Waiting Time DistributionsDate: 06/08/97 at 21:13:23 From: Nick Subject: probability I've tried talking to my teacher, but I need another person to explain waiting time distribution tables and how to solve them. Date: 06/09/97 at 12:48:50 From: Doctor Anthony Subject: Re: Probability Waiting time distributions are based on the exponential distribution, which in turn is derived from the Poisson distribution. Whilst the Poisson distribution governs the occurrence of random events in space or time, the intervals between such events is governed by the exponential distribution. Consider a problem such as the emission of radioactive particles. The number of emissions during any particular unit of time is governed by the Poisson distribution with mean L per unit time. We will consider the distribution of time intervals between successive emissions and find the probability that there is a time-interval of length t between successive emissions. We divide t into increments dt in length such that there is a small probability p of the occurrence of an emission during dt. q is the probability of no occurrence of an emission in that interval, where p+q = 1. dt is assumed small enough to make the probability of more than one emission in dt negligible. The probability, denoted by dP, that there are n incremental intervals between successive emissions is given by: dP = q^n*p That is n intervals with no emissions, and in the (n+1)th interval we have an emission. So dP = (1-p)^n*p. Now if there are L emissions per unit time, the mean number of emissions in time dt is L*dt. So using the formula, mean = number of trials times the probability of success at any trial, we get: L*dt = 1 x p So p = L*dt. This gives us dt = p/L. Writing n*dt = t, we also have dt = t/n. Equating expressions for dt we have: p/L = t/n p = Lt/n Then dP = (1 - Lt/n)^n*L*dt dP/dt = [1 - Lt/n]^n*L As n -> infinity, the expression in brackets tends to e^(-Lt) dP/dt = L*e^(-Lt) The righthand side is the exponential distribution, and is the probability density function for the interval between successive events, i.e. it is the probability that the time interval lies between t and t+dt. The mean is given by INT(0 to infinity)[L*e^(-Lt)*t*dt]. Integrating by parts gives Mean = 1/L. This result is what we should expect. If L is the expected number of events in unit time (from the original Poisson distribution), then we should also expect that the mean interval between events is 1/L. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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