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> Given real valued, integrable function f Of course that makes it possible. Note that it doesn’t say that f is simply measurable. By definition being integra
by LudwigNagasena 4y ago
> Given real valued, integrable function f
Of course that makes it possible. Note that it doesn’t say that f is simply measurable. By definition being integrable means E[|f|] < infinity. (And being square-integrable means E[f^2] < infinity, an even stronger condition).
Consider Y = Z/X where Z, X iid ~ N(0,1). Try to calculate E[Y|X], and try to minimize E[ (Y - f(X))^2 ]. Neither will make sense. I am sure both Rudin and Neveu were aware of this trivial result.
- graycat 4y ago> Of course that makes it possible. Note that it doesn’t say that f is simply measurable. Irrelevant. Again, once again, over again, yet again, for real valued random variable Y, E[Y^2] always exists. > By definition being integrable means E[|f|] < infinity. Wrong. Correcting you is a waste of my time.
- LudwigNagasena 4y agoYou are not correcting me, you are correcting Rudin :)