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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 257 34 "Nuclear decays: Poisson s
tatistics" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
388 "We consider sequences of nuclear decay data as collected by a Gei
ger counter. The data are shown to obey Poisson statistics, which repr
esents evidence for the fact that the events originate from independen
t spontaneous decays. In the limit of a large sample of data (very hig
h count rate) the distribution of events becomes Gaussian. A detailed \+
discussion can be found in John R. Taylor: " }{TEXT 256 33 "An Introdu
ction to Error Analysis" }{TEXT -1 185 ", chapters 10,11,12 (Universit
y Science Books 1982, 2nd ed. 1998). For readers who are less familiar
 with probability distributions we begin with a section on the binomia
l distribution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 8 "restart;" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 309 
21 "Binomial distribution" }}{EXCHG {PARA 0 "" 0 "" {TEXT 258 0 "" }
{TEXT -1 1666 "We begin with the binomial distribution to illustrate t
he statistics of a familiar experiment with several possible discrete \+
outcomes (throwing of dice, etc.), before proceeding with nuclear deca
ys. Any physics measurement (or indeed any measurement at all) results
 in a distribution of results (think of the repeated measurement of th
e period of a pendulum, or the measurement of the height of the female
 population in Canada, etc.). Usually we think that in physics measure
ments there should be a unique answer:  in the case of the pendulum pe
riod we might think that only one outcome is possible in the measureme
nt, while for the second example we clearly expect a distribution with
 a well-defined average and spread of data around the average. Neverth
eless, both situations have something in common: due to measurement er
rors we will obtain a distribution of results in the first case as wel
l (we can think of using an electronic stopwatch, and realize that at \+
some level of precision we will start to investigate our reaction time
). Even in automated experiments (using a light beam/photodetector set
-up) the numerical result will differ from measurement to measurement \+
when the readout is precise. The remaining question then is to underst
and the properties of the distribution of values in the limit that the
 number of measurements taken is very large, or infinite. This is the \+
main reason for studying the mathematical properties of distributions.
 The distribution will be able to tell us the probability for achievin
g a certain result in a particular measurement. The knowledge of the l
atter is crucial while assessing the validity of a certain measurement
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 342 "The
 distribution used most often in any experimental science is the Gauss
ian (or normal) distribution. It applies when a measurement is subject
 to many sources of small and random errors. It is a limiting distribu
tion in the sense that an experiment can be said to follow this distri
bution only in the limit of a large number of measurements." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 390 "The binomial d
istribution (and also the Poisson distribution discussed later) applie
s to finite sequences of data points. Let us illustrate it by the exam
ple of throwing three dice, and asking the question of finding the num
ber of times that a particular face appears (e.g., let us call an ace \+
when the face with six points appears). Obviously, we can have for a g
iven throw the outcome of " }{TEXT 19 10 "nu=0,1,2,3" }{TEXT -1 6 " ac
es." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 331 "T
his situation is easily figured out: the probability to throw one ace \+
with a single die is 1/6 as there are six faces, and we assume that th
e die is perfect. Whether a die is perfect or not, can later be assess
ed by observing whether its runs are following the correct distributio
n. The same is true for a random number generator." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 82 "Clearly the probability t
o throw three aces with three independet dies is given by" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "P3w3:=(1/6)^3; evalf(%);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 373 "This corresponds to a probability
 of about than 0.5 percent. Now we figure out the probability for two \+
aces. It should be the product of throwing two aces, and not throwing \+
a third one, i.e., we use the fact that not throwing an ace with one o
f the dies is given by the complement of 1/6 with 1. Naively, we might
 think that the answer would be given by the simple product:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "wrongP2w3:=(1/6)^2*(1-1/6); \+
evalf(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 379 "This answer corresp
onds to the case that the first two dice show aces, and the third one \+
does not. The problem is that unlike in the first case, where the thre
e dice were undistinguishable (they showed the same result), we can ha
ve three different configurations now: suppose we mark by A a face cal
led ace, and by B the 'no ace' face; then we can have with equal proba
bilities (" }{TEXT 267 1 "A" }{TEXT -1 2 ", " }{TEXT 266 1 "A" }{TEXT 
-1 2 ", " }{TEXT 265 1 "B" }{TEXT -1 4 "), (" }{TEXT 264 1 "A" }{TEXT 
-1 2 ", " }{TEXT 263 1 "B" }{TEXT -1 2 ", " }{TEXT 262 1 "A" }{TEXT 
-1 8 "), and (" }{TEXT 261 1 "B" }{TEXT -1 2 ", " }{TEXT 260 1 "A" }
{TEXT -1 2 ", " }{TEXT 259 1 "A" }{TEXT -1 215 "). The probability for
 any one of these three configurations is the one calculated above by \+
the simple product. Why is this so? One should think about the likelih
ood of the outcome of any such configuration, e.g., (" }{TEXT 270 1 "A
" }{TEXT -1 2 ", " }{TEXT 269 1 "B" }{TEXT -1 2 ", " }{TEXT 268 1 "A" 
}{TEXT -1 511 "). Its probability is given by the product of probabili
ties to dial an ace with the first die (1/6) times the probability of \+
not getting an ace with the second die (5/6), times the probability of
 getting one with the third die (1/6). This example shows that we need
 to be careful in probability theory, and that it is always easiest to
 find the probabilities for the most detailed (exclusive) information.
 The more inclusive information is obtained by summing over the possib
ilities that make up the statement. " }}{PARA 0 "" 0 "" {TEXT -1 47 "T
herefore, we have a probability of almost 7 %:" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 34 "P2w3:=3*(1/6)^2*(1-1/6); evalf(%);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 113 "Now we can figure out the remaining prob
abilities: we consider all the explicit possibilites for dialing one a
ce:" }}{PARA 0 "" 0 "" {TEXT -1 1 "(" }{TEXT 279 1 "A" }{TEXT -1 2 ", \+
" }{TEXT 278 1 "B" }{TEXT -1 2 ", " }{TEXT 277 1 "B" }{TEXT -1 4 "), (
" }{TEXT 276 1 "B" }{TEXT -1 2 ", " }{TEXT 275 1 "A" }{TEXT -1 2 ", " 
}{TEXT 274 1 "B" }{TEXT -1 4 "), (" }{TEXT 273 1 "B" }{TEXT -1 2 ", " 
}{TEXT 272 1 "B" }{TEXT -1 2 ", " }{TEXT 271 1 "A" }{TEXT -1 57 ") - t
his gives us the same multiplicity as with two aces:" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 34 "P1w3:=3*(1-1/6)^2*(1/6); evalf(%);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "Finally we have an easy one again:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "P0w3:=(5/6)^3; evalf(%)
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 93 "We can check that the probab
ilities sum to unity (otherwise these wouldn't be probabilities!)" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "P0w3+P1w3+P2w3+P3w3;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "A graphical representation of thes
e results can be obtained by vertical bars:" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 120 "plot([[0,P0w3*t,t=0..1],[1,P1w3*t,t=0..1],[2,P2w3
*t,t=0..1],[3,P3w3*t,t=0..1]],color=[red,blue,green,black],axes=boxed)
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 280 12 "Exercise B1:" }}{PARA 0 "" 
0 "" {TEXT -1 90 "Repeat the above steps with four dice, i.e., find th
e probabilites for 0, 1, 2, 3, 4 aces." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 88 "The general form of the binom
ial distribution is specified by the following parameters: " }}{PARA 
0 "" 0 "" {TEXT -1 36 "- the number of independent trials: " }{TEXT 
282 1 "n" }{TEXT -1 11 " (throwing " }{TEXT 281 1 "n" }{TEXT -1 15 " d
ice, tossing " }{TEXT 283 1 "n" }{TEXT -1 19 " coins, looking at " }
{TEXT 284 1 "n" }{TEXT -1 37 " electrons in an atomic shell, etc.)." }
}{PARA 0 "" 0 "" {TEXT -1 27 "- the number of successes: " }{TEXT 19 
2 "nu" }{TEXT -1 77 " (getting an ace, getting a head on a coin; havin
g an electron ionized, etc.)" }}{PARA 0 "" 0 "" {TEXT -1 30 "- the pro
bability of success: " }{TEXT 287 1 "p" }{TEXT -1 18 ", and of failure
: " }{TEXT 286 1 "q" }{TEXT -1 3 "=1-" }{TEXT 285 1 "p" }{TEXT -1 19 "
 in any one trial (" }{TEXT 288 1 "p" }{TEXT -1 34 "=1/6 for getting a
n ace on a die; " }{TEXT 289 1 "p" }{TEXT -1 130 "=1/2 for getting hea
ds on a coin; probability of ionization of an electron in a given atom
ic shell by an X ray of a given energy)." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 "The probability for " }{TEXT 19 2 
"nu" }{TEXT -1 14 " successes in " }{TEXT 290 1 "n" }{TEXT -1 53 " tri
als has to be proportional to the nu-th power of " }{TEXT 292 1 "p" }
{TEXT -1 29 ", and the (n-nu)-th power of " }{TEXT 291 1 "q" }{TEXT 
-1 136 ". The combinatorial factor that counts the number of possibili
tes can be figured out by using a mathematical identity. We know that \+
1 = " }{TEXT 294 1 "p" }{TEXT -1 1 "+" }{TEXT 293 1 "q" }{TEXT -1 39 "
, and can raise this expression to the " }{TEXT 295 1 "n" }{TEXT -1 
35 "th power. Examples are given below:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "expand(1^4=(p+q)^3);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "expand(1^4=(p+q)^4);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 60 "We recognize the following from our example discussed abo
ve:" }}{PARA 0 "" 0 "" {TEXT -1 34 "- the expansion of the expression \+
" }{TEXT 19 7 "(p+q)^n" }{TEXT -1 116 " involves the required powers o
f single-event probabilities for the desired success rates for all pos
sible outcomes;" }}{PARA 0 "" 0 "" {TEXT -1 119 "- the sum of the term
s on the RHS equals one, i.e., we can think of the identity as summing
 all possible probabilities;" }}{PARA 0 "" 0 "" {TEXT -1 6 "- for " }
{TEXT 296 1 "n" }{TEXT -1 8 "=3 (and " }{TEXT 297 1 "n" }{TEXT -1 130 
"=4 in the exercise) the binomial coefficients count the number of con
figurations for a given scenario as counted explicitly above." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 "Maple has
 a built-in function" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "bin
omial(4,2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 87 "but it is straight
forward to define the function known from the mathematical expansion:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "bc:=(n,nu)->n!/(nu!*(n-
nu)!);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "bc(4,2);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "Armed with this coefficient we can
 define the binomial distribution:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 27 "bc(n,nu)*p^nu*(1-p)^(n-nu);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 22 "BD:=unapply(%,p,n,nu);" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 55 "For a given case (given single-event probability value \+
" }{TEXT 299 1 "p" }{TEXT -1 28 " and fixed number of trials " }{TEXT 
298 1 "n" }{TEXT -1 37 ") we can graph the probabilities for " }{TEXT 
19 2 "nu" }{TEXT -1 11 " successes:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 9 "ps:=0.35;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
176 "plot([[0,BD(ps,4,0)*t,t=0..1],[1,BD(ps,4,1)*t,t=0..1],[2,BD(ps,4,
2)*t,t=0..1],[3,BD(ps,4,3)*t,t=0..1],[4,BD(ps,4,4)*t,t=0..1]],color=[r
ed,blue,green,brown,magenta],axes=boxed);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT 300 12 "Exercise B2:" }}{PARA 0 "" 0 "" {TEXT -1 64 "Explore the
 possibilites of the binomial distribution for fixed " }{TEXT 301 1 "n
" }{TEXT -1 44 "=4 by changing the single-event probability " }{TEXT 
19 2 "ps" }{TEXT -1 40 " in the above example. What happens for " }
{TEXT 19 6 "ps=0.5" }{TEXT -1 19 "? Explore also for " }{TEXT 302 1 "n
" }{TEXT -1 55 "=5. Graph the probabilities for all possible successes
 " }{TEXT 19 7 "nu=0..n" }{TEXT -1 47 " as a function of the single-ev
ent probability." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 77 "There are further properties that one should explo
re. The general expression " }{TEXT 19 13 "BD(ps, n, nu)" }{TEXT -1 
27 " gives the probability for " }{TEXT 19 2 "nu" }{TEXT -1 14 " succe
sses in " }{TEXT 19 1 "n" }{TEXT -1 51 " trials for given single-trial
 success probability " }{TEXT 19 2 "ps" }{TEXT -1 84 ". One can ask wh
at the average number for success is. This is obtained by averaging " 
}{TEXT 19 2 "nu" }{TEXT -1 37 " probabilistically, i.e., by summing " 
}{TEXT 19 2 "nu" }{TEXT -1 46 " with the probability as a weight coeff
icient." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
78 "We try a few examples (to anticipate the result of a mathematical \+
derivation)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "N:=5; sum(n
u*BD(p,N,nu),nu=0..N); simplify(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 51 "The average number of successes is simply given by " }{TEXT 
303 1 "n" }{TEXT -1 346 " times the single-event probability. Sometime
s this quantity is all that one is interested in. On the other hand on
e can argue that this is all that is required from a measurement, and \+
that the individual success probabilities can be reconstructed via the
 deduced single-trial probability (assuming that the experiment obeys \+
binomial statistics)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 51 "The standard deviation for the number of successes " }
{TEXT 19 2 "nu" }{TEXT -1 123 " can also be calculated. It is defined \+
as follows: sigma-squared is given as the average of the square of the
 deviation of " }{TEXT 19 2 "nu" }{TEXT -1 18 " from the average." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "nubar:=N*p;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "sigsq:=simplify(sum((nu-nubar)^2*BD
(p,N,nu),nu=0..N));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "We can gen
eralize this finding (try other values of " }{TEXT 304 1 "N" }{TEXT 
-1 89 "). It is also possible to show that this result is identical to
 the following difference:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
47 "simplify(sum(nu^2*BD(p,N,nu),nu=0..N))-nubar^2;" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 95 "The quantity sigma (or its square) measures the
 width of the distribution of probabilities for " }{XPPEDIT 18 0 "nu;
" "6#%#nuG" }{TEXT -1 275 " successes. The two results for the average
 and the width of the binomial distribution can be used to observe whi
ch parameters to choose in a Gaussian approximation to the binomial di
stribution. The latter becomes a valid approximation to the binomial d
istribution for large " }{TEXT 305 1 "n" }{TEXT -1 2 ")." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with
(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "N:=7; ps:=0.4;
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "P1:=plot([seq([nu,BD(ps
,N,nu)*t,t=0..1],nu=0..N)],color=blue):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 38 "nubar:=ps*N; sigma:=sqrt(N*ps*(1-ps));" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "P2:=plot(exp(-(nu-nubar)^2/(2*sigma
^2))/(sqrt(2*Pi)*sigma),nu=-0.1..8,color=red):" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 26 "display(P1,P2,axes=boxed);" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 153 "The comparison shows that the discrete binomial \+
distribution can be approximated reasonably by a continuous Gaussian d
istribution evaluated for discrete " }{TEXT 19 2 "nu" }{TEXT -1 31 " v
alues for moderate values of " }{TEXT 306 1 "n" }{TEXT -1 9 " already.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 742 "We c
an also understand now how the two seemingly different tasks can be pu
t on a common footing: examples of measurements where we obviously exp
ect a broad distribution (height of some species population with avera
ge and deviation the meaningful quantities), and others where we are s
eeking a 'single number' only. In the latter case the 'single number' \+
is given by the average of the distribution, and the deviation becomes
 a measure for the uncertainty in the measured result. While we would \+
like in an ideal world the uncertainty to be negligible compared to th
e 'single number', it is essential to determine the uncertainty from t
he distribution of repeated measurements in order to assess the qualit
y (accuracy) of the measured quantity." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 324 "To really understand the concept of \+
the probability distribution it is essential to analyze sequences of r
eal data. We encourage the reader to perform 'real' experiments with c
oins, dies, etc.. It is possible to analyze sequences produced with Ma
ple's random number generator (whose quality falls short of state-of-t
he-art)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "die:=rand(1..6)
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "die();" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 49 "We generate a sequence of tosses with thr
ee dies:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "M:=200; n:=3;" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "S1:=[seq([die(),die(),die
()],i=1..M)]:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "S1[1],S1[2
];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "We can analyze the presence
 of particular faces:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "F:=
6;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "DieAnalysis:=proc(S) \+
local nu,C,i,toss,isucc,j; global F,M,n,p_s;" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 44 "for nu from 0 to n do: C[nu]:=0; od: p_s:=0;" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "for i from 1 to M do: toss:=S[i]: i
succ:=0:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 " for j from 1 to n do:
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "  if toss[j]=F then isucc:=isuc
c+1; p_s:=p_s+1; fi;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 " od: C[isuc
c]:=C[isucc]+1;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "od: p_s:=evalf(p
_s/(M*n)); print(\"Single-event probability: \",p_s);" }}{PARA 0 "> " 
0 "" {MPLTEXT 1 0 35 "[seq(evalf(C[nu]/M),nu=0..n)]; end:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "DieAnalysis(S1);" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 15 "ps:=evalf(1/6);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 27 "[seq(BD(ps,n,nu),nu=0..n)];" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 28 "[seq(BD(p_s,n,nu),nu=0..n)];" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 78 "We appear to have results that are consis
tent with the binomial distribution. " }}{PARA 0 "" 0 "" {TEXT -1 323 
"We note that in addition to measuring directly from the sample the nu
-fold event probabilities we have the option to measure the single-eve
nt probability, and to generate nu-fold event probabilities from the b
inomial distribution. This allows one to 'beat' the statistics, i.e., \+
to generate predictions for unlikely events." }}{PARA 0 "" 0 "" {TEXT 
-1 211 "We leave it as an exercise to consider longer sequences of tos
ses to observe whether the consistency improves with the sample size. \+
We can also look at the distribution of frequencies of success for oth
er faces:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "F:=1;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "DieAnalysis(S1);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "[seq(BD(p_s,n,nu),nu=0..n)];" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "F:=2;" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 16 "DieAnalysis(S1);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "[seq(BD(p_s,n,nu),nu=0..n)];" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 337 "The scatter in the answers for the different faces de
monstrates that we have to be very careful when drawing conclusions fr
om measurements with small sample sizes (low number of repetitions)! E
ven the single-event probabilities vary substantially. The deviations \+
will shrink only slowly when the sample size is increased (typically w
ith " }{TEXT 19 9 "1/sqrt(M)" }{TEXT -1 3 " )." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 307 12 "Exercise B3:" }}{PARA 0 
"" 0 "" {TEXT -1 256 "Determine the single-event probabilities for two
 different faces. Determine whether the measurement from samples appro
aches the expected value of 1/6 when the sample size is increased by d
ifferent factors (e.g., 2, 10, 50). Is the approach consistent with " 
}{TEXT 19 9 "1/sqrt(M)" }{TEXT -1 11 " behaviour?" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 308 12 "Exercise B4:" }}{PARA 0 
"" 0 "" {TEXT -1 82 "Analyze the case where four dies (or some other n
umber) are tossed simultaneously." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 3 "" 0 "
" {TEXT 320 42 "Poisson distribution with chi-squared test" }}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 1402 "The Poisson distribution is discrete (a
s the binomial distribution) and applies to the following case: suppos
e a random process occurs, such as the decay of radioactive nuclei in \+
a sample. We cannot predict at what time any single nucleus will under
go a transition, but we do know that nuclei in the sample decay with s
ome rate, i.e., a certain number of events is expected per unit time. \+
If we pick some time interval, we expect on average a certain number o
f decayed atoms (a fractional number as it is an average). In reality,
 when we record time sequences, we obtain different numbers of decays \+
in such intervals. This is due to the fact that the number of decays i
s somehow distributed around the average. The Poisson distribution is \+
used in this situation as a simplified version of the binomial distrib
ution. It depends on some average rate, and a chosen time interval, an
d a decay is considered to be a success. The reason why we can simplif
y from the binomial distribution is that the number of trials is extre
mely large (a fraction of Avogadro's number), while the probability fo
r success (decay of an individual nucleus) is extremely small (of the \+
order of the inverse of Avogadro's number per unit time). Thus, we can
 ignore the factor for the probability of no successn raised to a huge
 power (it is practically one), and use a mathematical property for th
e binomial coefficient for large " }{TEXT 310 1 "n" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 37 "The Poisson distribution is given as:" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "PD:=(mu,nu)->exp(-mu)*mu^nu
/nu!;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 5 "Here " }{TEXT 19 2 "mu" }
{TEXT -1 71 " is the expected mean numberof counts in the time interva
l chosen, and " }{TEXT 19 2 "nu" }{TEXT -1 58 " labels the possibiliti
es of observing 0, 1, 2,... counts." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 525 "Note that in an actual experiment with \+
a Geiger-Mueller counter one does not observe all decays, but a fracti
on thereof (depending on the efficiency of the tube for particular for
ms of radiation; the count rate, which leads to a certain dead time fo
r the tube, the supply voltage on the tube, etc.). This does not alter
, however, the fact that one can determine the statistical nature of t
he radiation, as one simply looks at a subsample of the actual radiati
on sequence (one is also limited by the direction of observation)." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 156 "Why is m
u the mean (or average) count number for the chosen time interval? Let
 us calculate the probabilistic average number of observed decays (suc
cesses):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "mu:='mu':" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "simplify(sum(nu*PD(mu,nu),nu
=1..infinity));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "simplify
(sum(PD(mu,nu),nu=0..infinity));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
84 "Note that the sum started at one, as no contribution is made towar
ds the average by " }{XPPEDIT 18 0 "nu = 0;" "6#/%#nuG\"\"!" }{TEXT 
-1 108 ". The mathematical steps required in the calculation (done for
 us by Maple) are shown in J.R. Taylor's book." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 54 "The procedure when analyz
ing a sequence is as follows:" }}{PARA 0 "" 0 "" {TEXT -1 425 "First o
ne can determine the average count rate, by taking the total number of
 decays and divide by the total time interval (which is subdivided int
o a certain number of sub-intervals). Then one arrives at an average n
umber of decays per subinterval. Finally one forms a histogram by reco
rding the frequency of intervals with 0, 1 , 2, ... counts. This histo
gram is to be compared with the predictions of the PD for the given " 
}{TEXT 19 2 "mu" }{TEXT -1 112 " value. It is possible to re-analyze t
he same decay sequence by using a different choice of sub-interval len
gth." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 65 "L
et us look at the Poisson distribution for two cases, one where " }
{XPPEDIT 18 0 "mu < 1;" "6#2%#muG\"\"\"" }{TEXT -1 148 ", for which th
e number of zero-count intervals dominates, and another one, where one
 can see how the Gaussian distribution emerges as a limit again." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "mu:=0.6;" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 68 "plot([seq([nu,PD(mu,nu)*t,t=0..1],nu=0..10)],color=
blue,axes=boxed);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "mu:=3.4
;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "plot([seq([nu,PD(mu,nu)*t,t=0.
.1],nu=0..10)],color=blue,axes=boxed);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 291 "It would make sense to add to the graph an indication of
 the mu value just to show the location of the average number of decay
s per chosen interval for the asymmetric distribution. To find out whi
ch Gaussian distribution one can use in the limiting case, one needs t
o find out the deviation." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
57 "sigsq:=simplify(sum((nu-mu)^2*PD(mu,nu),nu=0..infinity));" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "Again, this can also be calculated
 from:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "simplify(sum(nu^2
*PD(mu,nu),nu=0..infinity)-mu^2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
65 "This result may appear to be puzzling: the deviation is given as \+
" }{TEXT 19 8 "sqrt(mu)" }{TEXT -1 77 "! Indeed, if one lists the aver
age with deviation as an interval one obtains " }{XPPEDIT 18 0 "[mu-sq
rt(mu), mu+sqrt(mu)];" "6#7$,&%#muG\"\"\"-%%sqrtG6#F%!\"\",&F%F&-F(6#F
%F&" }{TEXT -1 128 " . While using a larger count interval we do also \+
obtain a larger uncertainty, which seems improper. The fractional unce
rtainty " }{XPPEDIT 18 0 "sqrt(mu)/mu;" "6#*&-%%sqrtG6#%#muG\"\"\"F'!
\"\"" }{TEXT -1 53 ", however, is reduced for an increased time interv
al." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "mu:=8.3;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "P1:=plo
t([seq([nu,PD(mu,nu)*t,t=0..1],nu=0..20)],color=blue):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "sigma:=sqrt(mu);" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 79 "P2:=plot(exp(-(nu-mu)^2/(2*sigma^2))/(sqr
t(2*Pi)*sigma),nu=-0.1..20,color=red):" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 26 "display(P1,P2,axes=boxed);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 299 "Note that there is a systematic tendency for the Gaussian dist
ribution to underestimate the count probability to the left of the ave
rage. The Poisson distribution is asymmetric, and this asymmetry becom
es negligible only for sufficiently large average count probability fo
r the chosen time interval." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 389 "To properly understand the Poisson distributio
n one should analyze data runs from a Geiger counter. These are usuall
y computer-interfaced in undergraduate laboratories today, which has t
he drawback that the software performs all the data analysis, and the \+
students usually no longer manipulate the raw data. For this reason we
 use Maple to generate event sequences which are then analyzed." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 357 "A sequen
ce of events, corresponds to the sequence of clicks that one hears as \+
the Geiger counter picks up the radiation (either from natural backgro
und, i.e., mostly cosmic radiation, or from a small radioactive source
, such as a ceramic plate or tile that has been coated with a radiacti
ve layer containing, e.g., uranium oxide, to enhance the brightness).
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 150 "We u
se a random number generator to produce a time sequence of events; we \+
assume that a tick represents a second, and generate events at a given
 rate." }}{PARA 0 "" 0 "" {TEXT -1 224 "A random number generator for \+
integers in the interval [0,100] is used in conjunction with a given r
ate to decide in a time unit whether an observable decay occurs or not
. A decay is recorded as a one, a non-decay as a zero." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "rnd:=rand(0..100);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "rate:=1/10;" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 11 "tmax:=3000;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 112 "TS:=[]: for i from 1 to tmax do: if rnd() <= 100*rat
e then TS:=[op(TS),[i,1]]: else  TS:=[op(TS),[i,0]]: fi: od:" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "We graph a subset of the count dat
a:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "plot([seq(TS[i],i=1..
200)],style=point);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 302 "This sequ
ence could have come from a Geiger counter. Note the irregularity: the
re are stretches with no counts followed by segments where the counts \+
appear to cluster. This irregularity is noticable when listening to th
e clicks from a Geiger counter. Now we pick a time interval for the de
cay analysis:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "tint:=25; \+
# length of interval in seconds (time units)" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 73 "We can determine the expected average number of events \+
per time interval:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "nu_ba
r_exp:=tint*rate;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "Nt:=tm
ax/tint; # number of time intervals contained in sample" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 123 "Now we need to generate the histogram (f
requency distribution): for each time interval we find out how many ev
ents occured." }}{PARA 0 "" 0 "" {TEXT -1 157 "We begin by resetting t
he counters for the frequency. Assume that 10 events during the interv
al is the maximum (increase it, if the program below complains):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "Emax:=15;" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 54 "for icnt from 0 to Emax do: C[icnt]:=0; od:
 myrate:=0:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 240 "for i from 1 to Nt \+
do: icnt:=0: for j from 1 to tint do: if TS[(i-1)*tint+j][2]=1 then ic
nt:=icnt+1; myrate:=myrate+1; fi; od: if icnt <= Emax then C[icnt]:=C[
icnt]+1; else print(\"Increase Emax, please\"); fi; od: myrate:=evalf(
myrate/tmax);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "mymu:=myra
te*tint;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 138 "To compare the found
 results with the Poisson distribution we have to convert the probabil
ities calculated from the Poissonian to numbers." }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 72 "P2:=plot([seq([nu-0.05,Nt*PD(mymu,nu)*t,t=0..
1],nu=0..Emax)],color=red):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "P1:=
plot([seq([icnt+0.05,C[icnt]*t,t=0..1],icnt=0..Emax)],color=blue): dis
play(P1,P2,axes=boxed);" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 39 "We can make the following observations:" }}{PARA 0 "" 0 "
" {TEXT -1 200 "1) for the chosen moderate number of recordings (event
/no-event per second) we can find that the simulation does well on the
 event rate, and on the expected average number of events per time int
erval." }}{PARA 0 "" 0 "" {TEXT -1 156 "2) a reasonable result is obta
ined for the entire distribution; note that the Poissonian drawn for c
omparison is based on input from the data run (variable " }{TEXT 19 4 
"mymu" }{TEXT -1 368 " for the average number of events per time inter
val), and not the predetermined rate used to generate the sequence. Su
bstantial discrepancies occur for individual bins in the histogram. Th
is should not cause too much of an alarm: the Poisson distribution rep
resents a limiting probability distribution that should be reached in \+
the limit of many performed experiments." }}{PARA 0 "" 0 "" {TEXT -1 
237 "3) We should explore further the consistency of the data with a P
oisson distribution: either by considering larger data sets, or by per
forming a systematic test on the hypothesis (chi-squared test). The la
tter is attempted further below." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 311 12 "Exercise P1:" }}{PARA 0 "" 0 "" {TEXT 
-1 65 "Explore the above data by choosing different subinterval length
s " }{TEXT 19 4 "tint" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT 312 12 "Exercise P2:" }}{PARA 0 "" 0 "" 
{TEXT -1 94 "Repeat the above steps for cases where the decay rate is \+
increased and decreased respectively." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 423 "We now consider an interesti
ng additional check, namely a test of the hypothesis that the distribu
tion generated from the independent events is indeed compatible with a
 Poisson distribution. A chi-squared test of a hypothesis is a quantit
ative measure that essentially compares the data distributions as done
 in the above graph by summing the squares of the discrepancies, and t
hen turning the result into a useful measure." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 51 "In the graph above we hav
e used the data to deduce " }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 
91 " value for the Poissonian, and then generated a distribution of ex
pected values to observe " }{XPPEDIT 18 0 "nu = 0 .. infinity;" "6#/%#
nuG;\"\"!%)infinityG" }{TEXT -1 486 " events per time interval. (A lim
itation of the Poissonian is perhaps noticable here: our simulation, a
nd any Geiger counter sequence has a maximum number of observable even
ts, namely when the counter is clicking all the time, while the Poisso
nian distributes the probability over an infinite interval; this is th
e price to be paid for the replacement of a binomial by a Poisson dist
ribution; in practice this shouldn't be a problem, as the predictions \+
for high event counts are small)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 178 "We can calculate the chi-squared for the
 results obtained above (make sure that the statements that produce th
e graph above were executed) from the expected and observed results:" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "chi_sq:=add((Nt*PD(mymu,n
u)-C[nu])^2/(Nt*PD(mymu,nu)),nu=0..Emax);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 131 "Is this an acceptable value or not? This depends on the \+
number of degrees of freedom. Here we have a little problem: should we
 use " }{TEXT 19 4 "Emax" }{TEXT -1 195 ", which includes data where w
e have no counts and small expected probabilities, or not? After all, \+
there was some arbitrariness in choosing this variable, which acts as \+
a cut-off for the diagram." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 103 "The criterion for acceptability is that chi-squa
red be much less than the number of degrees of freedom " }{TEXT 314 1 
"d" }{TEXT -1 76 "; rejection of the hypothesis is given when chi-squa
red is much bigger than " }{TEXT 313 1 "d" }{TEXT -1 42 ". The questio
n is whether we can say that " }{TEXT 315 1 "d" }{TEXT -1 5 "=16 (" }
{TEXT 19 7 "=Emax+1" }{TEXT -1 2 ")?" }}{PARA 0 "" 0 "" {TEXT -1 82 "J
udging from the graph the right attitude should be to pick the 10 visi
ble points:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "chi_sq:=add(
(Nt*PD(mymu,nu)-C[nu])^2/(Nt*PD(mymu,nu)),nu=0..9);" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 515 "The conclusion based on the limited data set w
ould be that the data are not inconsistent with a Poisson distribution
, and that further testing is desirable. Further testing means to run \+
with bigger data sets. An important issue when performing these tests \+
is to choose reasonable time intervals (which depend on the actual cou
nt rate). The chi-squared comparison has to have a reasonable number o
f contributing data points (say six or more to account for two degrees
 of freedom to be subtracted as explained below)." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 86 "One should also be a bit \+
more precise when assessing the number of degrees of freedom " }{TEXT 
316 1 "d" }{TEXT -1 275 ". We have simply taken the number of non-negl
igible theoretical data points (we could have used the number of data \+
bins). It should be kept in mind that we really have one less independ
ent value, as the data were used to infer the average number of decays
 per time interval (" }{TEXT 19 4 "mymu" }{TEXT -1 245 "). There can b
e other constraints (usually the number of data points; in the case of
 a Gaussian distribution also its deviation) which are used to determi
ne a reduced number of degrees of freedom. A reduced chi-squared is de
fined as chi-squared/" }{TEXT 317 1 "d" }{TEXT -1 217 ", and for this \+
quantity the borderline between acceptance and non-acceptance of the h
ypothesis is the value of one. It is evident that for acceptance we ne
ed to obtain much less than one, and vice versa for rejection." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 124 "A health
ier attitude for dealing with the open-ended Poisson distribution woul
d be to call the bin neighbouring the largest " }{XPPEDIT 18 0 "nu;" "
6#%#nuG" }{TEXT -1 100 " value that still received non-zero counts to \+
be one bin for all events beyond the largest observed " }{XPPEDIT 18 
0 "nu;" "6#%#nuG" }{TEXT -1 34 " value. In our case that would be " }
{TEXT 19 4 "nu=9" }{TEXT -1 23 " (corresponding now to " }{XPPEDIT 18 
0 "9 <= nu;" "6#1\"\"*%#nuG" }{TEXT -1 30 ", and receiving 0 counts) a
nd " }{TEXT 318 1 "d" }{TEXT -1 132 " would be 10-2=8, i.e., a borderl
ine result [note that the precise value in the latter statement can ch
ange for repeated data runs]." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 9 "chi_sq/8;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "Our own test wi
th the code above produced the following interesting result:" }}{PARA 
0 "" 0 "" {TEXT -1 171 "- for data runs with up to 3000 time ticks (at
 a decay rate of 1/10 per tick) consistency with a Poissonian was obse
rved (reduced chi-squared observed between 0.4 and 1.4)" }}{PARA 0 "" 
0 "" {TEXT -1 184 "- for data runs with 5000 time ticks and more the P
oisson hypothesis was not confirmed (reduced chi-squared often around \+
1.5 rather than falling further below one in a consistent way)." }}
{PARA 0 "" 0 "" {TEXT -1 236 "This allows one to conclude that at the \+
level of random number sequences in the hundreds of thousands the pseu
do-random number generator has problems with correlations, i.e., the r
andom numbers are not truly independent from each other." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 241 "We can also concl
ude that it is hard to test a hypothesis, i.e., we have to run long si
mulations, and we have to know that the tools are accurate enough. Sho
rt simulations are often sufficient in order to obtain a qualitative u
nderstanding. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 319 12 "Exercise P3:" }}{PARA 0 "" 0 "" {TEXT -1 95 "Explore the
 chi-squared test of the Poisson distribution hypothesis with differen
t count rates." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 
1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }