These notes will supplement the textbook not replace what is there. defined for α >0

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1 Gamma Distribution These notes will supplement the textbook not replace what is there. Gamma Function ( ) = x 0 e dx 1 x defined for >0 Properties of the Gamma Function 1. For any >1 () = ( 1)( 1) Proof 1 x 1 -x - -x ( ) = x e dx = Lim x e + ( -1)x e dx 0 A = Lim A 1 A - -x + = ( ) ( -1)x e dx ( -1) 1 A x e 0 1 -x ( let u = x dv = e dx) Use integration by parts - -x du = ( -1)x dx v = -e The first term in the expression on the second line above has limit zero by application of L'Hospital' s rule.the result then follows by application of the definition of the Gamma function.. For any positive integer n (n) = (n 1)! Proof A 0 0 ( n) = ( n 1) ( n ) = ( n 1)( n ) ( n 3) =... = ( n 1)! A

2 3. 1 = π Pr oof let u = x then du = x dx and x = u 1/ -1/ 1 1 = = / x u x e dx 0 e du Now let w = u dw = du and u = w / Then 1 1 = w / = π e dw ( Why?) = π Gamma Distribution 1 1 fx (, β, ) = x e β ( ) x/ β x>0 Moments Let u = x / β du = dx / β n 1 n+ 1 x/ β 1 n n u ) x e u e du β ( ) β ( ) β E(X = dx = 0 0 n β ( n + ) = ( ) When n = 1 β( +1) β( ) µ = E(X) = = = β ( ) ( ) When n= β ( +) β ( + 1) ( ) E(X ) = = = β ( + 1) ( ) ( ) σ = β ( + 1) ( β) = β When n = 0 this corresponds to integrating the density function and the result is 1. Hence the Gamma distribution is indeed a bonafide pdf with mean β and variance β

3 Gamma and Weibull Distribution Examples 1. On average there is one attempt to dial into a university computer system every two minutes. What is the probability that : A. The time until the first dialup is less than.5 minutes? B. The time until the tenth dialup is between 18 and 38 minutes? C. There are at least 10 dialups during the first 5 minutes.? H*A*L à ExpA- x E â x 38 1 H*B*Là 18 H 10 * 9!L x 9 ExpA- x E â x ã ã 9 N@%D H*C*L1 - SumAExp@-1.5D*1.5 k k!, 8k, 0, 9<E

4 << Statistics DiscreteDistributions << Statistics ContinuousDistributions H*A*L ex = ExponentialDistribution@.5D ExponentialDistributionH0.5L CDF@ex,.5D H*B*L gam = GammaDistribution@10, D ChiSquareDistributionH0L CDF@gam, 38D - CDF@gam, 18D -Q H10, 0, 9L + QH10, 0, 19L N@%D H*C*Lpoi = PoissonDistribution@1.5D PoissonDistributionH1.5L 1 - CDF@poi, 9D Using Minitab 1A. MTB > CDF.5; SUBC> Exponential. Cumulative Distribution Function Exponential with mean = x P( X <= x) B. MTB > CDF 38 c1; SUBC> Gamma 10. MTB > CDF 18 c; SUBC> Gamma 10. MTB > let c3=c1-c MTB > print c3 Data Display C C.MTB > let c4=1

5 MTB > CDF 9 c5; SUBC> Poisson.5. MTB > let c4=1 MTB > cdf 9 c5; SUBC> poisson 1.5. MTB > let c6=c4-c5 MTB > print c6 Data Display C Special Cases of the Gamma Distribution 1. If = n β=1 Erlang. If =1 β = 1/ Exponential 3. If =ν/ β= Chi Square The exponential distribution has mean and standard deviation 1/. Why? Connection between the Exponential and the Poisson Distribution Let N(t) represent the number of happenings by time t for a Poisson process with mean t There are happenings per unit time. Let F(t) be the CDF for the time until the first happening. PNt [ ( ) = 0] = PT [ > t] = 1 PT [ t] = 1 Ft () = e t F(t) = 1 e t > 0 f(t) = e t, t > 0 t, t >0

6 Connections between the Gamma and Poisson Distribution with mean 1/. Let N(t) be as before and let T be the time until the n th happening. n 1 t PNt [ ( ) < n] = e / k! = P( T > t) = 1 G( t) k = 0 k k n 1 t te t Gt () = 1 e / k! = 1 e e... k = 0! 1 te 1 t te gt () = e + e + e 3! Terms keep cancelling until you obtain e g(t) = t n-1 n ( n 1)!, t>0 Illustration of Lack of Memory Property of Exponential Distribution The time to failure in thousands of hours of a machine part has pdf ft () = 1 3 e t / 3, t >0 A. What is the mean time to failure? The mean time to failure is 3000 hours. B. What is the probability the part lasts 3000 hours? 3 e 1 PT [ 3] = dt = e = C. If the part lasted 3000 hours what is the probability it lasts 6000 hours? Using the lack of memory property the answer is also.368. D. What is the median time to failure? 3 1 e =. 5 e 3 =. 5 t / 3 = ln. 5 t = 3ln. 5 =. 07 t = 070 hours

7 Weibull Distribution fx (, β, ) = β 1 ( x/ ) x e x > 0 β 0 otherwise Moments and Proof that Weibull is bonafide PDF 1 x Let u = (x / β). Then du = dx β EX n + n 1 x ( x/ β) n n/ n n = e dx = β u du= β + β When n = 0 you are integrating the density function to obtain 1. When n = 1 EX = β ( 1+ 1/ ) When n = EX = β ( 1+ / ) so σ = β ( ( 1+ / ) ( 1+ 1/ ) ).. A Weibull distribution for fan life has =1.053 and β=6710 hours. Find A. The median life. B. The mean life. C. Standard deviation of life. D. Fraction failing on 8000 hour warranty

8 H*A*L a_, b_d = 1 - Exp@-Ht bl a D 1 -ã -J t b Na Solve@F@t, 1.053, 6710D ==.5, td Solve::ifun : Inverse functions are being used by Solve, so some solutions may not be found. 88t << H*B*Lf@t_, a_, b_d = D@F@t, a, bd, t D ã -J t b Na a J t b Na-1 b m= à0 t * f@t, 1.053, 6710D â t H*C*L s= " Ù 0 t *f@t, 1.053, 6710D â t -m H*D*L F@8000, 1.053, 6710D H*A*Lwe = WeibullDistribution@1.053, 6710D WeibullDistributionH1.053, 6710L Quantile@we,.5D H*B*LMean@weD H*C*LStandardDeviation@weD CDF@we, 8000D

9 A. MTB > InvCDF.5; SUBC> Weibull Inverse Cumulative Distribution Function Weibull with first shape parameter = and second = P( X <= x) x E+04 D. MTB > CDF 8000; SUBC> Weibull Cumulative Distribution Function Weibull with first shape parameter = and second = x P( X <= x) 8.00E Let F(t) be the CDF of the time to failure, f(t) the CDF. The reliability R(t) = 1 F(t). P[Failure in [t,t + t] Survival to time t] Failure Rate = Lim t 0 t Ft ( + t) -F(t) ft ( ) = Lim = = Zt () t 0 R(t) t Rt () For the Weibull distribution 1 (/ t β) ( / β ) t e 1 Zt () = = ( / β ) t (/ t β) e If > 1 Z(t) is increasing.ifr(as part gets older it is more likely to fail) If = 1 Z(t) is constant CFR exponential distribution.. (Probability of failure independent of age) If < 1 Z(t) is decreasing DFR. (As part gets older it is less likely to fail.)