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Counting Random Events
Example: a measured rate of
1200 Hz = 1200/sec
would mean in 5 minutes we
should expect to count about
A. 6,000 events
B. 12,000 events
C. 72,000 events D. 360,000 events
E. 480,000 events F. 720,000 events
The Cosmic Ray Observatory Project
High Energy Physics Group
The University of Nebraska-Lincoln
Counting Random Events
Example: a measured rate of
1200 Hz = 1200/sec
would mean in 3 millisec we
should expect to count about
A. 0 events
B. 1 or 2 events
C. 3 or 4 events D. about 10 events
E. 100s of events F. 1,000s of events
1 millisec = 10-3 second
The Cosmic Ray Observatory Project
High Energy Physics Group
The University of Nebraska-Lincoln
Counting Random Events
Example: a measured rate of
1200 Hz = 1200/sec
would mean in 100 nanosec we
should expect to count about
A. 0 events
B. 1 or 2 events
C. 3 or 4 events D. about 10 events
E. 100s of events F. 1,000s of events
1 nanosec = 10-9 second
The Cosmic Ray Observatory Project
High Energy Physics Group
The University of Nebraska-Lincoln
Counting Random Events
The probability of a single COSMIC RAY passing
through a small area of a detector
within a small interval of time Dt is
p << 1
the probability
that none pass in
that period is
(1-p)1
While waiting N successive intervals
(where the total time is t = NDt )
what is the probability that we observe
exactly n events?
pn
n “hits”
The Cosmic Ray Observatory Project
???
× ( 1 - p )N-n
??? “misses”
N-n“misses”
High Energy Physics Group
The University of Nebraska-Lincoln
Counting Random Events
While waiting N successive intervals
(where the total time is t = NDt )
what is the probability that we observe
exactly n events?
P(n)
N!
= nn
! ( N -N
n)!
C
n
p
(1-p
N-n
)
From the properties of logarithms
you just reviewed
??? ln (1-p)
ln (1-p)N-n = (N-n)
ln x  loge x
e=2.718281828
The Cosmic Ray Observatory Project
High Energy Physics Group
The University of Nebraska-Lincoln
Counting Random Events
ln (1-p)N-n = (N-n) ln (1-p)
N ( N - 1) 2 N ( N - 1)( N - 2) 3
ln(1  x)  1  Nx 
x 
x 
2!
3!
N
and since p << 1
ln (1-p)  - p
ln (1-p)N-n = (N-n) (-p)
from the basic definition of a logarithm
this means
???? == ????
e-p(N-n)
(1-p)N-n
The Cosmic Ray Observatory Project
High Energy Physics Group
The University of Nebraska-Lincoln
Counting Random Events
P(n)
N!
=
n! ( N - n)!
pn ( 1 - p )N-n
P(n)
N!
=
n! ( N - n)!
-p(N-n)
n
p
e
If we have to wait
a large number of intervals, N, for a
relatively small number of counts,n
n<<N
P(n)
N!
= n! ( N - n)!
The Cosmic Ray Observatory Project
-pN
n
p
High Energy Physics Group
e
The University of Nebraska-Lincoln
Counting Random Events
P(n)
N!
=
n! ( N - n)!
-pN
n
p
e
And since
N!
 N (N - 1) (N - 2)  (N - n  1)
( N - n)!
N - (n-1)
 N (N) (N) … (N) = Nn
for n<<N
The Cosmic Ray Observatory Project
High Energy Physics Group
The University of Nebraska-Lincoln
Counting Random Events
P(n)
N!
=
n! ( N - n)!
-pN
n
p
e
n
N n -pN
P(n) =
p e
n!
n
( Np ) -Np
P(n) =
e
n!
The Cosmic Ray Observatory Project
High Energy Physics Group
The University of Nebraska-Lincoln
Counting Random Events
n
( Np ) -Np
P(n) =
e
n!
If the average rate of some random event is
p = 24/min = 24/60 sec = 0.4/sec
what is the probability of recording n events
in 10 seconds?
P(0) =
P(1) =
P(2) =
P(3) =
The Cosmic Ray Observatory Project
P(4) =
P(5) =
P(6) =
P(7) =
High Energy Physics Group
The University of Nebraska-Lincoln
Counting Random Events
(4)
P(n) =
n!
n
4
e
e-4 = 0.018315639
If the average rate of some random event is
p = 24/min = 24/60 sec = 0.4/sec
what is the probability of recording n events
in 10 seconds?
P(0) = 0.018315639
P(1) = 0.073262556
P(2) = 0.146525112
P(3) = 0.195366816
The Cosmic Ray Observatory Project
P(4) = 0.195366816
P(5) = 0.156293453
P(6) = 0.104195635
P(7) = 0.059540363
High Energy Physics Group
The University of Nebraska-Lincoln
Counting Random Events
n
( Np ) -Np
P(n) =
e
n!
Hey! What does Np represent?
The Cosmic Ray Observatory Project
High Energy Physics Group
The University of Nebraska-Lincoln
Counting Random Events
Another useful series we can exploit
2
3
4
x
x
x
ex  1 x    
2! 3! 4!
 xn
ex  
n!
n0


n 0
n 0
m, mean =  n P(n)  

 0 

n 1
e - Np
n
( Np ) n
n!
e - Np
n
( Np) n
n!
n=0 term
 0  e - Np


n 1
n
( Np ) n
n!
n / n! = 1/(???)
The Cosmic Ray Observatory Project
High Energy Physics Group
The University of Nebraska-Lincoln
Counting Random Events
2
3
4
x
x
x
ex  1 x    
2! 3! 4!
 xn
ex  
n!
n0
m, mean

e
- Np


n 1
 (Np ) e
- Np
( Np ) n
(n - 1)!


n 1
 (Np ) e
- Np
( Np)??
(n - 1)!


n 1
 (Np ) e
- Np
( Np ) n -1
(n - 1)!


m
(
N
p
)
let m = n-1
i.e., n(m=)!
m 0
The Cosmic Ray Observatory Project
High Energy Physics Group
The University of Nebraska-Lincoln
what’s this?
Counting Random Events
m, mean

e
- Np


n 1
 (Np) e- Np
( Np ) n
(n - 1)!


m 0
( Np) m
(m)!
m = (Np) e-Np eNp
m = Np
The Cosmic Ray Observatory Project
High Energy Physics Group
The University of Nebraska-Lincoln
Counting Random Events
m = Np
m e-m
n
P(n) =
n!
Poisson distribution
probability of finding exactly n
events within time t when the events
occur randomly, but at an average
rate of m (events per unit time)
The Cosmic Ray Observatory Project
High Energy Physics Group
The University of Nebraska-Lincoln
Counting Random Events
Recall: The standard deviation s is
a measure of the mean (or average)
spread of data away from its own
mean. It should provide an estimate
of the error on such counts.
N
1
2
s   ( xi - m )
N i 1
2
The Cosmic Ray Observatory Project
High Energy Physics Group
The University of Nebraska-Lincoln
Counting Random Events
The standard deviation s should provide
an estimate of the error in such counts
n - n
2
2
 n - 2n n  n
2
 n - 2n n  n
2
2
2
2
2
2
 n - 2n  n  n - n
2
n -m
2
The Cosmic Ray Observatory Project
High Energy Physics Group
The University of Nebraska-Lincoln
Counting Random Events
What is n2 for a Poisson distribution?


m -m
m
-m
n  n
e  n
e
n
!
(
n
1
)
!
n 0
n 1
2

n
2

n
first term in the series is zero
e
-m

m
n
(
n
1
)
!
n 1

n
factor out e-m which is independent of n
The Cosmic Ray Observatory Project
High Energy Physics Group
The University of Nebraska-Lincoln
Counting Random Events
What is n2 for a Poisson distribution?
n e
2
-m
 me

m
n
(
n
1
)
!
n 1
n

-m

Factor out a
m like before
n -1
m
n
(
n
1
)
!
n 1

Let j = n-1  n = j+1
 me
-m

m
( j  1)
(
j
)
!
j 0

The Cosmic Ray Observatory Project
High Energy Physics Group
j
The University of Nebraska-Lincoln
Counting Random Events
What is n2 for a Poisson distribution?
n  me
2
-m

m
( j  1)
( j) !
j 0
j

  mj
-m
 me  j

 j 0 ( j ) !

The Cosmic Ray Observatory Project
High Energy Physics Group

m

(
j
)
!

j 0


j
The University of Nebraska-Lincoln
Counting Random Events
What is n2 for a Poisson distribution?
n  me
2
-m

m
( j  1)
(
j
)
!
j 0
j

  mj
-m
 me  j

 j 0 ( j ) !

m 

(
j
)
!

j 0

j

This is just
n  me
-m
2

em again!
me
m

e
m

m m
2
The Cosmic Ray Observatory Project
High Energy Physics Group
The University of Nebraska-Lincoln
Counting Random Events
The standard deviation s should provide
an estimate of the error in such counts
n - n 
2
n -m
2
2
m m -m m
2
2
In other words
s
2=
m
s = m
The Cosmic Ray Observatory Project
High Energy Physics Group
The University of Nebraska-Lincoln