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