More of the derivation of the Poisson Distribution. Created by Sal Khan.
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- So I understand that if we know we have a binomial distribution, our expected value for any number of trials is np, but how do we know that some of these things even follow a binomial distribution? How do we know that the number of cars passing by in an hour is distributed binomially? Is it because, given any time interval, a car either does or does not pass, and there are no other options?
Edit: I think I understand it...if we separate the hour into an infinite number of intervals, even though the probability of a car passing in any single one of those intervals is infinitely low, over the course of an hour there will, on average, be a certain number of exact moments where a car DID pass by, AKA our expected value.(5 votes)
- Sometimes you know that things cannot be binomial because you know how they work. For instance if you are counting train carriages instead of cars you couldn't use binomial or poisson distribution because carriages don't come along at purely random intervals - they come in groups call trains.
The big picture here though is that in these early videos, Sal is presenting statistics to describe stuff. Armed with that knowledge, in the later videos he will create statistical tests for whether hypotheses are true. For example you can test the hypothesis that the arrival of cars follows a binomial probability distribution. The test I know of is called a Goodness of Fit test & Sal talks about it in a later video.(13 votes)
- Does this make sense for low n? I.e. if there are 9 cars that pass an hour then we use lambda = 9. We say this is equivalent to n*p, where n is the number and trials and p is the probability of a car passing in the number of trials. So say we only do 1 trial, then p = 9. This means there's a 900% chance of a car passing in one hour, lamba = n*p seems to make a little more sense when n is large (resulting in p less than 1). So is lambda = np a big assumption we make?(5 votes)
- I might be a little late to answer here as well, but I believe for a small number of trials, you would preferably use a RV with binomial distribution instead of a RV with Poisson distribution (since Poisson is for an infinite number of trials, ie: the interval of time between occurrences is negligible). Is this correct?(3 votes)
- why do we need the poisson process?? O-o(4 votes)
- Well, besides the traffic application Kahn exposed, we have several others. One of the most common is the telemarketing model which is basically a Poisson process. Another important example is the radioactive decay, there is a certain probability of a number of atoms to decay at every instant, which generates a Poisson process.(7 votes)
- this is not a question exactly about this topic but a combined one.
I am facing problems to identify a problem if it follows binomial or poisson or hypergeometric distribution.i get the question,have all the datas but cant find out which distribution i have to follow.are there any ways to identify them?
please anyone help.thanks in advance...(7 votes)
- why not that we could not conclude that when n approaches infinity, (1 + a/n) ^ n is just 1 (rather than e^a)?(2 votes)
- I got tripped up on this, too. Try plugging in an increasing n.
n=1, (1+(1/1))^1 = 2
n=2, (1+(1/2))^2 = 2.25.
n=3, (1+1/3))^3 = 2.37
You can see that as n increases, it doesn't near 0. That's because the n in the exponent outruns the 1/n.(3 votes)
- You are able to build all the bridges from our ignorance to your knowledge, and step by step you allow our minds to cross. You are truly great! Thank you for doing what you do best and sharing it!(3 votes)
- Hi Sal... at12:00you simplified lim<n->∞> (1-λ/n)^n as e^-λ and lim(n->∞> (1-λ/n)^-k as 1. Question is when you could simplify (1-λ/n) as (1-0) as n approaches ∞ in lim(n->∞)(1-λ/n)^-k.. why did not you simplify the lim(n->∞)(1-λ/n)^n as (1-0)^n which would also be 1 as n approaches ∞. which would have left us with further simplified value of (λ^k/k!) ?(2 votes)
- This is a good question. The lim (n->∞) of (1+1/n)^n is the definition of the number e. On the year 2007 it was calculated (by computers) with 10^11 decimals.(2 votes)
- Right, so if n= success per smaller interval
and we want to work out as n -> infinity then surely that means n happens an infinite amount of times in the interval?
and also how can x!/(x-k)!= x(x-1)(x-2)...[x-(k+1)]
if we did x=7 and k=3 that means its 7!/(7-3)!=210 7(7-1)(7-2)(7-3)[7-(3+1)] = 3024 how can they be equal?(2 votes)
- I'm not sure why he is defining n as the number of successes in an interval. I thought n was the number of intervals in an hour.
As the number of intervals (n) goes to infinity, the probability of success (p) in any given interval goes to zero. This happens in a way such that n times p is constant (ie, lambda).(2 votes)
- For the derivation of this poisson equation. I am seeing in a lot of other places the use of (lambda*(t)) instead of just lambda. Why would this be? I believed that time was not so much a factor after you initial assumptions.(2 votes)
- In the previous video, Sal made the assumption that lambda is constant (i.e. the probability that a car passes in a given interval doesn't change over time).
In general, this assumption need not be made. However, if we don't make this assumption, then lambda must be a function of time (i.e. lambda(t) ).(2 votes)
- Does anyone know where I can watch the differential calculus vifeo that is showing, when lim-->infinity (1 - λ/n) ^ n = e ^ - λ ?(1 vote)
- I'm not sure.
Have you seen the compound interest videos?
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 are 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 defined the random variable as the number of cars that pass in an hour at a certain point on a certain road. We measure a bunch, we sat out there a bunch of hours, and we got a pretty good estimate of this and we say it's lambda. And we said, OK, we wanted to model it as 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 in the last video we tried it out. We said, oh, well, what if we make this interval a minute and this is the probability of success per minute? We'd have maybe a reasonable description of what we're describing. But what if more than one car passes in a minute? And they 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. If we describes this as a binomial distribution with the limit as it approaches infinity, we could say that the probability that x is equal to some number-- so the probability that our random variable is equal to 3 cars in a particular hour, exactly 3 cars in an hour-- is equal to-- oh, we want to take the limit as it approaches infinity, right? The limit as n approaches infinity of n choose k. We're going to have k moments in time because n approaches infinity, these intervals become super super duper small. So these become moments in time. We're going to have close to an infinite number moments and this is the number of successful moments where cars pass. We have 3 moments where there was a success, where a car passed, and we had a total of 3 cars pass. Or 7 cars, 7 moments where it was true that a car passed, and we would have total 7 cars pass in the hour. So just finishing up with our binomial distribution, n moments, choose k successes times the probability of success. What's the probability of success? So this would be n. What's p equal to? p is equal to lambda divided by n, right? 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? Well, it's 1 minus the probability of success. And how many failures are we going to have? How many moments will not have a car pass? Well, we have total event 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 as n approaches infinity-- let me write out this binomial coefficient. That's n factorial over n minus k factorial times k factorial. Normally I write these the other way around, but it's the same thing. 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 exponents. 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. Let me simplify a little bit more. Let me swap spots with these two. You can kind of view them both as being in the denominator. So you can change the order of division or multiplication depending how you view it. So this is equal to the limit-- let me switch colors. The limit as n approaches infinity-- I don't like that color. Actually, let me just rewrite what we did in the last video. What is this thing right here? And we showed 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 in 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 these as 1, 2, 3-- all the way, there's going to be k terms here. We took care of that. We just rewrote that. And I said I would switch these two things around, so that's divided by n to the k times-- I'm just switching these-- lambda to the k over k factorial. And then, what do we have here? We'll 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 limit? If you take the limit, this is another property so you don't get too overwhelmed-- another property of limits. If I take the limit as x approaches anything, a of f of x times g of x. That's equal to the limit as x approaches a of f of x, times the limit as x approaches a of 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 this stuff up here. So first of all, what's this limit? Let me write this out. And let me pick a good color-- yellow. So we have the limit as n approaches infinity. So this thing up here, this n times n minus 1 times n minus 2-- all the way down to n minus k plus 1, what's it going to look like? It's going to be a polynomial right? We're multiplying a bunch of-- well really, we're multiplying a bunch of binomials and we're doing 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. It's going to be this big polynomial-- kth degree polynomial. And that's really all we need to know for this derivation. So it's going to be n to the k plus blah, blah, blah, blah, blah, blah, blah-- a bunch of other stuff. This thing when you multiply it out, over-- we have this n to the k, that's this part of it. Times the limit as-- well actually, we don't have to worry. This is a constant. So we could actually bring this out front. So we don't even have to write a limit. So times lambda to the k k over k factorial. There's no n 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 1 minus lambda over n to the minus k. All right, I know you can barely read this. So first of all, what's this limit? The limit as n approaches infinity of some polynomial where it's n to the kth power plus blah, blah, blah, blah. Where all of these other terms have a lower degree. This is the highest degree term. So you have n to 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 n to the k and you would get-- this would just be 1 plus 1 over n plus 1 over-- everything else would have a 1 over n in it, and this would just be a 1. And if you took the limit as you approach infinity, then all of these other terms would be zero and you'd be left with 1/1. But either way, you have the same degree in the top and the bottom, and their coefficients are the same, so the limit as n approaches infinity of this is 1, which is a nice simplification. So you end up with 1 times lambda k over k factorial. Now what's the limit as n approaches 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 you here, but instead of an a we have a 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 n approaches infinity? Let me write it a little bit neater. 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, lambda's a constant. As this approaches infinity, this term's going to approach 0. So you have 1 to the minus k. 1 to any power is 1, so that term becomes 1. So we have another 1 there. So there you have it. We're done. The probability that our random variable, the number of cars that passes in an hour, is equal to a particular number. You know, it's equal to 7 cars pass 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 showed that this is equal to lambda to the kth 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 vacuum-- if you have no context for it, you wouldn't guess that this is in any way related to the binomial theorem. I mean, it's got an e in there. It's got a factorial, but a lot of things have factorials in life, so not clear that that would make it a binomial theorem. But this is just the limit as you take smaller and smaller and smaller intervals, and the probability of success in each interval becomes smaller. But as you take the limit you end up with e. And if you think about it it makes sense because one of our derivations of e actually came out of compound interest and we kind of did something similar there. We took smaller and smaller intervals of compounding and over each interval we compounded by a 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, I'm the traffic engineer, and I figure out that on average, 9 cars passed per hour. And I want to know the probability that-- so this is my expected value. In a given hour, on average, 9 cars are passing. So I want to know the probability that 2 cars pass in a given hour, exactly 2 cars pass. That's going to be equal to 9 cars per hour to the 2'th power or squared, to the 2'th power. Divided by 2 factorial times e to the minus 9 power. 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 graphing calculator out. Well, I'll let you do that exercise to figure out what that is, but I'll see you in the next video.