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### Course: Statistics and probability > Unit 9

Lesson 9: Poisson distribution# Poisson process 2

More of the derivation of the Poisson Distribution. Created by Sal Khan.

## Want to join the conversation?

- 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) - 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!(6 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)

- 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?

https://www.khanacademy.org/economics-finance-domain/core-finance/interest-tutorial/cont-comp-int-and-e/v/continuously-compounding-interest-formula-e

https://www.khanacademy.org/math/algebra2/exponential_and_logarithmic_func/continuous_compounding(3 votes)

## Video transcript

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.