3 ms·
Actually, because differentiation is a limiting process, this relies on the bounded convergence theorem or monotone convergence theorem. However, I imagine this
by hexane360 5y ago
Actually, because differentiation is a limiting process, this relies on the bounded convergence theorem or monotone convergence theorem. However, I imagine this would hold in most distributions under consideration.
- contravariant 5y agoIndeed since expectation values make the constant functions integrable you'd have to do some pretty wild things before things goes wrong. In fact any bounded function (with bounded derivative) is going to be fine. I also wonder if differentiation doesn't have some nicer properties anyway, being a linear operation as well. Typically you can get quite far by showing things work for a sufficiently broad class of functions and just extending things from there. In particular you could use the Fourier transform, which trivially commutes with taking expectation values. You can always try something abhorrent like sin(e^(-1/x^2)) or something to make things go wrong though.