Yes. But it’s not deep; I recommend trying yourself before consulting the answer. It follows straightforwardly from the fact that the integral of x(dx/dt) is (x^2)/2. The rest is bookkeeping to eliminate edge cases.
I didn’t trouble to state the result with complete precision in the OP. For reference, here is an exact formulation (Theorem 2 of the linked note):
Let x be a differentiable real function. If the averages of x and dx/dt over the whole real line exist, and the correlation of x and dx/dt over the whole real line exists, then the correlation is zero.
I think precision would require you to state this in terms of a variable x and the function f(x). (EDIT: Sorry; please ignore this.)
If the averages of x and dx/dt over the whole real line exist,
This is a pretty harsh requirement! It will be true for constant functions, cyclic functions, symmetric functions, and maybe asymptotically-bounded functions. I don’t think you can say it’s true for y=x.
I think precision would require you to state this in terms of a variable x and the function f(x).
gjm has read the note I linked; I suggest you do the same. That is what a link is for.
This is a pretty harsh requirement!
Not particularly. The speed of a car, the temperature of a room, the height of an aircraft: such things are all around you. Stating the property of the whole real line is an idealisation, but Theorem 1 of the note treats of finite intervals also, and there is a version of the theorems for time series.
I don’t think you can say it’s true for y=x.
In keeping with the terminology established at the note I linked, I take this to mean x=t. Yes, it is not true of x=t. This does not have an average over the whole real line.
gjm has read the note I linked; I suggest you do the same. That is what a link is for.
I wish I hadn’t made my comment about precision, which was too nitpicking and unhelpful. But as long as we’re being snippy with each other:
To be excruciatingly precise: You just said you were being precise, then said “Let x be a differentiable real function.” That isn’t precise; you need to specify right there that it’s a function of t. If you’d said the link stated it precisely, that would be different.
I admit that I would have interpreted it correctly by making the most-favorable, most-reasonable interpretation and assuming x was a function of t. But, because of the sorts of things I usually see done with x and t, I assumed that x was a function of time, and the function of interest was some function of x(t), and I jumped to the conclusion that you meant to say “Let f(x) be a differentiable real function.” Which I would not have done had you in fact been precise, and said “Let x(t) be a differentiable real function.”
Sorry that I sounded dismissive. It’s a nice proof, and it wasn’t obvious to me.
I am uncomfortable with using Pearson correlation to mean correlation. Consider y=sin(x), dy/dx = cos(x). These are “uncorrelated” according to Pearson correlation, but given one, there are at most 2 possibilties for the other. So knowing one gives you almost complete info about the other. So calling them “independent” seems wrong.
Yes. But it’s not deep; I recommend trying yourself before consulting the answer. It follows straightforwardly from the fact that the integral of x(dx/dt) is (x^2)/2. The rest is bookkeeping to eliminate edge cases.
I didn’t trouble to state the result with complete precision in the OP. For reference, here is an exact formulation (Theorem 2 of the linked note):
Let x be a differentiable real function. If the averages of x and dx/dt over the whole real line exist, and the correlation of x and dx/dt over the whole real line exists, then the correlation is zero.
I think precision would require you to state this in terms of a variable x and the function f(x). (EDIT: Sorry; please ignore this.)
This is a pretty harsh requirement! It will be true for constant functions, cyclic functions, symmetric functions, and maybe asymptotically-bounded functions. I don’t think you can say it’s true for y=x.
He’s actually working with a variable t and the function x, whose value at a particular t is x(t). I don’t see anything wrong there.
Yep, sorry.
gjm has read the note I linked; I suggest you do the same. That is what a link is for.
Not particularly. The speed of a car, the temperature of a room, the height of an aircraft: such things are all around you. Stating the property of the whole real line is an idealisation, but Theorem 1 of the note treats of finite intervals also, and there is a version of the theorems for time series.
In keeping with the terminology established at the note I linked, I take this to mean x=t. Yes, it is not true of x=t. This does not have an average over the whole real line.
Full disclosure: actually I didn’t, I just inferred what the notation had to mean :-).
I wish I hadn’t made my comment about precision, which was too nitpicking and unhelpful. But as long as we’re being snippy with each other:
To be excruciatingly precise: You just said you were being precise, then said “Let x be a differentiable real function.” That isn’t precise; you need to specify right there that it’s a function of t. If you’d said the link stated it precisely, that would be different.
I admit that I would have interpreted it correctly by making the most-favorable, most-reasonable interpretation and assuming x was a function of t. But, because of the sorts of things I usually see done with x and t, I assumed that x was a function of time, and the function of interest was some function of x(t), and I jumped to the conclusion that you meant to say “Let f(x) be a differentiable real function.” Which I would not have done had you in fact been precise, and said “Let x(t) be a differentiable real function.”
Sorry that I sounded dismissive. It’s a nice proof, and it wasn’t obvious to me.
I am uncomfortable with using Pearson correlation to mean correlation. Consider y=sin(x), dy/dx = cos(x). These are “uncorrelated” according to Pearson correlation, but given one, there are at most 2 possibilties for the other. So knowing one gives you almost complete info about the other. So calling them “independent” seems wrong.
Sorry that I sounded dismissive. It’s a nice proof, and it wasn’t obvious to me.