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A pocketful of change…10 pennies!...are tossed up in the air.About how many heads (tails) do you expect when they land?
What’s the probability of no heads at all in the batch?
What’s the probability on all heads?
What’s the probability of finding just one head?
Relative Probability of N heads in 10 flips of a coin
0 1 2 3 4 5 6 7 8 9 10
12 dice are rolled.About how many 6s do you expect when they land?
What’s the probability of no sixes at all in the batch?
What’s the probability on all sixes?
What’s the probability of finding just one six?
0 1 2 3 4 5 6 7 8 9 10 11 12
Relative probability of getting N sixesin a toss of 12 dice.
Lo
g P
0 1 2 3 4 5 6 7 8 9 10 11 12
244140625
585937500
644531250
429687500
193359375
61875000
14437500
2475000
309375
27500
1650
60
1
/ 2176782336
The counts for RANDOM EVENTS fluctuate
•Geiger-Meuller tubes clicking in response to a radioactive source
•Oscilloscope “triggering” on a cosmic ray signal
Cosmic rays form a steady backgroundimpinging on the earth
equally from all directions
measured rates NOT literally CONSTANT
long term averages are just reliably consistent
These rates ARE measurably affected by•Time of day•Direction of sky•Weather conditions
You set up an experiment to observe some phenomena…and run that experiment
for some (long) fixed time…but observe nothing: You count ZERO events.
What does that mean?
If you observe 1 event in 1 hour of running
Can you conclude the phenomena has a ~1/hour rate of occurring?
Random events arrive independently •unaffected by previous occurrences•unpredictably
0 sec time
A reading of 1 could result from the lucky capture of an exceedingrare event better represented by a much lower rate
(~0?).
or the run period could have just missed an event (starting a moment too late or ending too soon).
A count of 1 could represent a real average
as low as 0 or as much as 2
1 ± 1
A count of 2
2 ± A count of 37
37 ±
1? ± 2?
at least a few?
A count of 1000
1000 ± ?
The probability of a single COSMIC RAY passingthrough a small area of a detector within a small interval of time t
can be very small:
p << 1
•cosmic rays arrive at a fairly stable, regular ratewhen averaged over long periods
•the rate is not constant nanosec by nanosec or even second by second
•this average, though, expresses the probability per unit time of a cosmic ray’s passage
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
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
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
The probability of a single COSMIC RAY passingthrough a small area of a detector within a small interval of time t
can be very small:
p << 1
for example (even for a fairly large surface area) 72000/min=1200/sec =1200/1000 millisec =1.2/millisec = 0.0012/sec =0.0000012/nsec
The probability of a single COSMIC RAY passingthrough a small area of a detector within a small interval of time t
can be very small:
p << 1
The probability of NO cosmic rays passingthrough that area during that interval t is
A. p B. p2 C. 2p
D.( p 1) E. ( p)
The probability of a single COSMIC RAY passingthrough a small area of a detector within a small interval of time t
can be very small:
p << 1
If the probability of one cosmic ray passing during a particular nanosec is
P(1) = p << 1the probability of 2 passing within the samenanosec must be
A. p B. p2 C. 2p
D.( p 1) E. ( p)
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”
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
ln (1p)N-n = ln (1p) ???ln (1p)N-n = (Nn) ln (1p)
ln (1p)N-n = (Nn) ln (1p)
and since p << 1
ln (1p)
4
x
3
x
2
xx)x1ln(
432
p
ln (1p)N-n = (Nn) (p)
from the basic definition of a logarithmthis means
e???? = ???? e-p(N-n) = (1p)N-n
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
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
P(n) = pn e-pN )!N( !
!N
nn
P(n) = pn e-pN !
N
n
n
P(n) = e-Np !
) N (
n
p n
P(n) = e-Np !
) N (
n
p n
Hey! What does Np represent?
Np
!4!3!21
432 xxxxex
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/(???)
, 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?
, mean
1
N
)!1(
)N(
n
np
n
pe
0
N
)!(
)N( )(N
m
mp
m
pp e
= (Np) eNp eNp
= Np
= 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)
P(n) = e4 !
) 4 (
n
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(4) =P(1) = P(5) = P(2) = P(6) = P(3) = P(7) =
e-4 = 0.018315639
0.0183156390.0732625560.146525112
0.195366816
0.195366816 0.156293453 0.104195635 0.059540363
Probability of Observing N Events When the Average Count Expected Should Be 4
0
0.05
0.1
0.15
0.2
0.25
0 1 2 3 4 5 6 7 8 9 10
Number of Events Counted
Pro
bab
ilit
y
Probability of Observing N Events When the Average Count Expected Should Be 8
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0 1 2 3 4 5 6 7 8 9 10
Number of Events Counted
Pro
bab
ilit
y
Probability of Observing N Events When the Average Expected Counts Should Be 1
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0 1 2 3 4 5 6 7 8 9 10
=1
=4
=8
Another abbreviation (notation):
mean, = x (the average x value)
i.e.
N1n nx
N
1x
3 different distributions
with the same mean
mean,
describe the spread in data by a calculation of
the average distance each individual data point is from the overall mean
(xi – )2
N-1
= i=1
N
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
22 )x(
or
for short
The standard deviation should provide an estimate of the error in such counts
222 2 nnnnnn
22
22222
22
2
2
n
nnnnn
nnnn
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
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
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 standard deviation should provide an estimate of the error in such counts
22
222
nnn
In other words
2 =
=
Assuming any measurement N usuallygives a result very close to the true
the best estimate of errorfor the reading
is
N
We express that statistical error in our measurement as
N ± N
Cosmic Ray Rate(Hz)
Time of day
1000
500
0
How many pages of text are there in the newHarry Potter and the Order of the Phoenix?
870
What’s the error on that number?
A. 0
B. 1 C. 2 D. /870 29.5E. 870/2 = 435
A punted football has a hang time of 5.2 seconds.What is the error on that number?
Scintillator is sanded/polished to a final thickness of 2.50 cm. What is the error on that number?
You count events during two independent runs of an experiment.
Run Events Counted Error
1
2
64
100
8
10
In summarizing, what if we want to
combine these results?
164 18 ???
But think: adding assumeseach independent experiment
just happened to fluctuate in the same way
Fluctuations must be random…they don’t conspire together!
1
2
64
100
8
10
164 18 ??
How different is combining these two experiments from running a single, longer uninterrupted run?
0 164 12.8Run Events Counted Error
?
8 + 10 12.8 2 2
8.12164
1
2
64
100
8
10
What if these runs were of different lengths in time?How do you compare the rates from each?
Run Elapsed Time Events Counted Error Rate
20 minutes
10 minutes 6.4 0.8 /min
5.0 0.5 /min??
??
velocity = d dt t
= ???d t
How do errors COMPOUND?
Can’t add d + t or even (d)2 + (t)2
• the units don’t match!
• it ignores whether we’re talking about km/hr , m/sec , mi/min , ft/sec , etc
How do we know it scales correctly for any of those?
That question provides a clue on how we handle these errors
vtx txv /
Look at:
or
taking derivatives:
vdttdvdx dtt
x
t
dxdv
2
t
dt
v
dv
x
dx
divide by: x vt vt
v x/t x/t
t
dt
x
dx
v
dv Fixesunits!
Though we still shouldn’t besimply adding the random errors.
And we certainly don’t expecttwo separate errors to magically
cancel each other out.
22
t
dt
v
dv
x
dx
22
t
dt
x
dx
v
dv
Whether multiplying or dividing, we add the relative errors in quadrature
(taking the square root of the sum of the squares)
What about a
rate background
calculation?
)()()( BRBRBBRR
butwe can’t guarantee thatthe errors will cancel!
The units match nicely!
22 )()( BR Once again:
108
6
4
3
2
1
-dE
/dx
[ MeV
·g-1cm
2 ]
Muon momentum [ GeV/c ]0.01 0.1 1.0 10 100 1000
1 – 1.5 MeVg/cm2
Minimum Ionizing:
-dE/dx = (4Noz2e4/mev2)(Z/A)[ln{2mev2/I(1-2)}-2] I = mean excitation (ionization) potential of atoms in target ~ Z10 GeV
The scinitillator responds to the dE/dx of each
MIP track passing through
A typical gamma detectorhas a light-sensitive
photomultiplier attachedto a small NaI crystal.
If an incoming particle initiates a shower,each track segment (averaging an interaction length)will leave behind an ionization trail with about the same energy deposition.
The total signal strength Number of track segments
Basically avg
MIPtracksmeasuredENE
Measuring energy in a calorimeter is a counting experiment governed by the statistical fluctuations expected in counting random events.
Since E Ntracks and N = N
we should expect E E
and the relative errorE E 1
E E E =
E = AE
a constant that characterizes the resolution
of a calorimeter