Why use such a sarnple space? If you requiremotivation for its study, why not consider the old fair-ground game of rolling acoin of diameter d orno aplane of squares of side I? Alternatively, consider the. meeting of friends problern whereby two friends agreeto rneet at a particular coffee house for lunch, arrivingat randorn within lhe lunch hour, and they agree towait only 15 rninutes. Both contexts pose obviousintcresting questions, and in each case lhe uni! squareis the correct sarnple space.

The mathematical structure of probabi!itydistributions that underlies Statistics is quite complexoStudents are bewildered by terrns such as pdf and cdf,mean and variance, and the rel.uionships betwcen theterms. Carefully choscn cxercises help, und it isparticularly instructive if a variety of such exerciseshave a common staning point which is easy tounderstand.

"Choosing a point at random in a square" is a sim pie,intuitive statistical experiment. With rnicro-computersin the classroom it can be dernonstrated easily. Weshow in this article just how much of the elernentarytheory of probability distributions can be explored Our sample space immediately provides two randornfr()ID. ",ithin this basic frarnework. At the same time; variables, X and 1'.As a first step students must answer. difficulties experienced by students are highlighted probability questions about X and Y by deterrniningin the hope that readers may respond with suggestions appropriate regions of the unit square and calculatingfor.overcoming.rheml, ._.. " ._~ar.e.ElS- .. . . _.

":...

~;':

$- PROBA,8!LlTIES ANO R.ANDOM ~~.VARI.ABLES

f; .f 4\'~THE UNIT SQUARE AS A ~.~'~)

62 Co~,,"~ ~ ~~ "" t-"'~~ o. ~o- y

oStudents find lhe: answers without too much difficultyand rnay be prornpred to enquire whether the fact that

can be answered easily using the independence cfX and Y. Of course, (e) can also be answered similarlyif we concentrate first on the complementary eventp(L> -t) = p(X> -t AND Y> -t). The two possiblecalculatibns of (e) are written below, together withappropriatc diagrarns.

answer (iii) = answer (ii) x answer (i)

is a freak ar not, (Hence to independence of randornvariables, )Toe. great thing about randorn variables is that they

hreed well! The usual arithmetical processes can beapplied to randorn variables on lhe same sarnple spaceto give birth tO more. Here are some examples andsome suggested caJculations, each of which shouldbeaccornpanied by a quick sketch of the appropriateregion of the sarnple space.

(a) s= X+ y p(s -t).p( X> +) == t

p( L

Parts (ii) and (iii) are mechanical and do not usuallypresent problems. However, (i) does!The use of a dummy variable (here I have used s)

seems to cause conceptual problems. Students are

happy to calculate p( X::; ), p( X s t), p( X::; )etc., but became mildly disconcerted with calculatingD(X < s) c O < < 1, _ lor _ s _ .. A related difficulty is that they very often fail tonote the phrase 'for ali relevant s'. Take for examplethe di stribution of S = X + Y. Here is a typicai solution.

2P(X + Y::;S)=T

But X + Y can take values in the range O to 2, andmoreover the nature of the calculation changes,dependent on s being less than or greater than 1. Hereis the ful! calculation .'

2P(X+ Y::;S)=T

O:O;s:O;l

s

P(X + y::; s) = 1 _ (2-2S)2l::;s::;2

"","A typicalreaction ofthis is "Row can you hav~ tW~answers to one question?"! Moreover the disquietcontinues when you suggest they should di fferen ti atethis cdf to find the pdf and hence calculare the meanand variance of S = X + Y. The integral has to be splitinto two integrais:

J 2

E(X + Y) = f s. s.ds +f s.(2 - s)ds = t +1= 1.O 1

(Of course this has to be thc answer asE(X + Y) = E(X) + E(Y) and to show thatE(X) = E(Y) = t is trivial!)It is worth continuing the exarnple to show that

Var(X + Y) =3; = Var(X) + Var(Y). (The equalityholds because X and Y are independent.) Vlorkingthrough similar ca1culations for lhe other randornvariables mentioned is also instructive. For exarnple:

P(L::; s) = 1- (I - s)2fL(s)=2(I-s)

O:O;sS;i0:0;.1':0;1

(f L (s) is the probability function of L).

It is worth pointing out that the graph of fL shows thatsmall values of L are more likely than large values,not surprising since L is lhe minimum of two randornquantities, Furtherrnore, E(L) = + and so E(U) rnustequal t.Why? because -

L+U=X+Y,

E(L) + E(U) = E(L + U) = E(X + Y) = I.

If time permits, make the student calculate Var(L) andVare U). Then it is easy to check that

) Var(L) + Vare U) :F Var(L + U) = Var(X + Y) = +.~. 0

Thus we have demonstrated that lhe iaw

does no! work for correlated variables L, U but doeswork for independent variables X, Y.Whal about M = XY? Calculation of lhe densiry of

M is straightforward.

P(M:O;S)=S+J' Ldx=s-s [agesS x

and eventually E(M) = t. This, of course, must bethe answer as X and Yare independent. hence E(XY)= E(X) x E(Y).

This leaves the ratio R = Y/X. Does E(R) = I?(Argument: E(R) "=" E(Y)IE(X) = 1.). A most ernphaticNo!

In fact, the expeeted value of R is infinite, which isnot difficult to prove providing the correct cdf iscalculated. (See the remarks for S = X + Y)

64

,-

P(Y/X::;s)=1-+_S

ls s c s=

A more interesting question is to calculate thedistribution of the waiting time of A (c all this H') andhence evaluate E(Vi'). A problern arises beeause VI isneither a diserete nor a eontinuous randorn variable -it takes the values O and 114 with positive probabilitywhereas individual values between O and 1/4 arepossible but have zero probability. So, such rnixcd-type random variables are not merely mathematicalartefacts: they do oceur in lhe simplest of probabilityspaees - thc unit square - inan uncontrived manner.The calculation of P(W ::; s) is straighrforward if

care is taken! Here are the steps with diagrams.

P(W = O) = P(X - t < Y < X)=7/32

1/s

This is a quick tour de force of the sort ofcaJculations that are possible using fairly obviousrandom variables generated in lhe unit square. Theex arnples are worthy exercises for student and teacherali ke. As i hope I have demonstrated, many aspects oflhe theory of distributions and expeeted values areillustrated, and particular emphasis is placed on thepath,

Probability ~ cdf ~ expected values,

which is possible because we siart with a simple butrich sample space - Lhe unit square.

~ A FiNAL EXAMPLE ~

For lhe fi nal example, return to the meeting of'friendsproblem. Let A and B be the two friends, and supposethat A arrives at time X (measured in hours, O ::; X::; 1)and B arrives at time r. The usual problem is tocalculare lhe probability that they actually meet.

. - Assurningthel S'minuie waitingtirne, th{s probabilityis sirnply given by

n(IX vI 1\ . (,,2 7r I - J < -) = I - -) -\ 4 4 16

Quote 9

1..4

P(W::; s) = P(X - + < Y < X + s)I 3 2 J 2=1-2("4) -2(1-s)

"'-"-"- -' PC W ::;+) := I

Below is a sketch of lhe curve P(W::; s), and a simpleway to calculate E(Vi') is to ca1culate lhe area of lheregion above this curve, bounded by y = 1 (see Sykes,1981) and lhe answer s i.

1..4

I Reference 1I Sykes, A.M. (]981). An Alternative Approach to II lhe Mean. Teaching Statistics, 3(3), 82-87. I

All models are wrong, some areuseful, C.E.? Box per M. Sandford.

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