Mathematical Methods for Physics and Engineering - Matematica.NET

PROBABILITYfrom which we obtainM X(t) ′ =λe t e λ(et−1) ,M X(t) ′′ =(λ 2 e 2t + λe t )e λ(et−1) .Thus, the mean and variance of the Poisson distribution are given byE[X] =M ′ X(0) = λ and V [X] =M ′′ X(0) − [M ′ X(0)] 2 = λ.The Poisson approximation to the binomial distributionEarlier we derived the Poisson distribution as the limit of the binomial distributionwhen n →∞and p → 0 in such a way that np = λ remains finite, where λ is themean of the Poisson distribution. It is not surprising, therefore, that the Poissondistribution is a very good approximation to the binomial distribution for largen (≥ 50, say) and small p (≤ 0.1, say). Moreover, it is easier to calculate as itinvolves fewer factorials.◮In a large batch of light bulbs, the probability that a bulb is defective is 0.5%. Forasample of 200 bulbs taken at random, find the approximate probabilities that 0, 1 and 2 ofthe bulbs respectively are defective.Let the random variable X = number of defective bulbs in a sample. This is distributedas X ∼ Bin(200, 0.005), implying that λ = np =1.0. Since n is large and p small, we mayapproximate the distribution as X ∼ Po(1), giving−1 1xPr(X = x) ≈ ex! ,from which we find Pr(X =0)≈ 0.37, Pr(X =1)≈ 0.37, Pr(X =2)≈ 0.18. For comparison,it may be noted that the exact values calculated from the binomial distribution are identicalto those found here to two decimal places. ◭Multiple Poisson distributionsMirroring our discussion of multiple binomial distributions in subsection 30.8.1,let us suppose X and Y are two independent random variables, both of whichare described by Poisson distributions with (in general) different means, so thatX ∼ Po(λ 1 )andY ∼ Po(λ 2 ). Now consider the random variable Z = X + Y .Wemay calculate the probability distribution of Z directly using (30.60), but we mayderive the result much more easily by using the moment generating function (orindeed the probability or cumulant generating functions).Since X and Y are independent RVs, the MGF for Z is simply the product ofthe individual MGFs for X and Y . Thus, from (30.104),M Z (t) =M X (t)M Y (t) =e λ1(et −1) e λ2(et −1) = e (λ1+λ2)(et −1) ,which we recognise as the MGF of Z ∼ Po(λ 1 + λ 2 ). Hence Z is also Poissondistributed and has mean λ 1 + λ 2 . Unfortunately, no such simple result holds forthe difference Z = X − Y of two independent Poisson variates. A closed-form1178