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I am not sure what you mean by moments, but in probability theory “the n-th moment” means E[X^n]; and you’ve clearly used them. In probability theory the Radon

I am not sure what you mean by moments, but in probability theory “the n-th moment” means E[X^n]; and you’ve clearly used them. In probability theory the Radon-Nikodym is used to justify existence of random variables. But it cannot be used to show that any variable has an expected value (ie has a moment, ie is integrable). The Cauchy distribution is a traditional counter example to such assertion. > For Y it is enough that it has an expectation That is false. Just look at your “simple example”. You arrive to E[Y^2]. An expectation is not enough. It should be clear intuitively because MSE is a minimization of the second moment of a certain rv, so you are bound to have at least some constraints for second-order relationships for the variables you’ve used to construct that certain rv. I think you may need to re-read Rudin and Neveau and refresh your knowledge. (Though I can say nothing about Neveau, I studied using the textbook by Shao.)

graycat 4y ago
For any real valued random variable, E[Y^2] always exists although it might be infinity. You mentioned moments -- as I recall some writing in statistics uses that word, but I never have. Your description of the Radon-Nikodym (R-N) result does not look right. Again, I don't want to write a lecture in graduate probability, but from memory the usual description of that result goes: Given real valued, integrable function f and sigma algebra s, there exists real value function g measurable with respect to sigma algebra s so that g integrates like f over all sets in s. So, with real valued random variable Y in place of function f, we let E[Y|s] be the function g. If s is the sigma algebra generated by real random variable X, then we write g(X) = E[Y|X] = E[Y|s] So the R-N result makes immediate the definition of E[Y|X] and shows that it exists. If you want a perfect, polished presentation, see the two references I gave, Rudin and Neveu. My main contribution here remains: f(X) = E[Y|X] minimizes E[ (Y - f(X))^2 ] and is a minimum variance, unbiased estimator of Y. That's a generalization of the start of this thread and a bit amazing.
- 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 ago
      You are not correcting me, you are correcting Rudin :)

graycat 4y ago

For any real valued random variable, E[Y^2] always exists although it might be infinity. You mentioned moments -- as I recall some writing in statistics uses that word, but I never have. Your description of the Radon-Nikodym (R-N) result does not look right. Again, I don't want to write a lecture in graduate probability, but from memory the usual description of that result goes: Given real valued, integrable function f and sigma algebra s, there exists real value function g measurable with respect to sigma algebra s so that g integrates like f over all sets in s. So, with real valued random variable Y in place of function f, we let E[Y|s] be the function g. If s is the sigma algebra generated by real random variable X, then we write g(X) = E[Y|X] = E[Y|s] So the R-N result makes immediate the definition of E[Y|X] and shows that it exists. If you want a perfect, polished presentation, see the two references I gave, Rudin and Neveu. My main contribution here remains: f(X) = E[Y|X] minimizes E[ (Y - f(X))^2 ] and is a minimum variance, unbiased estimator of Y. That's a generalization of the start of this thread and a bit amazing.

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 ago
    You are not correcting me, you are correcting Rudin :)