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Wouldn't you multiply like you do for functions as an inner product for hilbert space? Something like: $ f \cdot g = \int_\infty^\infty f(x)g(x)dx $ No? The
by Enginerrrd 6y ago
Wouldn't you multiply like you do for functions as an inner product for hilbert space?
Something like:
$ f \cdot g = \int_\infty^\infty f(x)g(x)dx $
No?
The schwartz inequality gives us that the product then has a value between 0 and 1. ...Not sure what it means if the value is less than one. What does orthogonality mean when it comes to probability density functions?
- sn41 6y agoUsually orthogonal means uncorrelated random variables (not necessarily independent). A good example is sine and cosine - they are uncorrelated in the interval 0 to pi. But clearly they are not independent.
- Mithriil 6y agoSine and cosine can be negative. Density functions cannot, and thus, orthogonality of the density functions implies that they don't have the same support with probability 1, or, said otherwise, the measure of the intersection of their support is 0.
- andi999 6y agoScalar product doesnt give you a function/density whatever, just a number (which can be important, but this doesnt adress the question)