value systems = local maxima of f, but only those higher than global average E f(x)
No. value systems = local maxima of f, period.
You said “if agents are better than random” in headlines. I take “than random” to mean ” > E f(x)”. So do you assume so or not?
What? Can’t parse.
Let me try again.
A—set of all local maxima x
B—set of all local maxima x, for which f(x) > E f(x)
In either case, avg(A) and avg(B) are arbitrary points, and we have no a priori reason to believe they will be special in any way, so E f(avg(A)) = E f(avg(B)) = E f(x). Is this right?
In case of set B—we assumed for all x \in B . f(x) > E f(x), so picking avg(B) which is essentially a random point makes it worse.
In case of set A—E f(x) for set of arbitrary local maxima is no better than set of arbitrary points, so E(f(x) | x \in A) = E(f(x)) = E f(avg(a)), so your entire argument fails.
B—set of all local maxima x, for which f(x) > E f(x)
f(x) is IC(x), and E is expected value?
I didn’t discuss anything corresponding to your B.
In either case, avg(A) and avg(B) are arbitrary points, and we have no a priori reason to believe they will be special in any way, so E f(avg(A)) = E f(avg(B)) = E f(x). Is this right?
No. You just said they’re local maxima, which is very special; and it would be surprising if either of their expectations were the same as E[f(x)].
In case of set B—we assumed for all x \in B . f(x) > E f(x), so picking avg(B) which is essentially a random point makes it worse.
This doesn’t describe what I wrote; but B is not a random set, so avg(B) is not a random point.
In case of set A—E f(x) for set of arbitrary local maxima is no better than set of arbitrary points, so E(f(x) | x \in A) = E(f(x)) = E f(avg(a)), so your entire argument fails.
No; local maxima (actually minima in this case, but same thing) are not arbitrary points. They’re locally maximal. Meaning they have a higher f(x) than all the points around them. So it’s impossible for the equality you just gave to hold, except when there are no local maxima (eg, a plane), or perhaps in some peculiar set constructed so that infinities make finite differences not matter. EDIT: Also add cases where a space is constructed so that local maxima are more common when f(x) is small, which is the type taw used in his reply. This is a large number of possible spaces. I believe it’s a minority of possible spaces, but it would take time to formalize that.
You say B is not a random set, but for arbitrary space and function, set of local maxima will behave essentially like a random set. It can as easily have average less or more than average of the whole space.
Here’s a really really simple example:
Space: [-1/2..+1/2], f(x) = cos(2.02 pi x).
There are 3 local maxima, x=-1/2, x=0, x=1/2
E f(x) = −0.009899
E(f(x) | x is local maximum) = −0.333 - so lower than space average
f(avg of local maxima) = f(0) = 1.
Another one:
Space: [-1/2..+1/2], f(x) = cos(3.03 pi x).
There are 3 local maxima, x=-1/2, x=0, x=1/2
E f(x) = −0.20987
E(f(x) | x is local maximum) = +0.3647 - so higher than space average
f(avg of local maxima) = f(0) = 1.
It is as trivial to construct sets with any relationship between E f(x), f(avg of local maxima), and E(f(x) | x is local maximum).
for arbitrary space and function, set of local maxima will behave essentially like a random set. It can as easily have average less or more than average of the whole space.
Not “easily”. Only if the space is constructed to have more local maxima when f(x) is small, or the whole space has areas where f(x) goes off to infinity with no local maximum, or has some other perversity of construction that I haven’t thought of.
Space: [-1/2..+1/2], f(x) = cos(2.02 pi x).
There are 3 local maxima, x=-1/2, x=0, x=1/2
You’re exploiting the boundaries, which you’ve chosen specially for this purpose. I admit that when I said this outcome was impossible, I was ignoring that kind of a space.
If, however, you consider the set of spaces where the lower and upper bounds can take on any two real values (with the upper bound > lower bound), you’ll find the average of the average of the local maxima is greater than the average of the average.
I concede that you can define spaces where the set of local maxima can be below average. But you are wrong to say that they are therefore like a random set.
Note that the fact that you can define spaces where the set of local maxima can be below the average in the space does not impact my proof, which never talked about the average in the space.
You keep trying to guess proper caveats, I can giving you trivial counterexamples.
This one has: range over entire R, values in [-11,+10], average value 0, global maximum 10, average of local maxima −2.6524 ?
Any function which is more bumpy when it’s low, and more smooth when it’s high will be like that. This particular one chosen for prettiness of visualization.
Any function which is more bumpy when it’s low, and more smooth when it’s high will be like that.
That’s what I just said:
Only if the space is constructed to have more local maxima when f(x) is small,
You are hyper-focusing on this as if it made a difference to my proofs. Please note my previous comment: It does not matter; I never talked about the average IC over all possible agents. I only spoke of IC over various recombinations of existing agents, all of which I assumed to have IC that are local minima. The “random agent” is not an agent taken from the whole space; it’s an agent gotten by recombining values from the existing agents.
Yes.
No. value systems = local maxima of f, period.
What? Can’t parse.
Yes.
Huh? Where did that come from?
You said “if agents are better than random” in headlines. I take “than random” to mean ” > E f(x)”. So do you assume so or not?
Let me try again.
A—set of all local maxima x
B—set of all local maxima x, for which f(x) > E f(x)
In either case, avg(A) and avg(B) are arbitrary points, and we have no a priori reason to believe they will be special in any way, so E f(avg(A)) = E f(avg(B)) = E f(x). Is this right?
In case of set B—we assumed for all x \in B . f(x) > E f(x), so picking avg(B) which is essentially a random point makes it worse.
In case of set A—E f(x) for set of arbitrary local maxima is no better than set of arbitrary points, so E(f(x) | x \in A) = E(f(x)) = E f(avg(a)), so your entire argument fails.
f(x) is IC(x), and E is expected value?
I didn’t discuss anything corresponding to your B.
No. You just said they’re local maxima, which is very special; and it would be surprising if either of their expectations were the same as E[f(x)].
This doesn’t describe what I wrote; but B is not a random set, so avg(B) is not a random point.
No; local maxima (actually minima in this case, but same thing) are not arbitrary points. They’re locally maximal. Meaning they have a higher f(x) than all the points around them. So it’s impossible for the equality you just gave to hold, except when there are no local maxima (eg, a plane), or perhaps in some peculiar set constructed so that infinities make finite differences not matter. EDIT: Also add cases where a space is constructed so that local maxima are more common when f(x) is small, which is the type taw used in his reply. This is a large number of possible spaces. I believe it’s a minority of possible spaces, but it would take time to formalize that.
You say B is not a random set, but for arbitrary space and function, set of local maxima will behave essentially like a random set. It can as easily have average less or more than average of the whole space.
Here’s a really really simple example:
Space: [-1/2..+1/2], f(x) = cos(2.02 pi x).
There are 3 local maxima, x=-1/2, x=0, x=1/2
E f(x) = −0.009899
E(f(x) | x is local maximum) = −0.333 - so lower than space average
f(avg of local maxima) = f(0) = 1.
Another one:
Space: [-1/2..+1/2], f(x) = cos(3.03 pi x).
There are 3 local maxima, x=-1/2, x=0, x=1/2
E f(x) = −0.20987
E(f(x) | x is local maximum) = +0.3647 - so higher than space average
f(avg of local maxima) = f(0) = 1.
It is as trivial to construct sets with any relationship between E f(x), f(avg of local maxima), and E(f(x) | x is local maximum).
Not “easily”. Only if the space is constructed to have more local maxima when f(x) is small, or the whole space has areas where f(x) goes off to infinity with no local maximum, or has some other perversity of construction that I haven’t thought of.
You’re exploiting the boundaries, which you’ve chosen specially for this purpose. I admit that when I said this outcome was impossible, I was ignoring that kind of a space.
If, however, you consider the set of spaces where the lower and upper bounds can take on any two real values (with the upper bound > lower bound), you’ll find the average of the average of the local maxima is greater than the average of the average.
I concede that you can define spaces where the set of local maxima can be below average. But you are wrong to say that they are therefore like a random set.
Note that the fact that you can define spaces where the set of local maxima can be below the average in the space does not impact my proof, which never talked about the average in the space.
How about this function?
You keep trying to guess proper caveats, I can giving you trivial counterexamples.
This one has: range over entire R, values in [-11,+10], average value 0, global maximum 10, average of local maxima −2.6524 ?
Any function which is more bumpy when it’s low, and more smooth when it’s high will be like that. This particular one chosen for prettiness of visualization.
That’s what I just said:
You are hyper-focusing on this as if it made a difference to my proofs. Please note my previous comment: It does not matter; I never talked about the average IC over all possible agents. I only spoke of IC over various recombinations of existing agents, all of which I assumed to have IC that are local minima. The “random agent” is not an agent taken from the whole space; it’s an agent gotten by recombining values from the existing agents.
If agents are, on average, better than the average random agent as defined. No requirement that each agent be better than random.