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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(%);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%%P3w3G#\"\"\"\"$;#" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+I'H'HY!#7" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 373 "
This corresponds to a probability of about than 0.5 percent. Now we fi
gure out the probability for two aces. It should be the product of thr
owing two aces, and not throwing a third one, i.e., we use the fact th
at not throwing an ace with one of 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(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%*wrongP2w3G#\"\"&\"$;#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+:[
\"[J#!#6" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 379 "This answer correspo
nds to the case that the first two dice show aces, and the third one d
oes not. The problem is that unlike in the first case, where the three
 dice were undistinguishable (they showed the same result), we can hav
e three different configurations now: suppose we mark by A a face call
ed ace, and by B the 'no ace' face; then we can have with equal probab
ilities (" }{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(%);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%%P2w3G#\"\"&\"#s" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+WWWWp!#6" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 113 "N
ow we can figure out the remaining probabilities: we consider all the \+
explicit possibilites for dialing one ace:" }}{PARA 0 "" 0 "" {TEXT 
-1 1 "(" }{TEXT 279 1 "A" }{TEXT -1 2 ", " }{TEXT 278 1 "B" }{TEXT -1 
2 ", " }{TEXT 277 1 "B" }{TEXT -1 4 "), (" }{TEXT 276 1 "B" }{TEXT -1 
2 ", " }{TEXT 275 1 "A" }{TEXT -1 2 ", " }{TEXT 274 1 "B" }{TEXT -1 4 
"), (" }{TEXT 273 1 "B" }{TEXT -1 2 ", " }{TEXT 272 1 "B" }{TEXT -1 2 
", " }{TEXT 271 1 "A" }{TEXT -1 57 ") - this gives us the same multipl
icity as with two aces:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "
P1w3:=3*(1-1/6)^2*(1/6); evalf(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
>%%P1w3G#\"#D\"#s" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+AAAsM!#5" }}}
{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(%)
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%P0w3G#\"$D\"\"$;#" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#$\"+Pq.(y&!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 93 "We can check that the probabilities sum to unity (otherwise the
se wouldn't be probabilities!)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 20 "P0w3+P1w3+P2w3+P3w3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "A graphical representation of
 these results can be obtained by vertical bars:" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 120 "plot([[0,P0w3*t,t=0..1],[1,P1w3*t,t=0..1],[2
,P2w3*t,t=0..1],[3,P3w3*t,t=0..1]],color=[red,blue,green,black],axes=b
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u7$Fahl$\"1dM7SbIeQFeu7$Fahl$\"1oX[#='\\cRFeu7$Fahl$\"1V:vcP(Q0%Feu7$F
ahl$\"1f#fUj`L9%Feu7$Fahl$\"1)4#orw!fC%Feu7$Fahl$\"1U'3`ALwL%Feu7$Fahl
$\"1+](ozJaV%Feu7$Fahl$\"1c0o3v.HXFeu7$Fahl$\"1H'H'H'H'HYFeu-Fft6&FhtF
(F(F(-%+AXESLABELSG6$%!GFiam-%*AXESSTYLEG6#%$BOXG-%%VIEWG6$%(DEFAULTGF
abm" 1 2 0 1 0 2 9 1 2 2 1.000000 45.000000 45.000000 0 }}}}{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 the probabili
tes 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 binomial d
istribution 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 " dice, to
ssing " }{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; having \+
an electron ionized, etc.)" }}{PARA 0 "" 0 "" {TEXT -1 30 "- the proba
bility 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 " i
n any one trial (" }{TEXT 288 1 "p" }{TEXT -1 34 "=1/6 for getting an \+
ace on a die; " }{TEXT 289 1 "p" }{TEXT -1 130 "=1/2 for getting heads
 on a coin; probability of ionization of an electron in a given atomic
 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 " trials
 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);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#/\"\"\",**$)%\"pG\"\"$\"\"\"F$*&)F(\"\"#F*%\"qGF$F)*&F(F$)F.F-F*F)*
$)F.F)F*F$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "expand(1^4=(p
+q)^4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/\"\"\",,*$)%\"pG\"\"%\"\"
\"F$*&)F(\"\"$F*%\"qGF$F)*&)F(\"\"#F*)F.F1F*\"\"'*&F(F$)F.F-F*F)*$)F.F
)F*F$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "We recognize the followi
ng from our example discussed above:" }}{PARA 0 "" 0 "" {TEXT -1 34 "-
 the expansion of the expression " }{TEXT 19 7 "(p+q)^n" }{TEXT -1 
116 " involves the required powers of single-event probabilities for t
he desired success rates for all possible outcomes;" }}{PARA 0 "" 0 "
" {TEXT -1 119 "- the sum of the terms on the RHS equals one, i.e., we
 can think of the identity as summing all possible probabilities;" }}
{PARA 0 "" 0 "" {TEXT -1 6 "- for " }{TEXT 296 1 "n" }{TEXT -1 8 "=3 (
and " }{TEXT 297 1 "n" }{TEXT -1 130 "=4 in the exercise) the binomial
 coefficients count the number of configurations for a given scenario \+
as counted explicitly above." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 29 "Maple has a built-in function" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 14 "binomial(4,2);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"\"'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 87 "but it is
 straightforward to define the function known from the mathematical ex
pansion:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "bc:=(n,nu)->n!/
(nu!*(n-nu)!);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#bcGR6$%\"nG%#nuG6
\"6$%)operatorG%&arrowGF)*&-%*factorialG6#9$\"\"\"*&-F/6#9%\"\"\"-F/6#
,&F1\"\"\"F6!\"\"\"\"\"!\"\"F)F)F)" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "bc(4,2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"'" }}}
{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);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#*&*(-%*factorialG6#%\"nG\"\"\")%\"pG%#nuGF)),&F)F)F+!\"
\",&F(F)F,F/F)\"\"\"*&-F&6#F,\"\"\"-F&6#F0\"\"\"!\"\"" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "BD:=unapply(%,p,n,nu);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%#BDGR6%%\"pG%\"nG%#nuG6\"6$%)operatorG%&arrow
GF**&*(-%*factorialG6#9%\"\"\")9$9&F4),&F4F4F6!\"\",&F3F4F7F:F4\"\"\"*
&-F16#F7\"\"\"-F16#F;\"\"\"!\"\"F*F*F*" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 55 "For a given case (given single-event probability value " 
}{TEXT 299 1 "p" }{TEXT -1 28 " and fixed number of trials " }{TEXT 
298 1 "n" }{TEXT -1 37 ") we can graph the probabilities for " }{TEXT 
19 2 "nu" }{TEXT -1 11 " successes:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 9 "ps:=0.35;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#psG$\"
#N!\"#" }}}{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=[red,blue,green,brown,m
agenta],axes=boxed);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 
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\"1#Hdz*))oE&*F37$Fahl$\"1r<$RYg6w*F37$Fahl$\"1+vy\"[:m(**F37$Fahl$\"1
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^cD)G04\"Ffp7$Fahl$\"1+++++v96Ffp-Fgt6&Fit$\")#)eqkF\\u$\"))eqk\"F\\uF
gam-F$6$7S7$$\"\"%F(F(7$F]bm$\"1<a8Za$4F$!#>7$F]bm$\"1Y64!*)ep6'Fbbm7$
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u)F,7$F]bm$\"1h!RX)=Nh!*F,7$F]bm$\"1\"Gj)\\y_l$*F,7$F]bm$\"1OMx)RjKq*F
,7$F]bm$\"1ek3\"Qt1+\"F37$F]bm$\"1PM-D`2L5F37$F]bm$\"1?h5-jVi5F37$F]bm
$\"18yI0c`%4\"F37$F]bm$\"16M1[&QZ7\"F37$F]bm$\"1\"RDeK4j:\"F37$F]bm$\"
1@0qJb<(=\"F37$F]bm$\"1.2s)e*[>7F37$F]bm$\"1<zCm=h]7F37$F]bm$\"1N'4,*)
QCG\"F37$F]bm$\"1iY9qB+98F37$F]bm$\"1D1CIf+V8F37$F]bm$\"1x1;$GZiP\"F37
$F]bm$\"1<a[X(yfS\"F37$F]bm$\"1y+[&oywV\"F37$F]bm$\"1sh(Gd>!o9F37$F]bm
$\"1++++]i+:F3-Fgt6&FitFjtF(Fjt-%+AXESLABELSG6$%!GFe[n-%*AXESSTYLEG6#%
$BOXG-%%VIEWG6$%(DEFAULTGF]\\n" 1 2 0 1 0 2 9 1 2 2 1.000000 
45.000000 45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 300 12 "Exercis
e 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. G
raph the probabilities for all possible successes " }{TEXT 19 7 "nu=0.
.n" }{TEXT -1 47 " as a function of the single-event probability." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "T
here are further properties that one should explore. The general expre
ssion " }{TEXT 19 13 "BD(ps, n, nu)" }{TEXT -1 27 " gives the probabil
ity for " }{TEXT 19 2 "nu" }{TEXT -1 14 " successes 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 what the average number f
or 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 coefficient." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 78 "We try a few ex
amples (to anticipate the result of a mathematical derivation)." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "N:=5; sum(nu*BD(p,N,nu),nu=0
..N); simplify(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"NG\"\"&" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,,*&%\"pG\"\"\"),&F&F&F%!\"\"\"\"%\"\"
\"\"\"&*&)F%\"\"#F+)F(\"\"$F+\"#?*&)F%F1F+)F(F/F+\"#I*&)F%F*F+F(F&F2*$
)F%F,F+F," }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$%\"pG\"\"&" }}}{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 pr
obability. Sometimes this quantity is all that one is interested in. O
n the other hand one can argue that this is all that is required from \+
a measurement, and that the individual success probabilities can be re
constructed via the deduced single-trial probability (assuming that th
e experiment obeys binomial statistics)." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 51 "The standard deviation for the num
ber of successes " }{TEXT 19 2 "nu" }{TEXT -1 123 " can also be calcul
ated. 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 " fr
om the average." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "nubar:=N
*p;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&nubarG,$%\"pG\"\"&" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "sigsq:=simplify(sum((nu-nuba
r)^2*BD(p,N,nu),nu=0..N));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&sigsq
G,&%\"pG\"\"&*$)F&\"\"#\"\"\"!\"&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
52 "We can generalize 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;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,&%\"pG\"\"&*$)F$\"\"#\"\"\"!\"&" }}}
{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 resul
ts for the average and the width of the binomial distribution can be u
sed to observe which parameters to choose in a Gaussian approximation \+
to the binomial distribution. The latter becomes a valid approximation
 to the binomial distribution for large " }{TEXT 305 1 "n" }{TEXT -1 
2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
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\"1+]PiP1^<Fbio$\"1F]irlS=AFb\\l7$$\"1++Dj(pd#>Fbio$\"12XJ*>5<X#Fb\\l7
$$\"1***\\P7$>,@Fbio$\"15w,g;chEFb\\l7$$\"1++vOgCqAFbio$\"1`!>'y>FJGFb
\\l7$$\"1+DJxAwBCFbio$\"1//:M'H4&HFb\\l7$$\"1]7`1^.:DFbio$\"1zm914T/IF
b\\l7$$\"1++vNzI1EFbio$\"1%3#y53tVIFb\\l7$$\"1+]P%*zi$o#Fbio$\"1+147g_
lIFb\\l7$$\"1*****H0[4w#Fbio$\"1)fY]\")4l2$Fb\\l7$$\"1DJqd)=f!GFbio$\"
1b!eIUuy2$Fb\\l7$$\"1\\iSi'*)3&GFbio$\"1XB`(4Nb2$Fb\\l7$$\"1u$4rYge*GF
bio$\"147B!G+&pIFb\\l7$$\"1*\\7=FJ3%HFbio$\"1cE\")G<zfIFb\\l7$$\"1\\i:
O[W?IFbio$\"1WM3+4rLIFb\\l7$$\"1****\\+%e+5$Fbio$\"1?Q\"RqDl*HFb\\l7$$
\"1+D\"ozbZF$Fbio$\"1X+b\"[4#yGFb\\l7$$\"1+vVI!46W$Fbio$\"1_0W=:^BFFb
\\l7$$\"1+]7y6o9OFbio$\"1wggz_?EDFb\\l7$$\"1+D1$fuSx$Fbio$\"1]'\\F#\\o
?BFb\\l7$$\"1**\\PC>+YRFbio$\"1hO]ts5#3#Fb\\l7$$\"1+DJ-heCTFbio$\"1;GZ
_>*e#=Fb\\l7$$\"1*\\(=aT/!G%Fbio$\"1K(f</)o.;Fb\\l7$$\"1+]74J%zW%Fbio$
\"1v'fgAJ;P\"Fb\\l7$$\"1++]e\"*R@YFbio$\"1(G\"\\iMrY6Fb\\l7$$\"1**\\ii
?4\"z%Fbio$\"1\"QIDIW(e%*F[o7$$\"1+D\"Gny_&\\Fbio$\"1F:BV$eRs(F[o7$$\"
1+]P5*zv8&Fbio$\"1*z,0CUI0'F[o7$$\"1****\\!o&Q,`Fbio$\"1A2Wvh,\"y%F[o7
$$\"1++v.LGwaFbio$\"1Q6b)Q4:l$F[o7$$\"1*\\7`amZj&Fbio$\"1uU1?@m:GF[o7$
$\"1++D*eI!3eFbio$\"1a)QI8(>$3#F[o7$$\"1*\\7=be5(fFbio$\"1LiV'RiNa\"F[
o7$$\"1+D1Z(p99'Fbio$\"1ETYo3P46F[o7$$\"1***\\(Qy23jFbio$\"1,@qjl,**yF
07$$\"1*\\7.2,D['Fbio$\"1Do?KBGQaF07$$\"1+++t8\\]mFbio$\"1^Yc/OfJPF07$
$\"1+](3s&GAoFbio$\"1E&)oyJ/&\\#F07$$\"1+D\"[_dE*pFbio$\"1$z7yDK]k\"F0
7$$\"1++]=:@\\rFbio$\"1^'3\"e5\"\\5\"F07$$\"1**\\Pi+kGtFbio$\"1:0E;u`x
oF,7$$\"1+++\"4B\"*[(Fbio$\"1[:3'[0#GWF,7$$\"1+D\"=ZJ-m(Fbio$\"1g)*)yH
')Gs#F,7$$\"1+v=DS+CyFbio$\"1)fq8d()=o\"F,7$$\"\")F($\"1Mh1b;2X)*F[^o-
Fgt6&FitFjtF(F(-%+AXESLABELSG6$%!GFehp-%*AXESSTYLEG6#%$BOXG-%%VIEWG6$;
$!\"\"F_ipF\\hp%(DEFAULTG" 1 2 0 1 0 2 9 1 2 2 1.000000 45.000000 
45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 153 "The comparison sh
ows that the discrete binomial distribution can be approximated reason
ably by a continuous Gaussian distribution evaluated for discrete " }
{TEXT 19 2 "nu" }{TEXT -1 31 " values for moderate values of " }{TEXT 
306 1 "n" }{TEXT -1 9 " already." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 742 "We can also understand now how the two s
eemingly different tasks can be put on a common footing: examples of m
easurements where we obviously expect a broad distribution (height of \+
some species population with average and deviation the meaningful quan
tities), and others where we are seeking a 'single number' only. In th
e latter case the 'single number' is given by the average of the distr
ibution, and the deviation becomes a measure for the uncertainty in th
e measured result. While we would like in an ideal world the uncertain
ty to be negligible compared to the 'single number', it is essential t
o determine the uncertainty from the distribution of repeated measurem
ents in order to assess the quality (accuracy) of the measured quantit
y." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 324 "To
 really understand the concept of the probability distribution it is e
ssential to analyze sequences of real data. We encourage the reader to
 perform 'real' experiments with coins, dies, etc.. It is possible to \+
analyze sequences produced with Maple's random number generator (whose
 quality falls short of state-of-the-art)." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 16 "die:=rand(1..6);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%$dieGR6\"6#%\"tGF&F&C%>%&_seedG-%%iremG6$,$F+\"-\"3p'>uU\"-*)******
****>8$F+,&-F-6$F3\"\"'\"\"\"F8F8F&6#F+F&" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 6 "die();" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"%" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "We generate a sequence of tosses w
ith three dies:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "M:=200; \+
n:=3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"MG\"$+#" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%\"nG\"\"$" }}}{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];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$
7%\"\"$\"\"%\"\"'7%\"\"&F$F&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "W
e can analyze the presence of particular faces:" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 5 "F:=6;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"FG
\"\"'" }}}{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);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6$Q;Single-event~probability:~6\"$\"++++]<!#5" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#7&$\"++++]b!#5$\"++++]OF&$\"+++++!)!#6\"\"!
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "ps:=evalf(1/6);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#psG$\"+nmmm;!#5" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 27 "[seq(BD(ps,n,nu),nu=0..n)];" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#7&$\"+Oq.(y&!#5$\"+BAAsMF&$\"+ZWWWp!#6$\"+K'H'HY!#
7" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "[seq(BD(p_s,n,nu),nu=0
..n)];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7&$\"+]i::c!#5$\"+]7GtNF&$\"
++vozv!#6$\"++]Pf`!#7" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 78 "We appea
r to have results that are consistent with the binomial distribution. \+
" }}{PARA 0 "" 0 "" {TEXT -1 323 "We note that in addition to measurin
g directly from the sample the nu-fold event probabilities we have the
 option to measure the single-event probability, and to generate nu-fo
ld event probabilities from the binomial 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 tosses to observe whether the consistenc
y improves with the sample size. We can also look at the distribution \+
of frequencies of success for other faces:" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 5 "F:=1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"FG\"\"
\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "DieAnalysis(S1);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6$Q;Single-event~probability:~6\"$\"+nmm
m9!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7&$\"++++]i!#5$\"+++++JF&$\"+
++++l!#6\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "[seq(BD(p_
s,n,nu),nu=0..n)];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7&$\"+.Py8i!#5$
\"+BA)R?$F&$\"+ZW%o]&!#6$\"+lH'\\:$!#7" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 5 "F:=2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"FG\"\"#" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "DieAnalysis(S1);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6$Q;Single-event~probability:~6\"$\"++++
+:!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7&$\"++++]g!#5$\"++++]MF&$\"+
++++X!#6$\"+++++]!#7" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "[se
q(BD(p_s,n,nu),nu=0..n)];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7&$\"+++D
Th!#5$\"+++D^KF&$\"+++]Pd!#6$\"++++vL!#7" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 337 "The scatter in the answers for the different faces demon
strates that we have to be very careful when drawing conclusions from \+
measurements with small sample sizes (low number of repetitions)! Even
 the single-event probabilities vary substantially. The deviations wil
l shrink only slowly when the sample size is increased (typically with
 " }{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 diffe
rent faces. Determine whether the measurement from samples approaches \+
the expected value of 1/6 when the sample size is increased by differe
nt 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 0 {PARA 3 "" 0 "
" {TEXT 320 42 "Poisson distribution with chi-squared test" }}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 1401 "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 success raised to a huge \+
power (it is practically one), and use a mathematical property for the
 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!;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#PDGR6$%#muG%#nuG6\"6$%)ope
ratorG%&arrowGF)*&*&-%$expG6#,$9$!\"\"\"\"\")F39%F5\"\"\"-%*factorialG
6#F7!\"\"F)F)F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 5 "Here " }{TEXT 
19 2 "mu" }{TEXT -1 71 " is the expected mean numberof counts in the t
ime interval chosen, and " }{TEXT 19 2 "nu" }{TEXT -1 58 " labels the \+
possibilities of observing 0, 1, 2,... counts." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 525 "Note that in an actual e
xperiment with a Geiger-Mueller counter one does not observe all decay
s, but a fraction thereof (depending on the efficiency of the tube for
 particular forms of radiation; the count rate, which leads to a certa
in dead time for the tube, the supply voltage on the tube, etc.). This
 does not alter, however, the fact that one can determine the statisti
cal nature of the radiation, as one simply looks at a subsample of the
 actual radiation sequence (one is also limited by the direction of ob
servation)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 156 "Why is mu the mean (or average) count number for the chosen ti
me interval? Let us calculate the probabilistic average number of obse
rved decays (successes):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "
mu:='mu':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "simplify(sum(n
u*PD(mu,nu),nu=0..infinity));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%#muG
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "simplify(sum(PD(mu,nu),
nu=0..infinity));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 84 "Note that the sum started at one, \+
as no contribution is made towards the average by " }{XPPEDIT 18 0 "nu
 = 0;" "6#/%#nuG\"\"!" }{TEXT -1 108 ". The mathematical steps require
d 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 analyzing a sequence is as follows:" }}{PARA 0 "" 
0 "" {TEXT -1 425 "First one can determine the average count rate, by \+
taking the total number of decays and divide by the total time interva
l (which is subdivided into a certain number of sub-intervals). Then o
ne arrives at an average number of decays per subinterval. Finally one
 forms a histogram by recording the frequency of intervals with 0, 1 ,
 2, ... counts. This histogram is to be compared with the predictions \+
of the PD for the given " }{TEXT 19 2 "mu" }{TEXT -1 112 " value. It i
s possible to re-analyze the same decay sequence by using a different \+
choice of sub-interval length." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 65 "Let us look at the Poisson distribution f
or two cases, one where " }{XPPEDIT 18 0 "mu < 1;" "6#2%#muG\"\"\"" }
{TEXT -1 148 ", for which the number of zero-count intervals dominates
, and another one, where one can see how the Gaussian distribution eme
rges 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);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#muG$\"\"'!\"\"" }}{PARA 13 "" 1 "" {GLPLOT2D 400 
300 300 {PLOTDATA 2 "61-%'CURVESG6#7S7$\"\"!F(7$F($\"1FK0%=`i>\"!#<7$F
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'COLOURG6&%$RGBGF(F($\"*++++\"!\")-%%VIEWG6$%(DEFAULTGFfcq" 1 2 0 1 0 
2 6 1 2 2 1.000000 45.000000 45.000000 0 }}}}{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));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&sigsqG$\"+,+++M!\"*" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 40 "Again, this can also be calculated from:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "simplify(sum(nu^2*PD(mu
,nu),nu=0..infinity)-mu^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"++++
+M!\"*" }}}{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 average with deviation as an i
nterval one obtains " }{XPPEDIT 18 0 "[mu-sqrt(mu), mu+sqrt(mu)];" "6#
7$,&%#muG\"\"\"-%%sqrtG6#F%!\"\",&F%F&-F(6#F%F&" }{TEXT -1 128 " . Whi
le using a larger count interval we do also obtain a larger uncertaint
y, which seems improper. The fractional uncertainty " }{XPPEDIT 18 0 "
sqrt(mu)/mu;" "6#*&-%%sqrtG6#%#muG\"\"\"F'!\"\"" }{TEXT -1 53 ", howev
er, is reduced for an increased time interval." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "mu:=8.3;" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "P1:=plot([seq([nu,PD(mu,nu)*t,t=0..
1],nu=0..20)],color=blue):" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#muG$
\"#$)!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "sigma:=sqrt(m
u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&sigmaG$\"+e?(4)G!\"*" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "P2:=plot(exp(-(nu-mu)^2/(2*s
igma^2))/(sqrt(2*Pi)*sigma),nu=-0.1..20,color=red):" }}{PARA 0 "> " 0 
"" {MPLTEXT 1 0 26 "display(P1,P2,axes=boxed);" }}{PARA 13 "" 1 "" 
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45.000000 45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 299 "Note th
at there is a systematic tendency for the Gaussian distribution to und
erestimate the count probability to the left of the average. The Poiss
on distribution is asymmetric, and this asymmetry becomes negligible o
nly for sufficiently large average count probability for the chosen ti
me interval." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 389 "To properly understand the Poisson distribution one should ana
lyze data runs from a Geiger counter. These are usually computer-inter
faced in undergraduate laboratories today, which has the 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 ge
nerate event sequences which are then analyzed." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 357 "A sequence of events, co
rresponds to the sequence of clicks that one hears as the Geiger count
er picks up the radiation (either from natural background, i.e., mostl
y cosmic radiation, or from a small radioactive source, such as a cera
mic plate or tile that has been coated with a radiactive layer contain
ing, e.g., uranium oxide, to enhance the brightness)." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 150 "We use a random numbe
r generator to produce a time sequence of events; we assume that a tic
k 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 rate to decide in
 a time unit whether an observable decay occurs or not. A decay is rec
orded as a one, a non-decay as a zero." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "rnd:=rand(0..100);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%$rndGR6\"6#%\"tGF&F&C%>%&_seedG-%%iremG6$,$F+\"-\"3p'>uU\"-*)******
****>8$F+-F-6$F3\"$,\"F&6#F+F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 11 "rate:=1/10;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%rateG#\"\"\"
\"#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "tmax:=3000;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%tmaxG\"%+I" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 112 "TS:=[]: for i from 1 to tmax do: if rnd() <= 10
0*rate 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);" }}{PARA 13 "" 1 "" {GLPLOT2D 691 299 299 
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\"$h\"F*F*7$$\"$i\"F*F*7$$\"$j\"F*F*7$$\"$k\"F*F(7$$\"$l\"F*F*7$$\"$m
\"F*F*7$$\"$n\"F*F(7$$\"$o\"F*F*7$$\"$p\"F*F*7$$\"$q\"F*F*7$$\"$r\"F*F
*7$$\"$s\"F*F*7$$\"$t\"F*F*7$$\"$u\"F*F*7$$\"$v\"F*F*7$$\"$w\"F*F*7$$
\"$x\"F*F*7$$\"$y\"F*F*7$$\"$z\"F*F*7$$\"$!=F*F*7$$\"$\"=F*F*7$$\"$#=F
*F*7$$\"$$=F*F*7$$\"$%=F*F*7$$\"$&=F*F*7$$\"$'=F*F*7$$\"$(=F*F(7$$\"$)
=F*F*7$$\"$*=F*F*7$$\"$!>F*F*7$$\"$\">F*F*7$$\"$#>F*F*7$$\"$$>F*F*7$$
\"$%>F*F*7$$\"$&>F*F*7$$\"$'>F*F*7$$\"$(>F*F*7$$\"$)>F*F*7$$\"$*>F*F*7
$$\"$+#F*F*-%'COLOURG6&%$RGBG$FE!\"\"F*F*-%+AXESLABELSG6$%!GFcam-%&STY
LEG6#%&POINTG-%%VIEWG6$%(DEFAULTGF[bm" 1 5 0 1 0 2 9 1 4 2 1.000000 
45.000000 45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 302 "This se
quence could have come from a Geiger counter. Note the irregularity: t
here are stretches with no counts followed by segments where the count
s appear to cluster. This irregularity is noticable when listening to \+
the clicks from a Geiger counter. Now we pick a time interval for the \+
decay analysis:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "tint:=25
; # length of interval in seconds (time units)" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%%tintG\"#D" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "We
 can determine the expected average number of events per time interval
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "nu_bar_exp:=tint*rate;
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%+nu_bar_expG#\"\"&\"\"#" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "Nt:=tmax/tint; # number of t
ime intervals contained in sample" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%#NtG\"$?\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 123 "Now we need to ge
nerate the histogram (frequency distribution): for each time interval \+
we find out how many events occured." }}{PARA 0 "" 0 "" {TEXT -1 157 "
We begin by resetting the counters for the frequency. Assume that 10 e
vents during the interval is the maximum (increase it, if the program \+
below complains):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "Emax:=1
5;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%EmaxG\"#:" }}}{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);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'myrateG$\"++++!4\"
!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "mymu:=myrate*tint;" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%mymuG$\"++++DF!\"*" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 138 "To compare the found results with the Po
isson distribution we have to convert the probabilities calculated fro
m 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)],col
or=red):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "P1:=plot([seq([icnt+0.0
5,C[icnt]*t,t=0..1],icnt=0..Emax)],color=blue): display(P1,P2,axes=box
ed);" }{TEXT -1 0 "" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 
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x$\"1\"QjiJ'p'y\"F_hv7$F^dx$\"1%4A3]Lh#=F_hv7$F^dx$\"1u.\"['HLr=F_hv7$
F^dx$\"18#GU!*f<\">F_hv7$F^dx$\"1+Le`M'[&>F_hv7$F^dx$\"1*30-0>h*>F_hv7
$F^dx$\"1+++BQXS?F_hvF``p-%+AXESLABELSG6$%!GFd]y-%*AXESSTYLEG6#%$BOXG-
%%VIEWG6$%(DEFAULTGF\\^y" 1 2 0 1 0 2 9 1 2 2 1.000000 45.000000 
45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "We can make the fo
llowing observations:" }}{PARA 0 "" 0 "" {TEXT -1 200 "1) for the chos
en moderate number of recordings (event/no-event per second) we can fi
nd that the simulation does well on the event rate, and on the expecte
d average number of events per time interval." }}{PARA 0 "" 0 "" 
{TEXT -1 156 "2) a reasonable result is obtained for the entire distri
bution; note that the Poissonian drawn for comparison is based on inpu
t from the data run (variable " }{TEXT 19 4 "mymu" }{TEXT -1 368 " for
 the average number of events per time interval), and not the predeter
mined rate used to generate the sequence. Substantial discrepancies oc
cur for individual bins in the histogram. This should not cause too mu
ch of an alarm: the Poisson distribution represents a limiting probabi
lity distribution that should be reached in the limit of many performe
d experiments." }}{PARA 0 "" 0 "" {TEXT -1 237 "3) We should explore f
urther the consistency of the data with a Poisson distribution: either
 by considering larger data sets, or by performing a systematic test o
n the hypothesis (chi-squared test). The latter is attempted further b
elow." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 311 12 
"Exercise P1:" }}{PARA 0 "" 0 "" {TEXT -1 65 "Explore the above data b
y choosing different subinterval lengths " }{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 step
s for cases where the decay rate is increased and decreased respective
ly." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 423 "We now consider an interesting additional check, namely a test
 of the hypothesis that the distribution generated from the independen
t events is indeed compatible with a Poisson distribution. A chi-squar
ed test of a hypothesis is a quantitative measure that essentially com
pares the data distributions as done in the above graph by summing the
 squares of the discrepancies, and then turning the result into a usef
ul measure." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 51 "In the graph above we have used the data to deduce " }{XPPEDIT 
18 0 "mu;" "6#%#muG" }{TEXT -1 91 " value for the Poissonian, and then
 generated a distribution of expected values to observe " }{XPPEDIT 
18 0 "nu = 0 .. infinity;" "6#/%#nuG;\"\"!%)infinityG" }{TEXT -1 486 "
 events per time interval. (A limitation of the Poissonian is perhaps \+
noticable here: our simulation, and any Geiger counter sequence has a \+
maximum number of observable events, namely when the counter is clicki
ng all the time, while the Poissonian distributes the probability over
 an infinite interval; this is the price to be paid for the replacemen
t of a binomial by a Poisson distribution; 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 c
alculate the chi-squared for the results obtained above (make sure tha
t the statements that produce the graph above were executed) from the \+
expected and observed results:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 66 "chi_sq:=add((Nt*PD(mymu,nu)-C[nu])^2/(Nt*PD(mymu,nu)),nu=0..Em
ax);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'chi_sqG$\"+>(4H]$!\"*" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 131 "Is this an acceptable value or no
t? This depends on the number of degrees of freedom. Here we have a li
ttle problem: should we use " }{TEXT 19 4 "Emax" }{TEXT -1 195 ", whic
h includes data where we have no counts and small expected probabiliti
es, or not? After all, there was some arbitrariness in choosing this v
ariable, which acts as a cut-off for the diagram." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 103 "The criterion for accept
ability is that chi-squared be much less than the number of degrees of
 freedom " }{TEXT 314 1 "d" }{TEXT -1 76 "; rejection of the hypothesi
s is given when chi-squared is much bigger than " }{TEXT 313 1 "d" }
{TEXT -1 42 ". The question is whether we can say that " }{TEXT 315 1 
"d" }{TEXT -1 5 "=16 (" }{TEXT 19 7 "=Emax+1" }{TEXT -1 2 ")?" }}
{PARA 0 "" 0 "" {TEXT -1 82 "Judging from the graph the right attitude
 should be to pick the 10 visible points:" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 63 "chi_sq:=add((Nt*PD(mymu,nu)-C[nu])^2/(Nt*PD(mymu,nu
)),nu=0..9);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'chi_sqG$\"+PzQQM!\"
*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 515 "The conclusion based on the
 limited data set would be that the data are not inconsistent with a P
oisson distribution, and that further testing is desirable. Further te
sting means to run with bigger data sets. An important issue when perf
orming these tests is to choose reasonable time intervals (which depen
d on the actual count rate). The chi-squared comparison has to have a \+
reasonable number of contributing data points (say six or more to acco
unt for two degrees of freedom to be subtracted as explained below)." 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 86 "One sho
uld 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-negligible theoretical data points (we could have used t
he number of data bins). It should be kept in mind that we really have
 one less independent value, as the data were used to infer the averag
e number of decays per time interval (" }{TEXT 19 4 "mymu" }{TEXT -1 
245 "). There can be other constraints (usually the number of data poi
nts; in the case of a Gaussian distribution also its deviation) which \+
are used to determine a reduced number of degrees of freedom. A reduce
d chi-squared is defined as chi-squared/" }{TEXT 317 1 "d" }{TEXT -1 
217 ", and for this quantity the borderline between acceptance and non
-acceptance of the hypothesis is the value of one. It is evident that \+
for acceptance we need to obtain much less than one, and vice versa fo
r rejection." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 124 "A healthier attitude for dealing with the open-ended Poisson d
istribution would be to call the bin neighbouring the largest " }
{XPPEDIT 18 0 "nu;" "6#%#nuG" }{TEXT -1 100 " value that still receive
d non-zero counts to be one bin for all events beyond the largest obse
rved " }{XPPEDIT 18 0 "nu;" "6#%#nuG" }{TEXT -1 34 " value. In our cas
e 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 re
ceiving 0 counts) and " }{TEXT 318 1 "d" }{TEXT -1 132 " would be 10-2
=8, i.e., a borderline result [note that the precise value in the latt
er statement can change for repeated data runs]." }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 9 "chi_sq/8;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$
\"+@\\)zH%!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "Our own test wit
h 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 "4 11 1 0" 0 }{VIEWOPTS 1 1 0 1 1 
1803 }