Transcript Statistics

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?
Relative probability
of getting N sixes
in a toss of 12 dice.
0
1
2
3
4
5
6
7
8
9
10
11
12
244140625 / 2176782336
585937500
644531250
429687500
Log P
193359375
61875000
14437500
2475000
309375
27500
1650
60
1
0
1
2
3
4
5
6
7
8
9
10
11
12
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 background
impinging 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
1
A reading of
could result from the lucky capture of an exceeding
rare 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 ± 1? ± 2?
A count of 37
37 ± at least a few?
A count of 1000
1000 ± ?
The probability of a single COSMIC RAY passing
through a small area of a detector
within a small interval of time Dt
can be very small:
p << 1
•cosmic rays arrive at a fairly stable, regular rate
when 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
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
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
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
The probability of a single COSMIC RAY passing
through a small area of a detector
within a small interval of time Dt
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/msec
=0.0000012/nsec
The probability of a single COSMIC RAY passing
through a small area of a detector
within a small interval of time Dt
can be very small:
p << 1
The probability of NO cosmic rays passing
through that area during that interval Dt is
A. p
D.( p - 1)
B. p2
C. 2p
E. ( 1 - p)
The probability of a single COSMIC RAY passing
through a small area of a detector
within a small interval of time Dt
can be very small:
p << 1
If the probability of one cosmic ray passing
during a particular nanosec is
P(1) = p << 1
the probability of 2 passing within the same
nanosec must be
A. p
D.( p - 1)
B. p2
C. 2p
E. ( 1 - p)
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”
???
× ( 1 - p )N-n
??? “misses”
N-n“misses”
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
)
N-n =
??? ln (1-p)
ln (1-p)N-n
(N-n)
ln (1-p)N-n = (N-n) ln (1-p)
2
3
4
x
x
x
ln( 1  x )  x 

2
3
4
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
P(n)
N!
=
n! ( N - n)!
pn ( 1 - p )N-n
P(n)
N!
=
n! ( N - n)!
pn e-p(N-n)
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)!
pn e-pN
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
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!
n
( Np
Np ) -Np
P(n) =
e
n!
Hey! What does Np represent?
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 / n! = 1/(???)
n
( Np ) n
n!
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
what’s this?
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
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)
(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
P(4) = 0.195366816
P(5) = 0.156293453
P(6) = 0.104195635
P(7) = 0.059540363
Probability of Observing N Events When the Average Expected Counts
Should Be 1
0.4
0.35
0.3
m=1
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
6
7
8
9
10
Probability of Observing N Events When the Average Count Expected Should Be 4
0.25
Probability
0.2
m=4
0.15
0.1
0.05
0
0
1
2
3
4
5
6
7
8
9
10
Number of Events Counted
Probability of Observing N Events When the Average Count Expected Should Be 8
0.16
0.14
0.12
m=8
Probability
0.1
0.08
0.06
0.04
0.02
0
0
1
2
3
4
5
6
Number of Events Counted
7
8
9
10
Another abbreviation (notation):
mean, m = x (the average x value)
i.e.
1 N
x   n 1 x n
N
3 different
distributions
with the
same mean
describe the spread in data
by a calculation of
the average distance
each individual data point
is from the overall mean
N
s=
(xi – m)2
i=1
N-1
mean, m
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
or
s  (x - m )
2
for short
2
The standard deviation s should provide
an estimate of the error in such counts
n - n
2
2
 n - 2n n  n
 n - 2n n  n
2
2
2
2
2
 n - 2n  n  n - n
2
 n -m
2
2
2
2
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
What is n2 for a Poisson distribution?
n e
2
-m
 me

m
n
(
n
1
)
!
n 1

-m

n
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

j
What is n2 for a Poisson distribution?

j
m
2
-m
n  me
 ( j  1) ( j) !
j 0
  mj
-m
 me  j

 j  0 ( j ) !


m

(
j
)
!

j 0


j
This is just
n  me
2
-m

m m
2
em again!
me
m

e
m

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
In other words
s 2= m
s = m
2
Assuming any measurement N usually
gives a result very close to the true
m
the best estimate of error
for the reading
is
N
We express that statistical error
in our measurement as
N ± N
1000
Cosmic
Ray
Rate
(Hz)
500
0
Time of day
How many pages of text are there in the new
Harry Potter and the Order of the Phoenix?
870
What’s the error on that number?
A. 0
B. 1
C.  2
D.  /870  29.5
E.  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
64
8
2
100
10
In summarizing,
what if we want to
combine these results?
164
 18
? ??
But think: adding assumes
each independent experiment
just happened to
fluctuate in the same way
Fluctuations must be random…they don’t conspire together!
How different is combining these two experiments
from running a single, longer uninterrupted run?
1
64
8
2
100
10
164
 18 ??
Run
Events Counted
Error
0
164
12.8
?
 8 2 + 10 2
12.8
164  12.8
What if these runs were of different lengths in time?
How do you compare the rates from each?
Run Elapsed Time Events Counted Error
8
1
10 minutes
64
Rate
6.4  0.8
?? /min
10
?? /min
5.0  0.5
2
20 minutes
100
How do errors COMPOUND?
d  Dd
velocity = t  Dt
Can’t add
Dd + Dt
=
or even
d
 ???
t
(Dd)2 + (Dt)2
• the units don’t match!
• it ignores whether we’re talking about
km/hr , m/sec , mi/min , ft/sec , etc
That question provides a clue
on how we handle these errors
How do we know it scales
correctly for any of those?
Look at:
x  vt
taking derivatives:
divide
by:
Fixes
units!
dx  tdv  vdt
x
vt
v  x/t
or
vt
dx dv dt


x
v
t
dx x
dv  - 2 dt
t t
v
x/t
x/t
dv dx dt

v
x
t
And we certainly don’t expect
Though we still shouldn’t be
simply adding the random errors. two separate errors to magically
cancel each other out.
Whether multiplying or dividing,
we add the relative errors in quadrature
(taking the square root of the sum of the squares)
2
dx
 dv   dt 
    
x
 v   t 
2
2
dv
 dx   dt 
    
v
 x  t 
2
What about a
rate - background
calculation?
( R  DR) - ( B  DB)  R - B  (DR - DB)
The units match nicely!
but
we can’t guarantee that
the errors will cancel!
Once again:
( DR)  ( DB)
2
2
-dE/dx = (4pNoz2e4/mev2)(Z/A)[ln{2mev2/I(1-b2)}-b2]
I = mean excitation (ionization) potential of atoms in target ~ Z10 GeV
-dE/dx [ MeV·g-1cm2 ]
10
8
6
4
Minimum
Ionizing:
3
2
1
0.01
1 – 1.5 MeV2
g/cm
0.1
1.0
10
100
Muon momentum [ GeV/c ]
1000
A typical gamma detector
has a light-sensitive
photomultiplier attached
to a small NaI crystal.
The scinitillator responds
to the dE/dx of each
MIP track
passing through
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
Emeasured  N tracks  EMIP
avg
Measuring energy in a calorimeter is a counting experiment governed
by the statistical fluctuations expected in counting random events.
Since E
 Ntracks
and DN
= N
we should expect DE
 E
and the relative error
DE
E

E
E
1
=
E
DE = AE
a constant that
characterizes the resolution
of a calorimeter