Confusion on Poisson and Binomial Distribution

[rahul77]

Hey guys,
Can anyone please explain the differences between binomial and poisson distribution.

THANK U>>>>>>>>>>>>>>>>>>>>>

 

[mathman]

Essentially the binomial distribution involves a finite number of possibilities. For example tossing a coin a fixed number of times and predicting the number of heads.

The Poisson distribution is concerned with the situation where the number of counts has no limit. An approximate example is radioactive decay counts. However, physically it is still a binomial, but Poisson is used where the mean is very small compared to the maximum possible.

 

[flying_young]

Hi, you may think of the Poisson distribution being used to describe the number of changes (i.e., radioactive particles that enter a particular region) for a fixed interval of time. From what I have learned, it is sometimes more simplistic to describe a process with Poisson distribution than with a Binomial distribution, even though in that particular case both could be used. Please correct me if I'm wrong. =)

 

[mathman]

When both apply, it is usually easier to work with Poisson. In the case of radioactive decay, the basic idea is that the mean (of the number decaying in a given unit of time) is very much smaller than the maximum possible (total number of atoms in the sample). Many other physical processes have this property, so the Poisson is used.

 

[flying_young]

This is very well-elaborated. Thank you greatly.

When both apply, it is usually easier to work with Poisson. In the case of radioactive decay, the basic idea is that the mean (of the number decaying in a given unit of time) is very much smaller than the maximum possible (total number of atoms in the sample). Many other physical processes have this property, so the Poisson is used.

 

[NoMoreExams]

Maybe I'm coming from a different background but Poisson is a counting process so for example if you talk about the number of things happening in an interval, you can think of Poisson as the first distribution (note that the interarrival times will be exponentially distributed). Binomial on the other hand is a simple yes or no process so to speak where each trial is independent. The way the distribution function for Binomial is developed is very intuitive to what it is (similarly with the geometric distribution).

At one point on our exams we had to look at a few results and decide if the underlying distribution was Binomial, Poisson or Negative Binomial. The way we did that was by look at the first 2 central moments. As such:

If E(X) = V(X) then we had Poisson simply from definition of Poisson parameter being both the mean and the variance
If E(X) > V(X) then we had Binomial since E(X) = np and V(x) = np(1-p) and (1-p) < 1 therefore we got our result
Finally if E(X) < V(X) then we had Negative Binomial since E(X) = rB and V(X) = rB(1+B)

Now obviously in the real world there's more than 3 choices and you don't always want to use a discrete distribution.

 

[mathman]

Maybe I'm coming from a different background but Poisson is a counting process so for example if you talk about the number of things happening in an interval, you can think of Poisson as the first distribution (note that the interarrival times will be exponentially distributed). Binomial on the other hand is a simple yes or no process so to speak where each trial is independent. The way the distribution function for Binomial is developed is very intuitive to what it is (similarly with the geometric distribution).

At one point on our exams we had to look at a few results and decide if the underlying distribution was Binomial, Poisson or Negative Binomial. The way we did that was by look at the first 2 central moments. As such:

If E(X) = V(X) then we had Poisson simply from definition of Poisson parameter being both the mean and the variance
If E(X) > V(X) then we had Binomial since E(X) = np and V(x) = np(1-p) and (1-p) < 1 therefore we got our result
Finally if E(X) < V(X) then we had Negative Binomial since E(X) = rB and V(X) = rB(1+B)

Now obviously in the real world there's more than 3 choices and you don't always want to use a discrete distribution.

The point I was trying to make is that when p is very small (the probability that a given atom will decay in the time interval - radioactive decay process), then the binomial can be approximated by the Poisson, since V(X) will be very close to E(X).

 

[NoMoreExams]

Well under certain conditions (such as n is large enough) they would approach Normal as well :) I guess we are just coming from different backgrounds. I know almost nothing of science especially decay processes.

 

[mathman]

Well under certain conditions (such as n is large enough) they would approach Normal as well :) I guess we are just coming from different backgrounds. I know almost nothing of science especially decay processes.

The binomial will approach normal for large n when p is not too small. It approaches Poisson when np is much smaller than n.

 

[NoMoreExams]

The binomial will approach normal for large n when p is not too small. It approaches Poisson when np is much smaller than n.

Really? I was under the impression that for sufficiently large n, both Binomial and Poisson will approach Normal (CLT?)

 

[mathman]

Really? I was under the impression that for sufficiently large n, both Binomial and Poisson will approach Normal (CLT?)

You are confusing two different situations. I'll illustrate by using the radioactive counting as an example.

When counting the number of decays in an interval, one can consider cases where the average number of decays per interval is "small" (for example around 10). Here the distribution can be considered Poisson, since the number of atoms in a sample is of the order 10²⁴.

However if you are considering a situation where the average number of decays is "large" (for example around 1000), then the Poisson will start looking like the normal.