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

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.

Log P /

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 sectime  A reading of 1 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 ± A count of ± 1? ± 2? at least a few? A count of ± ?

The probability of a single COSMIC RAY passing through 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 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

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 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 events C. 3 or 4 events D. about 10 events E. 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 events C. 3 or 4 events D. about 10 events E. 100s of events F. 1,000s of events 1200 Hz = 1200/sec Example: a measured rate of

The probability of a single COSMIC RAY passing through 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 = /  sec = /nsec

The probability of a single COSMIC RAY passing through 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 passing through that area during that interval  t is A. p B. p 2 C. 2p D.( p  1) E. (  p)

The probability of a single COSMIC RAY passing through 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 << 1 the probability of 2 passing within the same nanosec must be A. p B. p 2 C. 2p D.( p  1) E. (  p)

The probability of a single COSMIC RAY passing through a small area of a detector within a small interval of time  t 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 = N  t ) what is the probability that we observe exactly n events? × ( 1  p ) ??? ??? “misses” p n n “hits” × ( 1  p ) N-n N-n“misses”

While waiting N successive intervals (where the total time is t = N  t ) what is the probability that we observe exactly n events? P(n) = n C N p n ( 1  p ) N-n ln (1  p) N-n = ln (1  p) ??? ln (1  p) N-n = (N  n) ln (1  p)

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 ???? = ???? e -p(N-n) = (1  p) N-n

P(n) = p n ( 1  p ) N-n P(n) = p n e -p(N-n) P(n) = p n e -pN If we have to wait a large number of intervals, N, for a relatively small number of counts,n n<<N

P(n) = p n e -pN And since N - (n-1)  N (N) (N) … (N) = N n for n<<N

P(n) = p n e -pN P(n) = e -Np

Hey! What does Np represent? NpNp

, mean = n=0 term n / n! = 1/(???)

, mean let m = n  1 i.e., n =      0 N )!( )N( )(N m m p m p p e what’s this?

, mean  = (Np) e  Np e Np  = Np

P(n) = e  Poisson distribution probability of finding exactly n events within time t when the events occur randomly, but at an average rate of   (events per unit time)

P(n) = e  4 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 =

 =1  =4  =8

Another abbreviation (notation): mean,  = x (the average x value) i.e.

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  (x i –  ) 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. or for short

The standard deviation  should provide an estimate of the error in such counts

What is n 2 for a Poisson distribution? first term in the series is zero factor out e  which is independent of n

What is n 2 for a Poisson distribution? Factor out a  like before Let j = n-1  n = j+1

What is n 2 for a Poisson distribution? This is just e  again!

The standard deviation  should provide an estimate of the error in such counts In other words  2 =   = 

Assuming any measurement N usually gives a result very close to the true  the best estimate of error for the reading is N We express that statistical error in our measurement as N ± N

Cosmic Ray Rate (Hz) 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 88  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!

88   18 ?? How different is combining these two experiments from running a single, longer uninterrupted run?  12.8 Run Events Counted Error ?  2 2

8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   d t   t =  ??? dtdt 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

Look at: or taking derivatives: divide by: x vt v x/t Fixes units! Though we still shouldn’t be simply adding the random errors. And we certainly don’t expect 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)

What about a rate  background calculation? but we can’t guarantee that the errors will cancel! The units match nicely! Once again:

dE/dx [ MeV·g -1 cm 2 ] Muon momentum [ GeV/c ] – 1.5 MeV g/cm 2 Minimum Ionizing: -dE/dx = (4  N o z 2 e 4 /m e v 2 )(Z/A)[ ln {2m e v 2 /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 detector has a light-sensitive photomultiplier attached to 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 Measuring energy in a calorimeter is a counting experiment governed by the statistical fluctuations expected in counting random events.

Since E  N tracks and  N =  N we should expect  E   E and the relative error  E  E 1 E E  E  =  E = A  E a constant that characterizes the resolution of a calorimeter