# Re: Bayesian boxes and expectation value

```On Sat, 15 May 1999 [EMAIL PROTECTED] wrote:
> In a message dated 99-05-14 15:55:53 EDT, Jacques Mallah writes:
>       Ok, now you seem to think that the expected value for the other
>  box is exp((log(2m)+log(m/2))/2) given that the first box contains m and
>  given a 50% chance that the second box contains 2m.  Ok, that's
>  unconventional logic all right!  Weird conclusions from unrelated
>  assumptions. >>
>
>  I agree the conclusion is weird. However, As Wei Dai mentioned we need to
> revise the concept of probability in the context of the MW. The logarithmic
> distribution was just an example. IN FACT THE DISTRIBUTION CAN BE ANYTHING AS
> LONG AS IT SATISFIES THE EXPECTATION VALUE = m FOR THE SECOND BOX.. ```
```
You don't seem to understand:  that's NOT how to take an
expectation value.  It bears little resemblance to the formula for an
expectation value, regardless of what "the distribution of m" is.

> I didn't know I had "allies" and enemies. I thought we were all friends
> striving toward truth.

I'm not your enemy, any more than NATO is the enemy of the Serbian
people.  But I am your opponent in this debate.
Neither have you earned my friendship.

- - - - - - -
Jacques Mallah ([EMAIL PROTECTED])