Re: Bayesian boxes and expectation value

From: Gilles HENRI <Gilles.Henri.domain.name.hidden>
Date: Thu, 20 May 1999 09:50:54 +0200

>
>In the last four tests the lack of "reference" such as a boundary lets you
>float along what I feel to be a logarithmic scale. There does not appear to
>be any way to take a "FIX" as they say in the field of navigation. There does
>not appear to be any rational for switching. Or is there? One may argue that
>any distribution that my mind is capable of generating is centered around 1
>and on that basis there may be an extremely small advantage in switching when
>the number is below one and not switching when the number is above one. I
>could certainly counter this strategy by willingly biasing myself to generate
>distributions centered above one. And you knowing what I know will adapt your
>strategy too. In the limit the end result will be no advantage in switching.
>Certainly the simple minded expectation value of 0.5(2m + 0.5m) = 1.25m for
>switching does not make any sense. If you believe that it does make sense, I
>think I just have found a way to make a quick buck.
>
>George :-)

If you don't know the maximum value M, there is actually no rational for
switching. The simple minded expectation value 0.5(2m + 0.5m) = 1.25m is
not totally simple-minded: it evaluates correctly the expectation value of
the ratio between what you would find in the second box to what you found
in the first one. It is actually 1.25. However you should not use this
expectation value to conclude that it is better to take the second box,
because in the case where the second box contains 2m, it means also that
the value of m that you found in the first box is lower than the average
value 0.5*(m+2m) = 1.5m, and the opposite in the second case. The confusion
arises from the fact that m is not an expectation value of anything, and
summing 2m and 0.5m does not give an expectation value of what you can win,
because of course the expectation value of the ratio is not the ratio of
expectation values.
That's better seen if you call m0 the (unknown) amount which is really put
in the first box. Either you open the first box and you find m1=m0. The
second one contains 2*m1=2*m0. Or you open the second one and you find
m2=2*m0. The first one ontains 0.5*m2 = m0. The expectation value of the
content of the first box you opened is <m> = 0.5*(m1+m2) = 0.5*(m0+2*m0)=
1.5 m0. That of the other box is <m'> = 0.5*(2m1 + 0.5m2)=0.5*(2*m0+m0)=
1.5 m0, which is the same of course. You see more clearly that the mistake
comes from the fact that although m denotes always what you found in the
first box, its expectation value is not the same following which box you
assumed to have opened. However as I said, if you open systematically the
second box,and you calculate each time the ratio between box 2 and box 1,
you will indeed find an average value of 1.25. But it doesn't mean that you
will gain more than if you kept systematically the first box, because the
cases where the factor is 2 are just those where the value you found in the
first box is lower...

Gilles

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Gilles HENRI

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Received on Thu May 20 1999 - 01:06:54 PDT