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MTH/STA 561 GAMMA DISTRIBUTION, CHI-SQUARE DISTRIBUTION, AND EXPONENTIAL DISTRIBUTION In this section, we introduce the gamma, chi-square, Erlang, and exponential distributions. Gamma Distribution According to Theorem 1 of the Appendix, it follows at once that Z1 1 ( ) 1 y e y/ dy = 1 0 and thus we have the following de…nition: De…nition 1. The continuous random variable Y is said to have a gamma distribution with parameters and if and only if its density function of Y is de…ned by f (y; ; ) = 1 y ( ) 1 e y/ if > 0, > 0, and y > 0 elsewhere 0 A continuous random variable Y that follows a gamma probability distribution with parameter and is referred to as a gamma random variable with parameters and . The graphs of gamma distribution for …gure below. = 1; 2; 3; and 4 and 1 = 1 are sketched in the Poisson Distribution As a Special Case. If we let be a positive integer (that is, = 1; 2; 3: : : :) and = 1. Then gamma distribution has the form 1 y ( ) f (y; ) = for y 0 and 1 = 0; 1; 2; with parameter y 0. Theorem 1. 1 y e = 1 y ( e y 1)! , which coincides with the form of a Poisson distribution If Y has a gamma distribution with parameters 2 and = E (Y ) = and , then 2 = V ar (Y ) = : Proof Applying Theorem 1 of the Appendix in conjunction with the recursion formula (Proposition 3 ) of the Appendix, it follows that = E (Y ) = Z1 1 y ( ) 1 y y/ e Z1 1 dy = ( ) 0 1 = y/ y e dy 0 +1 1 ( + 1) = ( ) +1 ( )= ( ) and that E Y2 = Z1 1 y2 ( ) 1 y e y/ 1 dy = ( ) 0 = = Z1 +1 y e y/ dy 0 1 +2 ( ) ( + 1) 2 1 ( + 2) = +2 ( ) ( + 1) ( ) : Thus, 2 = E Y2 2 2 = Theorem 2. parameters and + = 2 ( + 1) 2 2 2 = 2 2 2 : The moment-generating function of a gamma random variable Y with is given by mY (t) = Proof 2 1 (1 for t < t) 1 By de…nition, mY (t) = E etY = Z1 ety y 1 1 ( ) y 1 0 = 1 ( ) Z1 e y/ dy = 1 ( ) Z1 0 e y/[ =(1 t)] dy 0 2 y 1 e y(1 t)/ dy In view of Theorem 1 of the Appendix with substituting = (1 1 mY (t) = ( ) 1 1 = (1 t) t) for , it follows that ( ) t for t < 1= Now di¤erentiating mY (t) with respect to t gives = E (Y ) = and E Y 2 d2 = 2 mY (t) dt d mY (t) dt = = (1 1 t) t=0 t=0 ( + 1) 2 (1 2 t) t=0 t=0 = = ( + 1) 2 Hence, 2 =E Y2 2 = ( + 1) 2 2 2 = 2 : Once the parameters and are speci…ed, the gamma distribution will be completely determined as illustrated in the example below. Example 1. In a certain city, the daily consumption of electric power, in millions of kilowatt-hours, is a random variable having a gamma distribution with mean = 4 and variance 2 = 8. What is the probability that on a given day the daily power consumption will exceed 12 millions of kilowatt-hours? 2 Solution. By Theorem 2, it follows that = = 4 and 2 = = 8. This implies that = 2 and = 2. Let Y stand for the daily power consumption. Then the desired probability is given by 1 P (Y > 12) = 2 2 (2) Z1 1 y 2 1 e y/2 dy = 4 12 Z1 ye y/2 dy: 12 By the integration by parts with u = y and dv = e y/2 dy, we obtain du = dy and v = 2e y/2 so that 8 9 Z1 = < 1 b P (Y > 12) = lim 2ye y/2 12 + 2 e y/2 dy ; 4 :b!1 12 1 2b 1 = lim b/2 + 24e 6 + 2 2e y/2 12 b!1 e 4 1 1 = 0 + 24e 6 4 0 e 6 = 28e 6 4 4 = 7e 6 = (7) (0:002479) = 0:017353; 3 where we have lim b!1 2b =0 e b/2 by L’Hôpital’s rule. Erlang distribution As seen in the preceding section, if we let Tr be the random variable representing the waiting time to occurrence of the rth event in a Poisson process with parameter , then the density function of Tr is r (r 1)! f (t; r; ) = tr 1 e t for t > 0 elsewhere 0 which is known as the Erlang distribution. It should be noted that the Erlang distribution is a special case of the gamma distribution with = r and = 1= . Then it follows from Theorems 2 and 3 that the mean and variance of the Erlang random variable are = r 2 and = r 2 and the moment-generating function is mY (t) = (1 1 t= )r for t < Chi-Square Distribution Another special case of the gamma distribution is obtained by letting = =2 and = 2, where is a positive integer. The probability density function so obtained is referred to as the chi-square distribution with degrees of freedom. De…nition 2. Let be a positive integer. The continuous random variable Y is said to have a chi-square distribution with degrees of freedom if and only if Y is a gammadistributed random variable with parameters = =2 and = 2; that is, the density function of Y is de…ned by ( 1 y ( =2) 1 e y/2 if > 0 and y 0 2 =2 ( 2 ) f (y; ) = 0 elsewhere Applying Theorems 1 and 2 with Theorem 3. = =2 and = 2 yields the following results. If Y has a chi-square distribution with = E (Y ) = 2 and 4 degrees of freedom, then = V ar (Y ) = 2 : The moment-generating function of a chi-square random variable Y with degrees of freedom is given by 1 1 mY (t) = for t < : =2 2 (1 2t) Exponential Distribution The gamma distribution with parameters 1 f (y; ) = = 1 and y/ e has the form if > 0 and y elsewhere, 0 0 which is easily seen to be an exponential distribution with parameters = 1= . By virtue of Theorems 1 and 2 with = 1, it follows that the mean, variance of an exponential random variable Y with parameter = 1= are, respectively, given by = E (Y ) = 2 and = 1= = V ar (Y ) = 2 = 1= 2 : and that the moment-generating function is given by mY (t) = or mY (t) = 1 1 for t < t 1 = 1 t= 1 for t < : t Appendix In order to de…ne the gamma probability distribution, we need to study a special function, called the gamma function. It is proved in books on advanced calculus that the integral Z1 1 t e t dt 0 exists for > 0 and that the value of the integral is a positive real number. The integral is called the gamma function of and we formally de…ne the function as follows. De…nition 1. The gamma function of ( )= is de…ned by Z1 0 for > 0. 5 t 1 e t dt In particular, the value of the gamma function at (1) = 1. Proposition 1. Proof = 1 is equal to one as shown below. By de…nition, (1) = Z1 e t dt = lim e b!1 t b 0 = 1: 0 The gamma function may be written in an alternative form as follows. For any Proposition 2. > 0, ( )=2 Z1 t2 1 e t2 dt: 0 Proof Let u = t2 . Then du = 2tdt so that ( )= Z1 u 1 e u du = 0 Z1 1 t2 e t2 (2t) dt = 2 0 Z1 t2 1 e t2 dt: 0 The following proposition presents a recursion formula for evaluating the gamma function. Proposition 3 (Recursion Formula). ( If > 1, then ( )=( 1 Proof If > 1, an integration by parts with u = t 1) t 2 dt and v = e t so that ( ) = lim 1 t b!1 e t b 0 +( 1) Z1 t = lim b!1 1 b eb +( 1) t 0 Repeated application of L’Hôpital’s rule shows that lim b!1 and by de…nition, Z1 t 2 1 b eb e t dt = 0 Hence, ( )=( 1) ( 1). 6 =0 ( 1) : 2 1). and dv = e t dt yields du = 2 e t dt 0 Z1 1) ( e t dt: When = n is a positive integer, repeated application of the recursion formula leads to the following result. Proposition 4. Proof For any positive integer n, (n) = (n 1)!. Repeated application of the recursion formula (Proposition 3 ) gives (n) = = = = = = = = (n (n (n 1) (n 1) 1) (n 2) (n 2) 1) (n 2) (n 3) (n (n (n (n (n 1) (n 1) (n 1) (n 1)!: 2) (n 2) (n 2) (n 3) 3) 3) 3) 2 (2) 2 1 (1) 2 1 The gamma function has been proved handy in evaluating certain improper integrals as demonstrated below. Proposition 5. For any > 0, Z1 e t dt = 1 1 : 0 Proof Let u = t . Then t = u1= and dt = (1= ) u(1= Z1 Z1 1 1 t e dy = u(1= ) 1 e u du = 0 ) 1 du so that 1 : 0 The value of the gamma function at = 12 can be shown to equal to useful in evaluating certain improper integrals. Proposition 6. 1 2 = p p , which is also . By the de…nition of the standard normal distribution, it follows that Z1 Z1 p 1 2 z 2 /2 p e dz = 1 or e z /2 dz = 2 : 2 1 1 p p p 2. Then dt = dz 2 or 2dt = dz so that Let t = z Proof p 2 = Z1 1 e z 2 /2 1 p Z dz = 2 e 1 7 t2 dt which implies that p = Z1 e t2 dy = 2 1 Z1 e t2 dy: 0 because the integrand is an even function. However, it follows from Proposition 2 that 1 2 Z1 e p : =2 t2 dt: 0 Consequently, 1 2 = Another interested result concerning gamma function is presented as follows. For any positive integer n, Proposition 7. n 2 n+1 2 = p (n 2n 1)! 1 : Proof Since either n or n+1 is a positive even integer, we may assume loss of generality that n is a positive even integer. Then n=2 is an positive integer. Repeated application of the recursion formula (Proposition 3 ) gives n 2 n 2 = n = n 2 1 != 2 n 2 1 4 2 n 2 4 2 : 2 2 2 2 1 and n+1 2 n 2 = = n 1 2 1 n 2 n 2 3 3 2 3 1 2 2 3 1 1 2 2 2 2 1 2 : Combining the preceding two equalities in conjunction with Proposition 6 gives n 2 n+1 2 = n 1 n 2 2 (n 1)! = 2n 1 2 n 3 n 2 1 2 = p Now let us review the famous Taylor’s Theorem. 8 2 (n 2n 4 4 3 2 1 2 2 2 2 1)! 1 : 1 2 Lemma 1 (Taylor’s Theorem). If f has continuous derivatives of order up to and including m + 1 in some neighborhood of x = a, then f (x) = m X f (k) (a) k! k=0 where Rm (x) = Zx a)k + Rm (x) (x f (m+1) (u) (x m! u)m du a is the remainder after the mth power of x Proof De…ne a. m X f (k) (u) g (u) = f (x) k! k=0 Then u)k : (x (4.2) f (0) (x) = f (x) f (x) = 0: 0! When g 0 (u) is computed using the product rule, the sum telescopes to just one term g (x) = f (x) 0 g (u) = = m X f (k+1) (u) k=0 m X k=0 k! m X f (k) (u) u) + (x (k 1)! k=1 k (x f (k+1) (u) (x k! k u) + m X1 u)k 1 f (k+1) (u) (x k! u)k f (m+1) (u) (x m! u)m du: k=0 f (m+1) (u) = (x u)m : m! Now with the fundamental theorem for calculus, we have g (a) = [g (x) Zx g (a)] = g 0 (u) du = a Zx a By (4:2), we also see that g (a) = f (x) m X f (k) (a) k! k=0 Hence, f (x) = m X f (k) (a) k=0 The function k! (x a)k + Zx (x f (m+1) (u) (x m! a ( ) de…ned by the integral of the form Z1 tn 1 e t dt 9 a)k : u)m du: is referred to as the incomplete gamma function. The relationship between the incomplete gamma function and the Poisson probability distribution can be found in the following result. Proposition 8. For any positive integer n, Z1 1 (n 1)! n 1 t t e dt = n 1 X k e : k! k=0 Proof Let f (x) = ex , then f (k) (x) = ex for all k. Applying Lemma 1 to f (x) with x = 0, a = and m = n 1, we have n 1 X e 1 = f (0) = k! k=0 = n 1 X e + k! k=0 = k n 1 X k + 1 (n 1)! k e k! k=0 Z0 + eu (n Z0 1 (n 1)! 1)! ( u)n Z ( u)n du 1 u e du tn 1 e t dt 0 where the last equality follows from the substitution of t = 1 (n 1 1)! Z n 1 t n 1 X t e dt = 1 u. Thus, k e k! k=0 0 (4.3) Note now that (n 1)! = (n) = Z1 n 1 t t e dt = 0 Z n 1 t Z1 t e dt + Z tn 1 e t dt 0 tn 1 e t dt which implies that Z1 so that 1 (n 1)! n 1 t t e dt = (n 1)! 0 Z1 n 1 t 1 t e dt = 1 (n 1)! Z tn 1 e t dt: 0 Therefore, it follows from (4:3) and (4:4) that 1 (n 1)! Z1 tn 1 e t dt = n 1 X k=0 10 k e : k! (4.4) By making the change of variable y = t, we obtain the following useful result. Theorem 1. For > 0 and Z1 > 0, 1 y e y/ dy = ( ): 0 Proof Let y = t (or t = y= ). Then dy = dt (or dt = dy= ). Also, y ! 1 as t ! 1 and y ! 0 as t ! 0. Thus, ( )= Z1 t 1 t e dt = 0 Hence, Z1 y 1 e y= 1 dy = 0 Z1 y 1 1 Z1 0 e y/ dy = 0 11 ( ): y 1 e y= dy: