somewhere near the beginning.

Probablistic Integrals

Filed under: Mathematics — Alex @ 11:44 pm 10/27/2006

I completed my functional analysis midterm today– didn’t do too well. One of the problems, to show that bounded sequences in the space C[a,b] which converge pointwise to to a function in the space converge weakly to the same limit, really stumped me. On the positive side, it gave me the first opportunity I’ve had to appreciate the usefulness of Radon measures– using the characterization of the dual space of C[a,b] as the set of Radon measures on the space, and the dominated convergence theorem, this result is obvious. On the other hand… I wasn’t able to do it using the tools we learned in class so far. The problem would be straightforward if I could see why the point evaluation functionals are dense in the dual space, but I wasn’t able to see why that should be the case; in fact, it just seems intuitively wrong to say that.

My 6 hour take home stochastic processes midterm is due on Tuesday, and that course has been bewildering me, so I decided to do a couple of practice problems from a probability theory problem book before I got started, just to get the juices flowing and fix techniques in my mind. The first problem was to find the distribution function and expectation of the random variable defined by the magnitude of the projection of a point uniformly randomly chosen on the perimeter of the unit circle onto the x-axis. The interesting thing to note is that if you calculate the expectation over the sample space, you get a simple integral:

\frac{1}{2\pi} \int_0^{2\pi} |\cos(\theta)|\;d\theta.

On the other hand, if you evaluate it in the image space using the distribution function, you get a more intimidating integral:

\displaystyle \int_0^1 \frac{2x}{\pi \sqrt{1 - x^2}}\;dx

This suggests that sometimes probabilistic interpretations can be used to reduce hairy integrals. Of course, finding a useful interpretation may be very hard. Even in this case, it’s hardly obvious that the second integral is really just the expectation of a relatively simple random variable.

Of course, the evaluation in the probability space is not always easier than the evaluation in the image space. For example, if we change the r.v. to the square root of the above r.v., then

\displaystyle \int_0^1 \frac{2}{\pi} \sqrt{\frac{x}{1-x^2}}\; dx = \frac{4}{\sqrt{\pi}} \frac{\Gamma(3/4)}{\Gamma(1/4)}

is a heck of a lot easier, I imagine, to calculate than

\displaystyle \frac{1}{2\pi} \int_0^{2\pi} \sqrt{|\cos(\theta)|}\:d\theta = \frac{1}{\pi} \left( (1+i) \left( (1-i) \text{EE}[2] + \text{EE}[\pi/4,2] - \text{EE}[3\pi/4, 2]\right)\right),

where EE is shorthand for the appropriate type of elliptic integral.

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