Mathematical Methods for Physics and Engineering - Matematica.NET

30.9 IMPORTANT CONTINUOUS DISTRIBUTIONSIn fact, almost all probability distributions tend towards a Gaussian when thenumbers involved become large – that this should happen is required by thecentral limit theorem, which we discuss in section 30.10.Multiple Gaussian distributionsSuppose X and Y are independent Gaussian-distributed random variables, sothat X ∼ N(µ 1 ,σ1 2)andY ∼ N(µ 2,σ2 2 ). Let us now consider the random variableZ = X + Y . The PDF for this random variable may be found directly using(30.61), but it is easier to use the MGF. From (30.114), the MGFs of X and YareM X (t) =exp ( µ 1 t + 1 2 σ2 1t 2) , M Y (t) =exp ( µ 2 t + 1 2 σ2 2t 2) .Using (30.89), since X and Y are independent RVs, the MGF of Z = X + Y issimply the product of M X (t) andM Y (t). Thus, we haveM Z (t) =M X (t)M Y (t) =exp ( µ 1 t + 1 2 σ2 1t 2) exp ( µ 2 t + 1 2 σ2 2t 2)=exp [ (µ 1 + µ 2 )t + 1 2 (σ2 1 + σ2)t 2 2] ,which we recognise as the MGF for a Gaussian with mean µ 1 + µ 2 and varianceσ 2 1 + σ2 2 . Thus, Z is also Gaussian distributed: Z ∼ N(µ 1 + µ 2 , σ 2 1 + σ2 2 ).A similar calculation may be performed to calculate the PDF of the randomvariable W = X − Y . If we introduce the variable Ỹ = −Y then W = X + Ỹ ,where Ỹ ∼ N(−µ 1 , σ 2 1 ). Thus, using the result above, we find W ∼ N(µ 1 −µ 2 , σ 2 1 + σ2 2 ).◮An executive travels home from her office every evening. Her journey consists of a trainride, followed by a bicycle ride. The time spent on the train is Gaussian distributed withmean 52 minutes and standard deviation 1.8 minutes, while the time for the bicycle journeyis Gaussian distributed with mean 8 minutes and standard deviation 2.6 minutes. Assumingthese two factors are independent, estimate the percentage of occasions on which the wholejourney takes more than 65 minutes.We first define the random variablesX =timespentontrain, Y = time spent on bicycle,so that X ∼ N(52, (1.8) 2 )andY ∼ N(8, (2.6) 2 ). Since X and Y are independent, the totaljourney time T = X + Y is distributed asT ∼ N(52 + 8, (1.8) 2 +(2.6) 2 )=N(60, (3.16) 2 ).The standard variable is thusZ = T − 603.16 ,and the required probability is given by()65 − 60Pr(T >65) = Pr Z> =Pr(Z>1.58) = 1 − 0.943 = 0.057.3.16Thus the total journey time exceeds 65 minutes on 5.7% of occasions. ◭1189