would mean in 5 minutes we should expect to count about 6,000 events B . 12,000 events
23
Ray Observatory Project High Energy Physics Group The University of Nebr Counting Random Events 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 1200 Hz = 1200/sec xample: a measured rate of
description
Example: a measured rate of. 1200 Hz = 1200 /sec. would mean in 5 minutes we should expect to count about 6,000 events B . 12,000 events C . 72,000 events D . 360,000 events E . 480,000 events F . 720,000 events. Example: a measured rate of. 1200 Hz = 1200 /sec. - PowerPoint PPT Presentation
Transcript of would mean in 5 minutes we should expect to count about 6,000 events B . 12,000 events
The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln
Counting Random Events
would mean in 5 minutes we should expect to count about A. 6,000 events B. 12,000
eventsC. 72,000 events D. 360,000 eventsE. 480,000 events F. 720,000 events
1200 Hz = 1200/sec
Example: a measured rate of
The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln
Counting Random Events
would mean in 3 millisec we should expect to count about A. 0 events B. 1 or 2 eventsC. 3 or 4 events D. about 10 eventsE. 100s of events F. 1,000s of events
1200 Hz = 1200/sec
Example: a measured rate of
1 millisec = 103 second
The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln
Counting Random Events
would mean in 100 nanosec we should expect to count about A. 0 events B. 1 or 2 eventsC. 3 or 4 events D. about 10 eventsE. 100s of events F. 1,000s of events
1200 Hz = 1200/sec
Example: a measured rate of
1 nanosec = 109 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 passingthrough a small area of a detector within a small interval of time t is
p << 1the probability
that none pass inthat period is
( 1 p ) 1
While waiting N successive intervals (where the total time is t = Nt ) what is the probability that we observe
exactly n events?
× ( 1 p )???
??? “misses” pn
n “hits”× ( 1 p )N-n
N-n“misses”
The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln
Counting Random Events
While waiting N successive intervals (where the total time is t = Nt )
what is the probability that we observe exactly n events?
P(n) = nCN pn ( 1 p )N-n )!N( !
!N
nn
From the properties of logarithmsyou just reviewed
ln (1p)N-n = ln (1p)
ln x loge xe=2.718281828
???ln (1p)N-n = (Nn) ln (1p)
The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln
Counting Random Events
ln (1p)N-n = (Nn) ln (1p)
and since p << 1
ln (1p)
32
!3
)2)(1(
!2
)1(1)1ln( x
NNNx
NNNxx N
p
ln (1p)N-n = (Nn) (p)
from the basic definition of a logarithmthis means
e???? = ???? e-p(N-n) = (1p)N-n
The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln
Counting Random Events
P(n) = pn ( 1 p )N-n
)!N( !
!N
nn
P(n) = pn e-p(N-n) )!N( !
!N
nn
P(n) = pn e-pN )!N( !
!N
nn
If we have to wait a large number of intervals, N, for a relatively small number of counts,n
n<<N
The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln
Counting Random Events
P(n) = pn e-pN )!N( !
!N
nn
1)n-(N 2)-(N 1)-(N N )!N(
!N
n
And since
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) = pn e-pN )!N( !
!N
nn
P(n) = pn e-pN !
N
n
n
P(n) = e-Np !
) N (
n
p n
The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln
Counting Random Events
P(n) = e-Np !
) N (
n
p n
If the average rate of some random event is p = 24/min = 24/60 sec = 0.4/secwhat is the probability of recording n events in 10 seconds?
P(0) = P(4) =P(1) = P(5) =P(2) = P(6) =P(3) = P(7) =
The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln
Counting Random Events
P(n) = e4 !
) 4 (
n
n
If the average rate of some random event is p = 24/min = 24/60 sec = 0.4/secwhat is the probability of recording n events in 10 seconds?
P(0) = 0.018315639 P(4) = 0.195366816P(1) = 0.073262556 P(5) = 0.156293453P(2) = 0.146525112 P(6) = 0.104195635P(3) = 0.195366816 P(7) = 0.059540363
e-4 = 0.018315639
The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln
Counting Random Events
P(n) = e-Np !
) N (
n
p 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
!4!3!2
1432 xxx
xex
0
! n
nx
n
xe
, mean = n
n
p
n
pn
ennn )N(
!)P(
0
N
0
n
n
pp
n
en )N(
! 0
1
N
n=0 termn
n
p pn
ne )N(
! 0
1
N
n / n! = 1/(???)
The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln
Counting Random Events
!4!3!2
1432 xxx
xex
0
! n
nx
n
xe
, mean
1
N
)!1(
)N(
n
np
n
pe
1
N
)!1(
??)N( )(N
n
p
n
pp e
1
1N
)!1(
)N( )(N
n
np
n
pp e
let m = n1i.e., n =
0
N
)!(
)N( )(N
m
mp
m
pp e
what’s this?
The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln
Counting Random Events
, mean
1
N
)!1(
)N(
n
np
n
pe
0
N
)!(
)N( )(N
m
mp
m
pp e
= (Np) eNp eNp
= Np
The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln
Counting Random Events
= Np
P(n) = e !n
n
Poisson distributionprobability of finding exactly n
events within time t when the eventsoccur randomly, but at an average rate of (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 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
iix
N 1
22 )(1
The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln
Counting Random Events
The standard deviation should provide an estimate of the error in such counts
222 2 nnnnnn
2
22222
22
2
2
n
nnnnn
nnnn
The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln
Counting Random Events
What is n2 for a Poisson distribution?
en
nen
nnn
n
n
n
10
22
! )1(!
first term in the series is zero
1! )1(
n
n
nne
factor out e which is independent of n
The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln
Counting Random Events
1
1
1
2
! )1(
! )1(
n
n
n
n
nn
nnn
e
e
What is n2 for a Poisson distribution?
Factor out a like before
Let j = n-1 n = j+1
0
! )()1(
j
j
jje
The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln
Counting Random Events
What is n2 for a Poisson distribution?
00
0
2
! )(! )(
! )()1(
j
j
j
j
j
j
jjj
jjn
e
e
The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln
Counting Random Events
What is n2 for a Poisson distribution?
00
0
2
! )(! )(
! )()1(
j
j
j
j
j
j
jjj
jjn
e
e
This is just
e again!
2
2
een e
The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln
Counting Random Events
The standard deviation should provide an estimate of the error in such counts
22
222
nnn
In other words
2 =
=
