Square Root of N Law of Random Counts

[tony873004]

"Square Root of N" Law of Random Counts

According to my lab manual, "Theory says that radioactive decay obeys statistics for which the standard deviation of the counts per time interval is equal to the square root of the mean number of counts for that interval, for cases when the mean is a large number."

I think they mean for cases where N is a large number.

In lab, taking counts from the geiger counter, I got an average of 616.56 counts per minute with a standard deviation (computed with Excel: stdev(my column of counts) of 30.74 from 25 sets of individual 1-minute counts.

But the sqrt(616.56) = 24.83, not 30.74. Thinking that it would reach about 30.74 if I had a larger N, I set up a Monte Carlo simulation on my computer. I generated N random gauissian numbers with a mean of 616.50, and a standard deviation of 30.74. But no matter how high I make N, even 1 million, I always get an answer of about 24.8 when I square the average of my random numbers.

Additionally, Googling for "Square Root of N Law of Random Counts" turns up nothing. Did the author of my lab manual just make this up?

Here's how I modeled it with the computer

Private Sub Form_Load()
   totalcounts = 0
   n = 1000000

   For k = 1 To n
       counts = Grnd(616.56, 30.74)
       totalcounts = totalcounts + counts
   Next k

   meancounts = totalcounts / n
   Label1.Caption = Sqr(meancounts)
End Sub

Function Grnd(Mean As Double, sd As Double) As Double
    Dim x1 As Double, x2 As Double, w As Double, y1 As Double, y2 As Double
    Do
        x1 = 2 * Rnd(1) - 1
        x2 = 2 * Rnd(1) - 1
        w = x1 * x1 + x2 * x2
    Loop Until w < 1
    
    w = Sqr((-2 * Log(w)) / w)
    y1 = x1 * w
    y2 = x2 * w
    Grnd = Mean + y1 * sd
End Function

 

[Hurkyl]

The manual is right; your counts should be given by a Poisson distribution. Remember that you aren't computing the true standard deviation; you're merely estimating it from a sample, and all of the various sorts of errors you're familiar with can apply.

 

[D H]

I generated N random gauissian numbers with a mean of 616.50, and a standard deviation of 30.74. But no matter how high I make N, even 1 million, I always get an answer of about 24.8 when I square the average of my random numbers.

Radioactive decay is not a Guassian process. Think about it this way: Suppose you create a sample of some radioactive substance such that at the moment of creation you have a pure sample. If radioactive decay was a Gaussian process, some of that sample would have before you created it!

Radioactive decay, as Hurkyl mentioned, is a Poisson process. The mean and variance of a Poisson distribution are equal to one another. The standard deviation is the square root of the variance, and hence your lab book's statement that "radioactive decay obeys statistics for which the standard deviation of the counts per time interval is equal to the square root of the mean number of counts for that interval." The qualification "for cases when the mean is a large number" pertains to your ability to estimate the mean, which improves (relatively speaking) as the count increases.

 

[tony873004]

Radioactive decay is not a Guassian process...

That's interesting, because under "Objectives" in our lab manual it says "Observe some properties of the Gaussian distribution".

Thanks for the replies. I'll have more questions on this in a few hours when I work on this in my lunch break.

 

[CRGreathouse]

That's interesting, because under "Objectives" in our lab manual it says "Observe some properties of the Gaussian distribution".

Well, with the help of the Central Limit Theorem, anything can help you observe the properties of the Gaussian distribution...

 

[tony873004]

I've turned in the lab, but I'm still interested in knowing this.

Another student asked the teacher in office hours and came away with the impression that if our N's were simply larger, that
"standard deviation of the counts per time interval is equal to the square root of the mean number of counts for that interval"
would be true. That was my first guess, prior to making the computer model, because as I plotted standard deviation computed after each run vs. sqr(mean counts) after each run, the lines did see to be converging towards each other, with slopes that gave me the impression I'd need about N=200 before they met.

But I find it strange that the number I computed with our lab data, and the number I generated using the N=1 million computer model gave virtually identical results (24.823 vs 24.830) for sqr(mean counts). Is this just a coincidence?

I still don't understand what a Poisson distribution means. The Wiki article doesn't quite do it for me.

So in my numerical experiment, was my mistake to use random Gaussian numbers, rather than Poisson numbers? What would I need to do in my numerical experiment to verify that stdev = sqrt(mean count)? Can I write a computer function to generate a Poisson distribution, similar to the "GRnd" function that I wrote (see 1st post) which generates random Guassian numbers?

-Thanks

 

[tony873004]

From the Wiki article:

Thus, the number of observed occurrences fluctuates about its mean λ with a standard deviation [tex]\sigma_{\kappa}=\sqrt{\lambda}[/tex].

So it seems I computed standard deviation wrong when I used [tex]\sigma = \sqrt {\frac{{\sum\limits_1^N {\left( {x_k - \bar x} \right)^2 } }}{{N - 1}}} [/tex].

But in this case, I don't see the point of this part of the lab. Of course if I compute my standard deviation as [tex]\sqrt\lambda[/tex] then it will be the same as the square root of my mean counts since lambda is my mean counts. It's as if they want me to show that [tex]\sqrt \lambda = \sqrt \lambda [/tex]. So how can I go wrong?