where N is normalisation; it seems that you suppose that =0 and <f’> finite, then. =0 follows from boundedness, but for the derivative it’s not clear. If <f’> on (a,b) grows more rapidly than (b-a), anything can happen.
This cannot happen. f is assumed bounded. Therefore the average of f’ over the interval [a,b] tends to zero as the bounds go to infinity.
The precise, complete mathematical statement and proof of the theorem does involve some subtlety of argument (consider what happens if f = sin(exp(x))) but the theorem is correct.
The standard form for correlation coefficient is
cov(x,y)=N(-)
where N is normalisation; it seems that you suppose that =0 and <f’> finite, then. =0 follows from boundedness, but for the derivative it’s not clear. If <f’> on (a,b) grows more rapidly than (b-a), anything can happen.
This cannot happen. f is assumed bounded. Therefore the average of f’ over the interval [a,b] tends to zero as the bounds go to infinity.
The precise, complete mathematical statement and proof of the theorem does involve some subtlety of argument (consider what happens if f = sin(exp(x))) but the theorem is correct.