Transcript Lecture 9

Physics 114: Lecture 9
Probability Density Functions
Dale E. Gary
NJIT Physics Department
Binomial Distribution

If you raise the sum of two variables to a power, you get:
( a  b) 0  1
(a  b)1  a  b
(a  b) 2  a 2  2ab  b 2
(a  b)3  a 3  3a 2b  3ab 2  b3
(a  b) 4  a 4  4a 3b  6a 2b 2  4ab3  b 4

Writing only the coefficients, you begin to see a pattern:
1
1
1
1
2
1
1
3
3
1
1
4
6
4
1
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Binomial Distribution

Remarkably, this pattern is also the one that governs the possibilities of
tossing n coins:
1
1
1
1
1
1
2
3
4
1
3
6
1
4
1
n
0
1
2
3
4
2n
1
2
4
8
16
n
n!
   (n choose x)
x !(n - x)!  x 
With 3 coins, there are 8 ways for them to land, as shown above.
In general, there are 2n possible ways for n coins to land.
How many permutations are there for a given row, above, e.g. how
many permutations for getting 1 head and 2 tails? Obviously, 3.
How many permutation for x heads and n - x tails, for general n and x?
Number of combinations in each row: C (n, x) 

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Probability

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With fair coins, tossing a coin will result in equal chance of 50%, or ½, of
its ending up heads. Let us call this probability p. Obviously, the
probability of tossing a tails, q, is q = (1 - p).
With 3 coins, the probability of getting any single one of the
combinations is 1/2n = 1/8th, (since there are 8 combinations, and each is
equally probable). This comes from (½) (½) (½), or the product of each
probability p = ½ to get a heads.
If we want to know the probability of getting, say 1 heads and 2 tails,
we just need to multiply the probability of any combination (1/8th) by the
number of ways of getting 1 heads and 2 tails, i.e. 3, for a total
probability of 3/8.
To be really general, say the coins were not fair, so p ≠ q. Then the
probability to get heads, tails, tails would be (p)(q)(q) = p1q2.
Finally the probability P(x; n, p) of getting x heads given n coins each of
n!
which has probability p, is P( x; n, p)   n  p x q n - x 
p x (1 - p)n - x .
 
x !(n - x)!
 x
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Binomial Distribution

This is the binomial distribution, which we write PB:
n!
p x (1 - p) n- x .
x !(n - x)!
Let’s see if it works. For 1 heads with a toss of 3 fair coins, x = 1, n = 3, p
= ½, we get
3! 1 1 1 2
PB (1;3, 12 ) 
 3 / 8.
1!(3 - 1)! 2 2
For no heads, and all tails, we get
3! 1 0 1 3
Note: 0!  1
PB (0;3, 12 ) 
2 2  1/ 8.
0!(3)!
Say the coins are not fair, but p = ¼. Then the probability of 2 heads
and 1 tails is:
3!
11 3 2
PB (2;3, 14 ) 
4 4  3  9 / 64  27 / 64.
2!(3 - 2)!
You’ll show for homework that the sum of all probabilities for this (and
any) case is 1, i.e. the probabilities are normalized.
PB ( x; n, p) 

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Binomial Distribution

To see the connection of this to the sum of two variables raised to a
power, replace a and b with p and q: ( p  q )0  1
( p  q )1  p  q
( p  q ) 2  p 2  2 pq  q 2
( p  q )3  p 3  3 p 2 q  3 pq 2  q 3
Since p + q = 1, each of these powers also equals one on the left side,
while the right side expresses how the probabilities are split among the
different combinations. When p = q = ½, for example, the binomial
1
1
triangle
becomes
1/
1/
1
1
2
2
1
2
1
2
1
/4
/4 1/4
1/
3/
3
1/
1
3
3
1
8
8 /8
8
1
4
6
4
1
4
6
4
1
/16 /16 /16 /16 1/16
 In MatLAB, use binopdf(x,n,p) to calculate one row of this triangle,
e.g. binopdf(0:3,3,0.5) prints 0.125, 0.375, 0.375, 0.125.

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Binomial Distribution
Let’s say we toss 10 coins, and ask how many
heads we will see. The 10th row of the
triangle would be plotted as at right.
 The binomial distribution applies to yes/no
cases, i.e. cases where you want to know the
probability of something happening, vs. it not
happening.
 Say we want to know the probability of
getting a 1, rolling five 6-sided dice. Then p =
1/6 (the probability of rolling a 1 on one die),
and q = 1 – p = 5/6 (the probability of NOT
rolling a 1). The binomial distribution applies
to this case, with PB(x,5,1/6). The plot is
shown at right.
Binomial PDF for 10 coins
0.4
>> binopdf(0:5,5,1/6.)
ans = 0.4019 0.4019 0.1608 0.0322 0.0032 0.0001
Binomial Distribution

0.3
0.2
0.1
0
0
2
4
6
8
10
x
PDF for rolling a 1 on five 6-sided dice
Binomial Distribution
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
Number of 1's rolled
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4
5
Binomial Distribution Mean
Let’s say we toss 10 coins N = 100 times.
Then we would multiple the PDF by N, to find
out how many times we would have x number
of heads.
 The mean of the distribution is, as before:
1
 n
m  lim   xi    xPB ( x; n, p)
N  N

 x 0
Binomial PDF for 10 coins
40
0.4
Binomial Distribution

30
0.3
0.2
20
m
0.1
10
000
2
4
6
8
10
x
n
n!
p x (1 - p) n - x  np
x !(n - x)!
x 0
 For 10 coins, with p = ½, we get m = np = 5.
 For 5 dice, with p = 1/6, we get m = np = 5/6.
x
PDF for rolling a 1 on five 6-sided dice
Binomial Distribution
0.5
0.4
0.3
0.2
0.1
0
0
m
1
2
3
Number of 1's rolled
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4
5
Binomial Standard Deviation
The standard deviation of the distribution is
the “second moment,” given by the variance:
n
2
2
1
2
  lim    xi - m      x - m  PB ( x; n, p )
N  N

 x 0
n
n!
2
 x - m
p x (1 - p ) n - x  np (1 - p )
x !(n - x)!
x 0
For 10 coins, with p = ½, we get
  np(1 - p)  2.5  1.58
 For 5 dice, with p = 1/6, we get
Binomial Distribution

Binomial PDF for 10 coins
40
0.4

30
0.3
0.2
20
m
0.1
10
000
2
4
6
8
10
x

Binomial Distribution
  np(1 - p)  25 / 36  0.83
PDF for rolling a 1 on five 6-sided dice
0.5
0.4
0.3

0.2
0.1
0
0
m
1
2
3
Number of 1's rolled
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4
5
Summary of Binomial Distribution

The binomial distribution is PB: PB ( x; n, p) 

The mean is m  np.

The standard deviation is   np(1 - p)
n!
p x (1 - p) n- x .
x !(n - x)!
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Poisson Distribution
An approximation to the binomial distribution is very useful for the case
where n is very large (i.e. rolls with a die with infinite number of sides?)
and p is very small—called the Poisson distribution.
 This is the case of counting experiments, such as the decay of radioactive
material, or measuring photons in low light level.
 To derive it, start with the binomial distribution with n large and p << 1,
but with a well defined mean m = np. Then
1 n!
PB ( x n; n, p 1) 
p x (1 - p) n - x .
x ! (n - x)!
n!
 n x . because x is small, so most of the terms cancel
 The term
(n - x)!

leaving a total of x terms each approximately equal to n.
 This gives
1
mx
x
n- x
PB ( x n; n, p 1)   np  (1 - p) 
(1 - p) - x (1 - p) n .
x!
x!
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Poisson Distribution
Now, the term (1 – p)-x  1, for small p, and with some algebra we can
show that the term (1 – p)n  e-m.
 Thus, the final Poisson distribution depends only on x and m, and is defined
as
m x -m
PP ( x; m ) 
e .
x!
 The text shows that the expectation value of x (i.e. the mean) is

n
x  x
mx
e- m m.
x!
Remarkably, the standard deviation is given by the second moment as
x 0

n
  (x - m)   (x - m)
2
2
2
mx
e - m  m.
x!
These are a little tedious to prove, but all we need for now is to know that
the standard deviation is the square-root of the mean.
x 0

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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
m
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
m
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 describe further next time.
February 12, 2010