# Poisson dist. with small numbers

1. Feb 16, 2010

### penguindecay

Dear Physicists,
I have a poisson dist with a mean at 0.00107. I tried that usual SQRT(mean) for the standard deviation but of course I got an sigma larger than my actual plot. Can someone point me to some text or the right theory?

Cheers

L

2. Feb 16, 2010

### Bob S

The sqrt(N) does not work with uncertainty distributions that are not Gaussian, which means N less than say 100. The Poisson distribution is the correct distribution to use in your case, but it is limited to integer values; 0, 1, 2, etc.. So in your case, about 99.9% of the time you get zero, and 0.107% of the time you get 1.

Bob S

3. Feb 16, 2010

### uart

How many data samples do you have? That's a really low mean, only 1/1000 expected hits per sample. The poisson distribution for that will be very close to just 99.9% for zero events, 0.1% for one event and close to zero probabilty for >1 events. Is that roughly what your data looks like (Yes/No)? If no then give some more information about your data, and it's probably not Poisson(0.001).

4. Feb 16, 2010

### uart

Hi Bob, the Poission distribution is discrete, but the parameter ($\lambda$) is not limited to integer values.

Also the standard deviation of the poisson distribution is $\sqrt{\lambda}$.

In this case $\lambda \simeq 0.001$ so $\sigma = \sqrt{0.001} \simeq 0.0316$.

Even if you approximate the distribution with just two points, P(0)=0.999 and P(1)=0.001, you get a very good approximation to match both mean (lambda) and stdev (sigma). In this case you still get,
lambda = 0.001
sigma = sqrt( 0.999*(0-.001)^2 + 0.001*(1-.001)^2) = sqrt(0.000999) = 0.0316

I suspect the OP might not have enough data points, he'll need many thousand of points before he's likely to get a good statistical representation of the system.

5. Feb 16, 2010

### penguindecay

Hi, just to clarify, I have a poisson dist. that has it's peak at about 0.0001 and a mean value of 0.00107, there are 43194 entries. Sorry I got some numbers mixed up at the start.

Bascically I've been counting decay times of the same events. Thanks, I tried the usual root(mean) but the dist. only goes from 0 to 0.02, the root(0.001) gets me 0.03 as a sigma. Am I going about it the wrong way?

6. Feb 16, 2010

### Staff: Mentor

I don't know what you mean by this. The Poisson distribution is a discrete distribution, so it will go to some integer number, not to a fraction of an integer. With such a low probability I would assume that it would only go from 0 to 1 with 2 only if you sampled hundreds of thousands of samples, or maybe even more.

In any case, those numbers seem correct to me. If you have 43194 samples with a mean of .00107 that probably means that you have about 46 1's and the rest 0's. The standard deviation of that is indeed about 0.03.

7. Feb 17, 2010

### uart

Ok so you're recording the "time until event", that's a continuous distribution so it's definitely not Poisson.

If you record the number of events in a given time interval (possible values are 0, 1, 2, 3 etc events) then you get a Poisson distribution, in your case (time until event) you should be looking at the exponential distribution (which is of course a continuous distribution).

BTW. For an exponential distribution the mean and the standard deviation (sqrt variance) are equal. So it does seem compatible with your data.

See http://en.wikipedia.org/wiki/Exponential_distribution

8. Feb 17, 2010

### penguindecay

Hi, thanks for all the information you have given me. I think I should explain my graph and data in detail. I have a plot running the decay time from 0 to 0.02 in intervals of 0.001. I'm binning into these intervals the events that fit into the bin width (of 0.001). I get (I assume) a poisson dist. I assmed it's a poisson because I'm doing a counting exp?

It is not a exp dist. as the event in question has many interactions before losing energy, in short the decay time I record is from the incident event to the last event that has an energy above a threshold, Thanks

9. Feb 17, 2010

### uart

Are you sure it's not exponential. You said it has a peak in the first bin and decays from there, that's consistent with exponential.

What about the std-dev, have you calculated it from the data. Is it close to the mean? If so that's also consistent with exponential.

Finally what does it look like, does it look exponential?

10. Feb 17, 2010

### uart

Here's another interesting fact. If the distribution was exponential then the expected number of samples greater than some time "t" would equal n/u exp(-t/u)

Putting your numbers, n=43194, u=0.00107 gives t=0.019 the lowest "t bin" for which the expected number (of samples greater than "t") is less than one. In other words, it gives a likely approx figure for the largest "t" value you would expect to see in that experiment, given the number of samples you took. Now what did you say before, your largest "t" is 0.02. It seems to fit the data pretty well, yes?

11. Feb 17, 2010

### penguindecay

Thanks, I'll ask my teacher about this. But the thing looks like a poisson!! definitely not a decay slope like an exp. It's definitely a hill shape with equal areas on each side of the peak. Thanks again everyone!!

12. Feb 17, 2010

### uart

Ok but previously you said :
That puts the peak in the very first bin! So could you please clarify which of those two statements is the correct one. I cant see how they can both be true.

13. Feb 17, 2010

### Staff: Mentor

There should be no confusion about Poisson v. Exponential. If it is a continuous distribution then it cannot be Poisson, if it is a discrete distribution then it cannot be exponential.

You mentioned that you have 43194 data points. Are they integers (e.g. 0 and 1) or are they real numbers (e.g. 0.00015)?

Last edited: Feb 17, 2010