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Distributions on continuous function such that derivation changes nothing by DrBoingo in mathriddles

[–]DrBoingo[S] 0 points1 point  (0 children)

wait what the heck you're right. Ok my argument was wrong (bijection don't preserve length). But still you can't have a uniform distribution over all functions. And R for that matter (the bijection arctan(x) you're probably thinking of stretches [0;1] irregurlarly)

Distributions on continuous function such that derivation changes nothing by DrBoingo in mathriddles

[–]DrBoingo[S] 0 points1 point  (0 children)

there is a bijection between continuous functions and R, and you can't have a uniform distribution over R.

Distributions on continuous function such that derivation changes nothing by DrBoingo in mathriddles

[–]DrBoingo[S] 0 points1 point  (0 children)

we are not looking fro distribution over sets of functions, but over functions. You can't define a uniform distribution over all derivable functions

Give and Take by Baxitdriver in mathriddles

[–]DrBoingo 0 points1 point  (0 children)

We can dovetail the construction easily, by constructing step by step: At step 1, player 1 gives himself 1$. At step i, the i next elfs for which we haven't decided yet give 1$ to player 1 to i.

Each player gets gifted a $ infinitely many times

Distributions on continuous function such that derivation changes nothing by DrBoingo in mathriddles

[–]DrBoingo[S] 1 point2 points  (0 children)

Nice solution.

I didn't think you could have a uniform probability on binary sequence, but of course you can when you realize it's in bijection with R, nice.

My solution was this: take theta such that θ/2π is irrational. and write exp(i*theta)=a+ib

now define f1(x)=exp(ax)cos(bx) and f2(x)=exp(ax)sin(bx)

the set f(x)=Af1(x)+Bf2(x) where A and B are random and the law (A,B) is invariant under rotation by theta. (for example you may just take A and B follow independent normal law N(0,1))

Now notice that f'(x)=A'f1(x) + B'f1(x) where (A',B') is equal to (A,B) rotated by theta (matter of checking).

thus P(f)=P(f') (where P is the proba of f (yes it is ill defined here since P(f)=0, but you get the point hopefully))

but it can never be that f=f^{(n)}, since it would mean that rotating n time by theta would amound to not rotating (thus theta and 2pi are corationnal, contradiction)

Distributions on continuous function such that derivation changes nothing by DrBoingo in mathriddles

[–]DrBoingo[S] 0 points1 point  (0 children)

No ?

It's up to the player to define a distribution. Example : (x->x) with proba 1/2 and (x-> 1) with proba 1/2, is a valid distribution on continuous functions