Transcript Lecture 10

Physics 114: Lecture 10
PDFs Part Deux
Dale E. Gary
NJIT Physics Department
Binomial & Poisson Distributions

The binomial distribution is PB ( x; n, p) 

The mean is   np.

n!
p x (1  p) n x .
x !(n  x)!
MatLAB: binopdf(x,n,p)
Use for yes/no statistics
The standard deviation is   np(1  p)

The Poisson distribution is PP ( x;  ) 

The mean is

The standard deviation is    .
x  .
x
x!
e  .
MatLAB: poisspdf(x,)
Use for counting statistics
February 18, 2010
Example 2.3

Some students measure some background counts of cosmic rays. They
recorded numbers of counts in their detector for a series of 100 2-s
intervals, and found a mean of 1.69 counts/interval. They can use the
standard deviation formula from chapter 1, which is
1
s 2   ( xi  x )2 ,
N
to get a standard deviation directly from the data. They do this and get s =
1.29. They can also estimate the standard deviation by   1.69  1.30.
 Now they change the length of time they count from 2-s intervals to 15-s
intervals. Now the mean number of counts in each interval will increase.
Now they measure a mean of 11.48, which implies   11.48  3.17, while
they again calculate s directly from their measurements to find s = 3.39.
 We can plot the theoretical distributions using MatLAB poisspdf(x,mu),
e.g. poisspdf(0:8,1.69) gives
ans = 0.1845 0.3118 0.2635 0.1484 0.0627 0.0212 0.0060 0.0014 0.0003
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Example 2.3, cont’d
Probability of counts per interval
Poisson Distribution for mean 1.69
0.3
0.25

0.2
0.15
0.1

0.05
0
0
2
4
6
8
Number of Counts
Probability of counts per interval
Poisson Distribution for mean 11.48
The plots of the distributions is shown
for these two cases in the plots at right.
 You can see that for a small mean, the
distribution is quite asymmetrical. As
the mean increases, the distribution
becomes somewhat more symmetrical
(but is still not symmetrical at 11.48
counts/interval).
 I have overplotted the mean and
standard deviation. You can see that
the mean does not coincide with the
peak (the most probable value).

0.35
0.12
0.1
0.08

0.06
0.04

0.02
0
0
5
10
15
20
Number of Counts per 15 s
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25
Example 2.3, cont’d

Here is the higher-mean plot with the equivalent Gaussian (normal
distribution) overlaid.
Poisson Distribution for mean 11.48
Probability of counts per interval
0.15
0.1
0.05
0
0
5
10
15
20
25
Number of Counts per 15 s

For large means (high counts), the Poisson distribution approaches the
Gaussian distribution, which we will now describe further.
February 18, 2010
Gaussian or Normal Distribution

We will simply give the expression for the Gaussian distribution without
derivation. Note that it is the limiting case of the Poisson distribution
(counting statistics) as the mean  becomes large.
 1  x   2 
1
PG ( x;  ,  ) 
exp   
 .
2

 2
 
 

Unlike the binomial and Poisson distributions, which are defined only for
integer values, the Gaussian distribution is continuous. That means it is a
probability density function (pdf), and to get the probability that a value will
fall between two values of x, you have to multiply by the bin width dx, or in
the limit of infinitesimal bin widths, integrate:
x2
P( x1  x  x2 )   PG ( x;  ,  )dx.
x1

Here, the mean  and standard deviation  are part of the definition of the
distribution, so are not defined separately in terms of other parameters.
February 18, 2010
Characteristics of the Gaussian
As always, the Gaussian pdf is normalized so that the area under the curve
is unity (i.e. the integral from  to  is 1). The exponential itself has unit
amplitude at x =  (i.e. exp(0) = 1), and if you do the integral of the
exponential you will find that it is  2 . Therefore, you have to divide by
this factor to normalize the integral.
 As you did in the first homework, you can determine the full width at half
maximum for the Gaussian distribution by finding where the function falls to
½ its amplitude:
 1  x 2  1
exp      
 2     2

2 x   2 2 ln 2.

A good way to think about the standard deviation is that “most” values will
lie within 1 of the mean. The actual percentage for 1 is 65%. If you go
to 2, it is 95%. If you go to 3, it is 99.7%. We can think about the
effect of this using a star image.
February 18, 2010
Other Distributions
There are many other distributions that are met with in various
circumstances, some phenomenological, and some based on theory. An
interesting one is the Lorentzian distribution (or Cauchy distribution), which
describes the shape of spectral lines in gases or plasmas such as the Sun.
1
/2
PL ( x;  , ) 
.
  x   2    / 2 2
 This has a middle part that looks a little like a gaussian (the so-called line
core), but the parts far from the mean (called the line wings) decrease
slowly.
0.4
Gaussian ( =1)
0.35
 Although the mean is , the standard
Lorentzian (=2.354)
0.3
deviation of this distribution is undefined
0.25
(i.e. the second moment is undefined)
0.2
because of how slowly the distribution
0.15
falls off (its integral is infinite).
0.1
Normalized distributions

0.05
0
0
1
2
3
4
5
x
6
7
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