NetNews Usenet Archive 1992 #23/NN_1992_23.iso/spool/sci/math/13277

Text File | 1992-10-15 | 3.5 KB | 71 lines

Newsgroups: sci.math
Path: sparky!uunet!snorkelwacker.mit.edu!galois!riesz!jbaez
From: [e-mail] (John C. Baez)
Subject: Re: Probably you know the probability ;-)
Message-ID: <[e-mail]>
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Organization: MIT Department of Mathematics, Cambridge, MA
References: <[e-mail]> <[e-mail]> <stephen.719167970@mont>
Date: Thu, 15 Oct 92 22:42:53 GMT
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In article <stephen.719167970@mont> [e-mail] (Stephen Montgomery-Smith) writes:
>In <[e-mail]> [e-mail] (Surya N Kavuri ) writes:
>

>>  All the points on the rod are equally likely to be breakage points.
>>
>
>It seems to me that one would break the rod into lengths (a,b,c) such
>that all admissible triplets (i.e. a+b+c = l) are equally likely.  I
>think that if you pick the break points uniformly along the rod that
>you get a different distribution.

Let me be nitpicky because I think it is useful in this sort of problem.
Saying "all points on the rod are equally likely" or "all admissible
triplets are equally likely" is essentially meaningless.  Practically
any probability distribution on the space of triplets that you will
dream of will have the property that all triplets are equally likely,
because the probability of any given triplet occurring is ZERO.  In
other words, I was trying to get you to tell me the probability measure
on the space of triplets and you replied by telling me that the measure
of each triplet is the same as the measure of any other.  This does not
specify a measure on the space of triplets!!  (It only says that the
measure, whatever it is, is nonatomic.)

Now I know this is not what you MEANT to say.  You probably meant to say
that the break points are uniformly distributed according to Lebesgue
measure on the interval.  That seems to be what Stephen is suggesting at
the end there.  But it pays to be clear in this sort of thing because
there are many famous examples of where you get messed up unless you
realize that in posing a probability problem of this sort you must
specify a *probability measure* on the space of outcomes.

If you don't know measure theory I'm afraid the above may simply be
gibberish.  Let me give 2 probability measures that might be reasonable
choices.  (These seem to be the 2 Stephen is considering.)

First let me specify the space of outcomes.  Instead of what Stephen
does, I will specify the outcomes as follows.  Start with a stick given
by the unit interval [0,1].  Make the first break at x and the second at
y.  The outcome is a pair (x,y) of numbers in [0,1].  We are thus
seeking a measure on [0,1]^2, the space of such ordered pairs.

We can:

1) Say that x and y are both randomly chosen in a uniform way from
[0,1], that is, both chosen according to Lebesgue measure on [0,1].
The resulting probability measure on [0,1]^2 is thus  dx dy.   If we
want to compute the chance that something happens and it happens when
the pair (x,y) is in the set S, we compute integral_S dx dy.

2)  Say that x is randomly chosen in a uniform way from [0,1], and then
y is randomly chosen in a uniform way from [0,x].  Thus the second break
will always be to the left of the first.  We thus get a measure on
the triangle {(x,y): 0 <= y <= x <= 1}.  This measure will be
dx dy/x.  This will give very different answers.

I can think of lots of other ways to do things and none is "right" until
a model of the breaking is given - or else you can just pick one!