3 ms·
Thinking about this more... first of all it should have been, where P(q)=="probability of q": alpha = E[E[P(alpha|data)|given data]] Now we can simplify. Say
by nonbel 11y ago
Thinking about this more... first of all it should have been, where P(q)=="probability of q":
alpha = E[E[P(alpha|data)|given data]]
Now we can simplify. Say x=alpha, y=data, z="given data", the E operator can be an arbitrary function f, and applying function f twice is denoted as f#. We get:
x=f#(P(x|y)|z))
We know P(a|b)=P(a & b)/P(b) and p(a & b)=P(a)P(b), so
x=f#{[P(z)P(x)P(y)]/P(y)}=f#{P(z)P(x)}
Denote the inverse of f# as F, then:
F(x)=P(z)P(x)
Now, z was "given data" which has probability 1. So,
F(x)=P(x)
In other words, there is method to your madness.
- graycat 11y agoJust routine conditional expectation. For random variables X and Y (under very meager assumptions, E[X|Y] is the almost sure unique random variable that integrates like X over all events in the sigma algebra generated by Y. In particular E[X] = E[E[X|Y]] Details follow from the Radon-Nikodym theorem in, say, Rudin, Real and Complex Analysis where Rudin give the von Neumann proof. Details are also in the standard texts on graduate probability by Loeve, Neveu, Breiman, Chung, and a few more.
- nonbel 11y ago>"alpha is the expected value of p-value" Well, you were correct about this above. You may be onto something here in terms of "actual use".