FW: Observation selection effects

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-----Original Message-----
From: Stathis Papaioannou [mailto:[EMAIL PROTECTED]
Sent: Friday, October 15, 2004 12:35 AM
To: [EMAIL PROTECTED]
Subject: RE: Observation selection effects```
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Brent Meeker wrote:

QUOTE-
It's not wrong - I just don't think it addresses the paradox.  To
resolve the paradox you must explain why it is wrong to reason:

I've opened one envelope and I see amount m.  If I keep it my gain
is m.  If I switch my expected gain is 0.5*m/2 + 0.5*2m = 1.25m,
therefore I should switch.
-ENDQUOTE

The reason this is wrong is that "m" is not a constant for the
purposes of
the probability calculation. I open one envelope and I see amount
m. I know
that one envelope contains x and the other 2x, but I don't know
whether m=x
or m=2x at this point. Pr(m=x)=0.5 and Pr(m=2x)=0.5. So my
expected gain if
I keep the envelope is 0.5*x + 0.5*2x = 1.5x. My expected gain if
I switch
is the same.

The obvious response to the above is to point out that "m" IS a
constant,
and if I keep the first envelope my gain is exactly m, whatever
"m" may be
in terms of "x". This is true, but I need to know what m is in
terms of x if
I am to go on to calculate the probabilities in the situation
where I
switch:

(a) If I happen to have opened the envelope first which makes m=2x
(with
Pr=0.5), then there is a 100% probability that the second envelope
contains
0.5m=x and 0% probability that the second envelope contains
2m=4x - so that
my expected gain is 0.5x (or 0.25m) from this half of the decision
tree.

(b) If I opened the envelope first which makes m=x (with Pr=0.5),
then if I
switch there is a 100% probability that the second envelope
contains 2m=2x
and 0% probability that it contains 0.5m=x - so that my expected
gain is x
(or m) from this half of the decision tree.

Now, to get the final expected gain from switching, I add up the
result from
(a) and (b). If I add up the x's I get 0.5x + x = 1.5x, the same
kept the first envelope. If I add up the m's, I get 0.25m + m =
1.25m, which
seems to be greater than the m I would get if I kept the first
envelope, as
per your analysis above. Which should it be for comparison
purposes, 1.5x or
1.25m? I think you can see that whereas x is a constant, m is one
amount in
(a) and a different amount in (b). You can't add up the m's in the
two parts
of the decision tree as if they were the same.

Stathis Papaioannou

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