somewhere near the beginning.

A cool problem in measure theory

Filed under: Mathematics — Alex @ 5:45 pm 4/9/2007

On the first stochastic controls homework we have the following bonus problem (which is apparently now superbonus, since the instructor sent out an email saying we aren’t expected to be able to do this rigorously ),

Suppose that we have a sequence X_1, \ldots which are i.i.d. Gaussian random variables with mean zero and unit variance under P, and such that X_1 - a_1, X_2 - a_2, \ldots are i.i.d. Gaussian random variables with mean zero and unit variance under Q. Give a necessary and sufficient condition on the non-random sequence a_1, a_2, \ldots such that Q \ll P. In the case that Q \ll P, give the corresponding Radon-Nikodym derivative. If Q \not\ll P, find an event A so that P(A) = 0 but Q(A) \neq 0. in theory, how would you solve the hypothesis testing problem when Q \ll P? How about when Q \not\ll P?

(Here I’m assuming that Q = \bigotimes_{i=1}^\infty Q_i where Q_i = \mathcal{N}(a_i, 1), and similarly for P.)

I’m stumped on it. I have the intution that the sequence of means a_1, \ldots should be bounded, and I can construct an example where the fact that the a_1, \ldots aren’t bounded is crucial in establishing that Q \not\ll P. I can’t establish either the necessity or sufficiency, however. Aaargh!

For the necessity, given an unbounded sequence of means, I’ve been trying to construct a sequence A_i of sets such that Q(A) = \prod Q_i(A_i) > 0, but P(A) = \prod P_i(A_i) = 0, using the fact that the ‘mass’ of the Q_i are moving away from the origin. The easiest way of doing this that occured to me is to take the sets W_i = [-w_i, w_i] such that under the Gaussian measure \mu(W_i) = e^{-1/i^2}, shift them, and define A = \times_{i=1}^\infty (a_i +W_i) so Q(A) = \prod Q_i(W_i + a_i) = \prod \mu(W_i) = e^{-\pi^2/6} > 0, but these sets don’t move away from the origin fast enough to get P(A) = 0. Stumped there.

For the sufficiency, given a bounded set of means, I’ve been attempting to use the facts
 Q(A) = \inf \left\{ \sum_{i=1}^\infty Q(A_i) \,:\, A \subset \cup_{i=1}^\infty A_i \right\},
and likewise for P(A), where the inf is taken over the \{A_i\} in the algebra of finite Borel cylinders. Then if I could show that Q(A) goes to zero fast enough when P(A) does, just for the finite Borel cylinders, then I’d be done. But that fast enough is eluding me.

Any hints?

Update I got the necessity; it was actually trivial. My complaint above can be avoided by just taking an (absolutely) increasing subsequence of means which move away from the origin fast enough, doing the construction on it, and ignoring the other means by setting the other A_i = \R… Now on to sufficiency.

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