binomial and poisson

Binomial and Poisson PhL 3.5.03 1. The 1D drunk. This is outlined in Reif's brown book on stat mech. The drunk makes N steps, R of them to the right and L of them to the left. What is the probability that he does R steps to the right, assuming that the probability of a step to the right is r and to the left is = 1-r. First, here is the answer: P(R) = rR N-R The binomial coefficient factor is the number of ways of doing R steps to the right out of N, that certainly seems clear. For each step to the right, we get a factor of r. But for each step to the left, we get a probability of . Each step is a random independent event, so we are supposed to multiply probabilities in this way. So this result seems pretty reasonable. By the way, if he takes R steps to the right, his distance from the start is R*dx - L*dx = dx(R-L) which is dx*(R - (N-R)) = dx*(2R-N) = dx*D. So you could set D = 2R-N and replace R in the formula to get the probability of ending up distance D steps to the right. 2. Trials and successes. We are supposed to think of the above example in a more general way as N trials in which there were R successes where the probability of a single success is r, see M&M page 438. 3. Radiation from an isotope. At first this problem seems quite unrelated. We have a Geiger counter which measures 3 clicks per second on the average from a random radioactive source. What is the probability of having 5 clicks in a 1 second interval? If we divide time into msec, then we have 1000 intervals for each second, and the probability of a click in one of these intervals is r = .003. The probability of having 5 clicks is P(5) = (.003)5 (.997)995 Why is this so? In order to have 5 clicks, we assume (with slight inaccuracy) that this means we have to have a click in each of 5 of our 1 msec intervals. That is, we ignore cases of 2 or more clicks in the same interval (for the moment). The first factor is the ways to pick 5 intervals. For each way, the rest is the probability of having 5 clicks and 995 non-clicks. Each click here is like a step of the drunk, an independent event having no correlation to the last click. With the drunk, the events might occur at some even rate like 1 step per second, so we could add the element of time to the drunk problem. Then the "events" of the problem (the steps) occur one after the other at a fixed rate. This notion of time is completely irrelevant and has nothing to do with time in the radiation problem. For the drunk, there were N trials, and R is the number of successful steps to the right. For radiation, there are 1000 trials, and 5 is the number of successful clicks. Each trial in the second case is the elapse of 1 msec, and it is a successful trial if a click occurs in that 1 msec. For the drunk, each trial is a step, regardless of how long it takes to make that step. Let's now restate the solution to the radiation problem as follows. Let R0 be the mean number of particles to radiate in one second, so R0 is going to be the mean of our distribution. In the above example, R0 was 3 and R was 5. So we can say P(R) = (R0/N)R ( 1 - R0/N)N-R where N is the number of divisions in the interval, and R is the number of clicks of interest, and as noted, R0 is the mean number of clicks during the whole integral. If we can take the limit as N , then we have corrected the inaccuracy mentioned above. To do this limit, consider these facts: NR /R! // just write it out! lim (1 + c/n)n = ec => ( 1 - R0/N)N e-Ro and ( 1 - R0/N)-R 1-R = 1 So we have P(R) NR/R! (R0/N)R e-Ro * 1 = (Ro)R e-Ro / R! Normally the mean number of success events is written as and not Ro, so we have P(R) = R e- / R! and this is the famous Poisson Distribution! 4. Confirm that <R> = : This is pretty easy! <R> = !Syntax Error, I R P(R) = e- !Syntax Error, IR R/ R! = e- !Syntax Error, IR R/ R! e- !Syntax Error, I R-1/ (R-1)! = e- !Syntax Error, IR-1/ (R-1)! = e- !Syntax Error, IR/ (R)! = e- e = 5. Restate things in terms of a car traffic problem. (a) A freeway onramp gets 5 cars per minute average. What is the probability of getting no cars in a given minute? P(R) = R e- / R! = the mean = 5 cars/minute. P(0) = 50 e-5 / 0! = e-5 = 0.7% (b) What is the probability of getting 10 cars in a given minute P(10) = 510 e-10 / 10! = 9765625 * .00673 / 3628800 = .018 = 1.8% (c) what is the probability of getting the mean in a minute? It is about 17%. Here is a plot of P(R) versus R for mean being 5: The chances of 20 cars appearing in a minute is very small. 6. Restate in terms of Sklar's ALOHA discussion. What is the probability that we will see K messages start in a time interval T, if the mean number of transmitted message starts is t messages/sec. The mean number of message starts in this time interval will be t T. So put this in for the mean in the formula to get P(K) = (t T)K e-t T / K! Now when anyone sends a message, it lasts time . This message will avoid collision if there are no message starts between before this message starts and after this message ends. This is a window of width 2. It this window is clean, you get no collision and the message will be a success. What is the probability of getting zero messages in this window? P(0) = (t 2)0 e-t 2 / 0! = e-t 2 This is the probability of success in sending a message and it must equal /t where is the mean number of messages that are successes per second. This results in = t e-2t y = x e-2x which you can plot as I just did in Maple. This is a normalized success rate as a function of the attempt rate, and it peaks at a lousy 18% which is pretty surprising. Here it is from maple So this is the "pure ALOHA" curve, see Sklar and Sklar notes for more.