The robot doesn’t care about your irrelevant technicality, it cares about maximum entropy. e^0 is exponential, but it is not the maximum entropy distribution with a given mean. In this case, it is
Pr(X=x_k) = C*r^(x_k) for k = 1,2,3,4
where the positive constants C and r can be determined by the requirements that the sum of all the probabilities must be 1 and the expected value must be 2.5.
Just because I didn’t actually write the formula doesn’t mean it doesn’t exist or you can replace it with any formula you like. So if the robot works as described, this is what the robot will update its expected probabilities to upon learning that the mean is 2.5, and not 2.5*e^0 because it would be convenient for you.
This is why many of us are terrified of the Singularity, because the author of a program seldom anticipates its actual result. What’s even more terrifying is that this should have been obvious to you, as you gave an example as a strength of your idea of when the mean was 3.0. Why are you upset that I pointed out the consequences when the mean was 2.5? Instead of acknowledging the fact, you entirely forget what you told your robot to do and blurt out a misleading technicality?
Allow me to explain less snarkily and more directly than Manfred.
As you correctly observe, the maximum-entropy probability on {1,2,3,4} with any given mean is one that gives k (k=1,2,3,4) probability A.r^k for some A,r, and these parameters are uniquely determined by the requirement that the probabilities sum to 1 and that the resulting mean should be the given one.
In the particular case where the mean is 2.5, the specific values in question are A=1/4 and r=1.
This distribution can be described as exponential, if you insist—but it also happens to be the same uniform distribution that’s maximum-entropy without knowing the mean.
So the inconsistency you seemed to be suggesting—of an entropy-maximizing Bayesian robot choosing one distribution on the basis of maxent, and then switching to a different one on having one property of that distribution confirmed—is not real. On learning that the mean is 2.5 as it already guessed, the robot does not switch to a different distribution.
It just occurred to me that I really ought to check whether I ought to check that it was in fact different rather than going, “what are the odds that out of all the possibilities that equation happens to be the uniform distribution?”. Guess I should have done that before posting.
It also occurs to me now, that I didn’t even have to calculate out the equation (which I thought was too much effort for a “someone is wrong on the internet”), and could just plug in the values … and in fact I even already did that when finding the uniform distribution and its mean.
This post sponsored by “When someone is wrong on the internet, it’s sometimes you”
Pr(X=xk) = C*r^(xk) for k = 1,2,3,4 where the positive constants C and r can be determined by the requirements that the sum of all the probabilities must be 1 and the expected value must be 2.5.
r^x = e^kx, where e^k=r. So. Would you like to wager whether r=1?
e^0 is also exponential.
e^0 is not the exponential you told your robot to choose in that situation.
The robot doesn’t care about your irrelevant technicality, it cares about maximum entropy. e^0 is exponential, but it is not the maximum entropy distribution with a given mean. In this case, it is Pr(X=x_k) = C*r^(x_k) for k = 1,2,3,4 where the positive constants C and r can be determined by the requirements that the sum of all the probabilities must be 1 and the expected value must be 2.5.
Just because I didn’t actually write the formula doesn’t mean it doesn’t exist or you can replace it with any formula you like. So if the robot works as described, this is what the robot will update its expected probabilities to upon learning that the mean is 2.5, and not 2.5*e^0 because it would be convenient for you.
This is why many of us are terrified of the Singularity, because the author of a program seldom anticipates its actual result. What’s even more terrifying is that this should have been obvious to you, as you gave an example as a strength of your idea of when the mean was 3.0. Why are you upset that I pointed out the consequences when the mean was 2.5? Instead of acknowledging the fact, you entirely forget what you told your robot to do and blurt out a misleading technicality?
Allow me to explain less snarkily and more directly than Manfred.
As you correctly observe, the maximum-entropy probability on {1,2,3,4} with any given mean is one that gives k (k=1,2,3,4) probability A.r^k for some A,r, and these parameters are uniquely determined by the requirement that the probabilities sum to 1 and that the resulting mean should be the given one.
In the particular case where the mean is 2.5, the specific values in question are A=1/4 and r=1.
This distribution can be described as exponential, if you insist—but it also happens to be the same uniform distribution that’s maximum-entropy without knowing the mean.
So the inconsistency you seemed to be suggesting—of an entropy-maximizing Bayesian robot choosing one distribution on the basis of maxent, and then switching to a different one on having one property of that distribution confirmed—is not real. On learning that the mean is 2.5 as it already guessed, the robot does not switch to a different distribution.
[EDITED: minor tweaks for clarity.]
It just occurred to me that I really ought to check whether I ought to check that it was in fact different rather than going, “what are the odds that out of all the possibilities that equation happens to be the uniform distribution?”. Guess I should have done that before posting.
It also occurs to me now, that I didn’t even have to calculate out the equation (which I thought was too much effort for a “someone is wrong on the internet”), and could just plug in the values … and in fact I even already did that when finding the uniform distribution and its mean.
This post sponsored by “When someone is wrong on the internet, it’s sometimes you”
r^x = e^kx, where e^k=r. So. Would you like to wager whether r=1?