Learn about Poisson Process 2

Khan Academy Presents: More of the derivation of the Poisson Distribution.

Learn about Poisson Process 2 I think we now have all the tools we need to move forward. So just to review a little bit of the last video, we said we’re trying to model out the probability distribution of how many cars might pass in an hour. And the first thing we did is we sat at that intersection and we found a pretty good expected value of our random variable and this random variable just to go back to the top. We define the random variable as a number of cars that passed in an hour at a certain point, on a certain road. And we said that we measured a bunch we sat out there bunch of hours and we got a pretty good estimate of this and we say it’s lambda. And we said, okay, we want to model it’s a binomial distribution, so if this is a binomial distribution then this lambda would be equal to the number of trials times the probability of success per trial. And so if we could view a trial as an interval of time. This is the total number of successes in an hour. And so this would be success in a smaller interval and this would be the probability of success in that smaller interval. And on the last video, we tried it out we said, oh, what if we make this interval a minute and this is the probability of success per minute we’d have or maybe a reasonable description of what we’re describing, what if more than one car passes in a minute and we said, oh, let’s make this per second and this is the probability of success per second. But then we still have the problem more than one car could pass in a second very easily. So what we wanted to do is we want to take the limit as this approaches infinity and then see what kind of formula we get from the math gods. So if we describe this as a binomial distribution with the limit as it approaches infinity we could say that the probability that x is equal to some numbers or the probability that our random variable is equal to—I don’t know three cars in a particular hour, exactly three cars in an hour is equal to—we want to take the limit as it approaches infinity. The limit is n approaches infinity of n choose k, we’re going to have k moments in time and approaches infinity, this intervals becomes super, super duper small so this become moments in time. So we have—we’re going to have an infinite number of moments and this is the number of successful moments where cars passed. So we have three moments where cars—where there were a success, where car passed then we had a total of three cars passed or 7 cars, 7 moments where it was true that a car passed and we would have total 7 cars passed in the hour. So just finishing up with our binomial distribution and moments choose k successes times the probability of success. What’s the probability of success? We said, if this is—so this could be—if n is, so this will be n, what’s P equal to? P is equal to lambda divided by n, n times p is lambda so let me just write that down. P is equal to lambda divided by n. I just rearranged this up here. So our probability of success is lambda times n and we’re saying what’s the probability that we have k successes and then what’s the probability that we have a failure that’s 1 minus the probability of successes and how many failures we’re going to have, how many moments will not have a car pass? We have total of end moments and k of them were successes so we’ll have n minus k failures, let’s see what we can do with this. So this is equal to—let me rewrite it all and I’ll change colors. The limit is n approaches infinity—let me write out this binomial coefficient that’s n factorial over n minus k factorial times k factorial—let me write this the other way around but it’s the same—times lambda to the k, using my exponent properties over n to the k and then this expression right here, I can actually separate out the—this is the same thing as 1 minus lambda over n to the n times 1 minus lambda over n to the minus k. You have the same base you could add the exponents and you would get this up here. And now let me simplify a little bit more. Let me swap spots with these two, they’re both—you can kind of use them both just being in the denominator so you can change the order of division or multiplication depending how you do it. So this is equal to the limit—let me switch colors—the limit as n approaches infinity—I don’t like that color. Let me just rewrite what we did in the last video, what is this thing right here? And we show that at the end of the last video n factorial divided by n minus k factorial. It was n times n minus 1 times n minus 2 all the way to n minus k plus 1. If this was 7 over 7 minus 2 factorial we would have 7 times 6 and 6 is one more than 7 minus 2 so that’s where we got that. We did that on the last video if you’re getting confused. And we also said that there’s going to be exactly k terms here. So if you counted this as one, two, three all the way there’s going to be k terms here. And so that, we took care of that we just rewrote and I said I would switch these two thins around so that’s divided by n to the k times—I’m just switching this, lambda to the k over k factorial and then what do we have here? We have—oh I can just rewrite that. This is continuing the same line. 1 minus lambda over n to the n times 1 minus lambda over n to the minus k. Now, we can take the limit, so what happens when we take the—so just if you take the limit, this is another property that so you don’t get too overwhelmed. Another property of limits if I take the limit a x approaches anything a of f of x times g of x that’s equal to the limit as x approaches of f of x times the limit as x approaches g of x. So we could take each of these limits in the product and then multiply them and then we’ll be all set. So let’s do that and I want to leave the stuff up here. So first of all, what’s this limit? Let me write this out, let me pick a good color, yellow. So we have the limit as n approaches infinity. So this thing up here, these, n times n minus 1 times n minus2 all the way down to n minus k plus 1, what’s going to look like? It’s going to be a polynomial. We’re multiplying a bunch of, well, really we’re multiplying a bunch of binomials, we’re going it k times. So the largest degree term is going to be n to the k. it’s going to be n to the k plus something times n to the k minus 1 is going to be this big kind of binomial—this big polynomial k degree polynomial and that’s really all we need to know for this derivation is going to be n to the k plus blah, blah, blah, a bunch of other stuff. This thing when you multiply it out over—we have this into the k so we just—this part of it times the limit as—well, actually we don’t have to worry, this is a constant so we can actually bring this out front so we don’t even have to write the limit. So times lambda to the k over k factorial there’s no end here so this is a constant with respect to n times the limit as n approaches infinity of 1 minus lambda over n to the n times one minus lambda over n to the minus k. I know you can barely see through it, read this. So first of all, what’s this limit? The limit is an approach of infinity of some polynomial words into the k power plus blah, blah, blah where all of these other terms have a lower degree. This is the highest degree term. So you have entered the k in the numerator and you have n to the k in the denominator so the highest degrees are the same, the coefficients are 1 so this limit is 1. Another way you could do it, you could divide the numerator and the denominator by enter the k and you would get—this would be 1 over—this would just be 1 plus 1 over--n plus 1 over everything else will have a one over n in it and this would just be 1. And as you take the limit as you approach infinity then all of these other terms would be zero you got left with 1 over 1. But either way you have the same degree in the top and the bottom and their coefficients are the same so the limit is approach as infinity. This is one which is a nice simplification so you end up with 1 times lambda k over k factorial. Now, what’s the limit as an approach as infinity of this thing right here? 1 minus lambda over n to the n. Well, in the last video we showed that it would be—I’ll write it right here. That the limit as n approaches infinity of 1 plus a over n to the n is equal to e to the a. that’s exactly what we have here but instead of an a we have a minus lambda, minus lambda so this is going to be equal to e to the minus lambda, we have a minus lambda instead of an a. And then finally, what’s the limit as in approaches infinity, let me write a little bit neither of one—I’m just rewriting this term 1 minus lambda over n to the minus k power. What happens as n approaches infinity? Well, this term like lambda is a constant, is this approaches infinity this term is going to approach zero so you have 1 to the minus k, one to any power is one so that term becomes one. So we have another one there. So there you have it, we’re done. The probability that our random variable, the number of cars that passed in an hour is equal to particular number, it’s equal to 7 cars passed in an hour is equal to the limit as n approaches infinity of n choose k times—well, we said it was lambda over n to the k successes times 1 minus lambda over n to the n minus k failures and we just show that this is equal to lambda to the k power over k factorial times e to the minus lambda. And that’s pretty neat because when you just see it in kind of a vaccum, you have no context for it, you would guess that this is in a way relate to the binomial theorem I mean there’s kind of an e in there, there’s kind of factorial but a lot of things have factorials in life. So not clear to that would make it a binomial theorem but this is just a limit as you take smaller and smaller intervals and the probability success in each interval becomes smaller but as you take the limit you end up with e. And if you think about it makes sense because one of our derivations actually came out of compound interest so kind of needed something similar there. We took smaller, smaller, smaller intervals of compound again over each interval we compounded by much smaller number and when you took the limit you got e again. And that’s actually where that whole formula up here came from to begin with. But anyway, just so that you know how to use this thing so let’s say that I were to go out on the traffic engineer and I figure out that on average 9 cars pass per hour and I want to know the probability that—so this is my expected value right in the given hour on average 9 cars are passing. So I want to look probability that two cars passed in a given hour, exactly two cars passed. That’s going to be equal to 9 cars per hour to the twoth power or squared, of the twoth power divided by 2 factorial times e to the minus 9 powers. So it’s equal to 81 over 2 times e to the minus 9 power and let’s see maybe I should just get the graphic calculator out there. Well, I’ll let your do that exercise to figure what that is but I’ll see you in the next video.