probability Distribution

1 116 Chapter 4. Continuous Variables and Their Probability Distributions (ATTENDANCE 7) The Gamma Probability DistributionThe continuousgammarandom variableYhas densityf(y) ={y 1e y/ ( ),0 y < ,0,elsewhere,where the gammafunctionis defined as ( ) = 0y 1e ydyand its expected value (mean), variance and standard deviation are, =E(Y) = , 2=V(Y) = 2, = V(Y).One important special case of the Gamma , is the continuouschi squarerandom vari-ableYwhere = 2and = 2; in other words, with densityf(y) ={y 22e y22 /2 ( 2),0 y < ,0,elsewhere,and its expected value (mean), variance and standard deviation are, =E(Y) = , 2=V(Y) = 2 , = V(Y).Another important special case of the Gamma , is the continuousexponentialrandomvariableYwher e = 1; in other words, with densityf(y) ={1 e y/ ,0 y < ,0,elsewhere,and its expected value (mean), variance and standard deviation are, =E(Y) = , 2=V(Y) = 2, =.}}}

2 Exercise (The Gamma Probability Distribution ) Distribution .(a) Gamma function8, ( ).8 The gammafunctionis a part of the gammadensity. There is no closed form expression for thegamma function except when is an integer. Consequently, numerical integration is required. Wewill mostly use the calculator to do this 6. The Gamma Probability Distribution (ATTENDANCE 7)117i. ( ) = 1e ydy=(choose one) (i) (ii) (iii) (iv) GAMFUNC ENTER ENTER (again!) ENTERii. ( ) (choose one) (i) (ii) (iii) (iv) GAMFUNC ENTER ENTER (again!) ENTERiii. Notice ( ) = ( 1) ( 1) = ( ) ( ) other words, in general, ( ) = ( 1) ( 1), >1(i)True(ii)Falseiv. (1) = (choose one) (i)0(ii) (iii) (iv) GAMFUNC ENTER ENTER 1 ENTERv. (2) = (2 1) (2 1) = (i)0(ii) (iii) (iv) (3) = 2 (2) = (i)1(ii)2!(iii)3!(iv)4!.vii. (4) = 3 (3) = (i)1(ii)2!(iii)3!(iv)4!.viii. In general, ifnis a positive integer, (n) = (n 1)!

3 (i)True(ii)False(b)Graphs of Gamma the graphs in Figure (4)(5)(6)(a) increasing a, b = 3 (b) increasing a, b = 5 f(y)y020(1)(2)(3) (y)Figure : Gamma densitiesi. Match Gamma density, ( , ), to graph, (1) to (6).( , ) =(1,3)(2,3)(3,3)(1,5)(2,5)(3,5)graph(1)F or ( , ) = (1,3), for example,PRGM GAMGRPH ENTER ENTER 1 ENTER 3 ENTER 20 ENTER ENTER118 Chapter 4. Continuous Variables and Their Probability Distributions (ATTENDANCE 7)ii. As and grow larger, Gamma density becomes (choose one)(i)more symmetric.(ii)more As and grow larger, center (mean) of Gamma density(i)decreases.(ii)remains the same.(iii) As and grow larger, dispersion (variance) of Gamma density(i)decreases.(ii)remains the same.(iii)increases.(c) Gamma Distribution : area under Gamma If ( , ) = (1,3),P(Y < ) =F( ) (choose one)(i) (ii) (iii) (iv) GAMDSTR ENTER ENTER 1 ENTER 3 ENTER ENTERii.

4 If ( , ) = (1,3),P(Y > ) = 1 P(Y ) = 1 F( ) (i) (ii) (iii) (iv) , PRGM GAMDSTR ENTER ENTER 1 ENTER 3 ENTER ENTER then subtract result from 1!iii. If ( , ) = (2,5),P( < Y < ) =P(Y ) P(Y ) (i) (ii) (iii) (iv) PRGM GAMDSTR ENTER ENTER 2 ENTER 5 ENTER ENTER then subtract PRGM GAMDSTR ENTER ENTER 2 ENTER 5 ENTER ENTER(d)Mean, variance and standard deviation of Gamma If ( , ) = (2,5), =E(Y) = = (2)(5) = (choose one)(i)10(ii)11(iii)12(iv) If ( , ) = ( , ), =E(Y) = (choose one)(i) (ii) (iii) (iv) If ( , ) = (2,5), 2=V(Y) = 2= (2)(5)2= (choose one)(i)40(ii)50(iii)60(iv) If ( , ) = ( , ), = 2= ( )( )2 (choose one)(i) (ii) (iii) (iv) Distribution again: time to fix the time,Y, to fix a car isapproximately a Gamma with mean = 2 hours and variance 2= 2 hours2.(a)What are and ?Since = = 2, 2= 2= ( ) = 2 = 2,then =22= 1 and also = =21= (choose one)(i)2(ii)3(iii)4(iv) 6.

5 The Gamma Probability Distribution (ATTENDANCE 7)119(b) In this case, the Gamma density,f(y) ={y 1e y/ ( ),0 y < ,0,elsewhere,is given by (choose one)(i)f(y) ={ye y,0 y < ,0,elsewhere,(ii)f(y) ={ye y (3),0 y < ,0,elsewhere,(iii)f(y) ={y2e y/222 (1),0 y < ,0,elsewhere.(c) What is the chance of waitingat hours?Since ( , ) = (2,1),P(Y < ) =F( ) (choose one)(i) (ii) (iii) (iv) GAMDSTR ENTER ENTER 2 ENTER 1 ENTER ENTER(d)P(Y > ) = 1 P(Y ) = 1 F( ) (choose one)(i) (ii) (iii) (iv) PRGM GAMDSTR ENTER ENTER 2 ENTER 1 ENTER ENTER from one.(e) What is the 90th percentile waiting time; in other words, what is that timesuch that 90% of waiting times are less than this time?IfP(Y < ) = , then (choose one)(i) (ii) (iii) (iv) GAMINV ENTER ENTER 2 ENTER 1 ENTER Distribution : waiting time to McDonalds in Westville,waiting time to order (in minutes),Y, follows a chi square Distribution .}}}}

6 (a) graphs in Figure If = 4, the Probability of waiting less than minutes isP(Y < ) =F( ) (choose one)(i) (ii) (iii) (iv) DISTR 2cdf(0, ,4).ii. If = 10,P( < Y < ) = (choose one)(i) (ii) (iii) (iv) DISTR 2cdf( , ,10).120 Chapter 4. Continuous Variables and Their Probability Distributions (ATTENDANCE 7) (Y < ) = ? ( < Y < ) = ? f(y)f(y)yy2 4 6 8 102 4 6 8 10(a) chi-square with 4 df(b) chi-square with 10 dfFigure : Chi square probabilitiesiii. Chance of waiting timeexactly3 minutes, say, iszero,P(Y= 3) = 0.(i)True(ii)False(b) graphs in Figure (b) Chi-square with 10 df(a) Chi-square with 4 df72nd percentile?72nd percentile? 4 6 8 10 2 4 6 8 10 f(y)f(y)yyFigure : Chi square percentilesi.

7 If = 4 andP(Y < ) = , then (choose one)(i) (ii) (iii) (iv) CHI2 INV ENTER 4 ENTER ENTERii. If = 10 andP(Y < ) = , then (choose one)(i) (ii) (iii) (iv) CHI2 INV ENTER 10 ENTER ENTERiii. The 32nd percentile for a chi-square with = 18 df, is(i) (ii) (iii) (iv) CHI2 INV ENTER 18 ENTER ENTERiv. The 32nd percentile is that waiting time such that 32% of the waitingtimes are less than this waiting time and 68% are more than this time.(i)True(ii) square Distribution 6. The Gamma Probability Distribution (ATTENDANCE 7)121(a) If = 3, the chi square density,f(y) ={y 22e y22 /2 ( 2),0 y < ,0,elsewhere,is given by (choose one)(i)f(y) ={12e y2 (2),0 y < ,0,elsewhere,(ii)f(y) ={12 (2)e y2,0 y < ,0,elsewhere,(iii)f(y) ={y12e y223/2 (32),0 y < ,0,elsewhere,(b) If = 3,P(Y < ) (choose one)(i) (ii) (iii) (iv) DISTR 2cdf(0, ,3).(c) If = 3,P( < Y < ) (choose one)(i) (ii) (iii) (iv) DISTR 2cdf( , ,3).}}}}

8 (d) If = 3, =E(Y) = = (choose one)(i)3(ii)4(iii)5(iv)6.(e) If = 3, 2=V(Y) = 2 = (choose one)(i)3(ii)4(iii)5(iv)6.(f) If = 3 andP(Y < ) = , then (choose one)(i) (ii) (iii) (iv) CHI2 INV ENTER ENTER 3 ENTER ENTER(g) A chi square with = 3 degrees of freedom is a Gamma with parameters( , ) = ( 2,2) = (choose one)(i)(02,2)(ii)(12,2)(iii)(22,2)(iv)(32,2). : waiting time for waiting times for emails followan exponential Distribution ,f(y) ={1 e y/ ,0 y < ,0,elsewhere,122 Chapter 4. Continuous Variables and Their Probability Distributions (ATTENDANCE 7)(a) If = 2, the chance of waiting at most minutes is9P(Y ) =F( ) = y/2dy=[ e y/2] 1 e 12( )=(i) (ii) (iii) (iv) (b) If =13,P(Y ) =F( ) = 3ydy=[ e 3y] 1 e 3( )=(i) (ii) (iii) (iv) (c) If =15,P(Y < ) =F( ) = 1 e 5( ) (circle one) (i) (ii) (iii) (iv) is also possible to use: PRGM EXPDSTR ENTER 1/5 ENTER ENTER(d) If =13, the chance of waiting at least minutesP(Y > ) = 1 F( ) = 1 (1 e (3)( ))=e (3)( ) (i) (ii) (iii) (iv) (e) If =13,P( < Y < ) =F( ) F( ) =e (3)( ) e (3)( ) (i) (ii) (iii) (iv) (f) If =13andP(Y < ) = , then (choose one)(i) (ii) (iii) (iv) EXPINV ENTER ENTER 1/3 ENTER ENTER(g)Expectation and = 2, = = (choose one) (i)2(ii)3(iii)4(iv) =13, = = (choose one) (i)12(ii)13(iii)14(iv) = 2, 2= 2= (choose one) (i)2(ii)3(iii)4(iv) =13, 2= 2= (choose one) (i)13(ii)15(iii)17(iv) again: battery the Distribution of the lifetimeof camera flash batteries,Y, is exponential, with mean (average) lifetime of = =13.}

9 (a) The chance batteries last at least 10 hours isP(Y >10) = 1 F(10) = 1 (1 e 3(10)) =(choose one) (i)e 10(ii)e 20(iii)e 30(iv)e the Gamma and chi-square distributions, it is fairly easy to integrate exponential 7. The Beta Probability Distribution (ATTENDANCE 7)123(b) The chance batteries last at least 15 hours, given that they have alreadylasted at least 5 hours isP(Y >15|Y >5) =P(Y >15, Y >5)P(Y >5)=P(Y >15)P(Y >5)=1 (1 e 3(15))1 (1 e 3(5))=(choose one) (i)e 10(ii)e 20(iii)e 30(iv)e 40.(c) In other words10,P(Y >15|Y >5) =P(Y >10).This is an example of the memoryless property of the exponential.(i)True(ii)False(d)Implicati on of memoryless property of exponential: (Y > s+t|Y > t) =P(Y > s);s, t 0and alsoP(Y > s+t|Y > t) =P(Y > s+t, Y > t)P(Y > t)=P(Y > s+t)P(Y > t)then, combining the last two equations,P(Y > s+t)P(Y > t)=P(Y > s)orP(Y > s+t) =P(Y > t)P(Y > s)ButP(Y > a) =e a , and soe s+t =e s e t (i)True(ii) The Beta Probability DistributionThebetarandom variableY, with parameters >0 and >0, has densityf(y) ={y 1(1 y) 1B( , ),0 y 10,elsewhere,10 The chance a battery lasts at least 10 hours or more, is the same asthe chance a battery lasts atleast 15 hours, given that it has already lasted 5 hours or more.}

10 Thisis kind of surprising, becauseit seems to imply the battery s life starts fresh after 5 hours, asthough the battery forgot aboutthe first five hours of its 4. Continuous Variables and Their Probability Distributions (ATTENDANCE 7)and Distribution function, called anincomplete beta function,F(y) = y0t 1(1 t) 1B( , )dt=Iy( , ),whereB( , ) = 10y 1(1 y) 1dy= ( ) ( ) ( + ),and its expected value (mean), variance and standard deviation are, =E(Y) = + , 2=V(Y) = ( + )2( + + 1), = V(Y).Exercise (The Beta Probability Distribution ) Distribution .(a)Graphs of beta the graphs in Figure (1)(2)(3)(a) increasing a, b = 3 y501(4)(5)(6)(b) increasing a, b = 5 f(y)f(y)3 Figure : Beta densitiesi. Match beta density, ( , ), to graph (1) to (6).(a, b) =(1,3)(2,3)(3,3)(1,5)(2,5)(3,5)graph(1)F or ( , ) = (1,3), for example,PRGM BETAGRPH ENTER ENTER 1 ENTER 3 ENTER 3 ENTERii.