Could you show the form of a simple 2,3, or n dimensional SDE and then provide a simple intuitive explanation of what the terms are describing in the sense of local probability flows? If so, could you give a real world example where you might think you know how the local probability should be flowing and thereby just write down the corresponding SDE to model the situation?

My thought has been that the first order term is describing the local linear drift while the second order term is describing the local shape of expansion of "point masses" of probabilty. I was never able to get a teacher to say something like this though. All I could ever get to such requests for an intuitive concept was the intricate technical definition with laborious proof of the appropriate stochastic limit to the sequence of appropriate discrete jump paths. Or something like that.

Thanks,

PairTheBoard

I would have put Courant and UBC over Columbia and Stanford on Jason's original list but otherwise agree with it. Prior to hiring Berger, I would have said that UCLA was noticeably below the other schools on his list and that it didn't really have a good culture for young probabilists. Noam is both very strong and energetic, which helps, but unless you are interested in the sort of research that he and Marek do I wouldn't go there.

SHELDON ROSS!!!!!!!!!

Characteristic functions are probably as good as anything. I would have just said that the law of (X_i,X_j) (that is, the probability measure on R^2 induced by (X_i,X_j)) is just the product measure m*m, where m is the law of X_1. For each \varepsilon, there exists a Borel set A such that

P(|X_i - X_j| > \varepsilon) = P((X_i,X_j) in A) = m*m(A),

and this does not depend on i or j.

You want to be careful here that you are not assuming the existence of any moments. You could also say that, by independence, F(x) = P(X_1 <= x) satisfies F(x) = (F(x))^2. Hence, for each x, F(x) = 0 or F(x) = 1. Since F is nondecreasing and right-continuous, there exists a constant c such that F(x) = 0 for x < c and F(x) = 1 for x >= c.

ok, now i feel silly

what drew you to math as opposed say physics, engineering or computer science?

Here is a simple 2-dimensional SDE:

dX = m(X) dt + s(X) dB.

The solution, X, will be a process taking values in R^2; B is a 2-dimensional Brownian motion. The function m maps R^2 to R^2, and the function s maps a point in R^2 to a 2 x 2 matrix. The above equation is shorthand for the pair of equations

dX_1 = m_1(X) dt + s_{11}(X) dB_1 + s_{12}(X) dB_2,
dX_2 = m_2(X) dt + s_{21}(X) dB_1 + s_{22}(X) dB_2.


(Think of X, m(X), and dB as 2 x 1 column vectors; dt is a scalar; and s(X) is a 2 x 2 matrix.) In turn, these are shorthand for the integral equations

X_i(t) = X_i(0) + \int_0^t m_i(X(s)) ds
+ \int_0^t s_{i1}(X(s)) dB_1(s)
+ \int_0^t s_{i2}(X(s)) dB_2(s),

where the last two integrals are Ito integrals.

I am not sure what you mean by the above term in bold. Does this refer to the way the probability density function of the solution changes with time? Under suitable assumptions on m and s, the solution X(t) has a density function f(x,t) that satisfies the Kolmogorov forward equation (also called the Fokker–Planck equation).

Personally, I do not find this to be a very intuitive way of thinking about the dynamics described by the SDE. I prefer to think about the paths of X, rather than the densities f(x,t). Informally, I would write

dX/dt = m(X) + s(X) W(t),

where W(t) = dB/dt. The "white noise" term W(t) exists only as a generalized function, but you can think of it informally as a stationary, R^2-valued process that gives an infinitesimal random push at time t to X. The push has no preferred direction and is independent of the pushes given prior to time t. The usual example is a pollen grain on the surface of still water being pushed by collisions with the surrounding water molecules.

In this case, the term m(X) gives a deterministic drift, and the term s(X) serves to modify the magnitude of the noise term and, possibly, give it a preferred direction. (Here, "direction" does not mean a vector, but a line through the origin. The modified noise term will still have mean 0.) In this sense, the term s(X) is changing the shape of the density of the noise term. But this is all completely informal. Rigorously, the noise term is not even a random variable (in the usual sense), so there is no density whose shape can change.

There should be some nice examples in Oksendal that illustrate what I described above. If not, you should be able to cook something up yourself. Maybe you could model a pollen grain on a gently drifting pool of water whose temperature varies spatially.

Would you rather live forever or die?

Originally, as an undergrad, it was probably just the fact that math was what I perceived as my strongest subject. But I did always enjoy my math classes much more than any others, so I am sure that was a factor on some level.

Personally, I have a hard time understanding a lot of physicists, engineers, and computer scientists. Maybe it's the skeptic in me, but I need to see the rigor before I feel I truly understand. I had a math professor in grad school who was originally a physicist. He was an older guy and said he was considering going back into physics after all these years. He explained to us that he originally left physics for math so that he could figure out what all those physicists were actually talking about. Well, now he knew, so he was ready to return.

I also like the flexibility I have. I can work on mathematical models in any field of my choice. Not only that, I am free to play with the models in any way I like, as long as it is mathematically interesting. And all of my research can be done with a pen and paper. That I like.

I expect the reaction of most math people will be "lol".

ummmm... no

I've seen some Twilight Zones where the guy gets a seemingly amazing gift, then there's a twist, and the so-called gift becomes a fate worse than death. So out of risk-aversion, I think I would rather keep reality as it is and die the death I've got coming to me.

However, had you asked:

I think I'd go for it.

This is x-post from the probability forum but it's pretty quiet so i'll try if you could answer me...

I was in a live tournament, and when the cards were being dealt one guy had his second card flip over and it was a Q. After he had been given new card, he checked them and said "i have a Q, but i dont know which of these cards was the first one, so it's 50-50 that i would've gotten a pair of queens". Now i thought his logic was false and it would be more than 50% to get the pair, as when the first cards were dealt there were four queens in the deck and on the second card only three. But i couldnt prove it better than that and have no idea where to start if i had to count the correct odds for catching the pair, so what's the best way to do it?

Thanks,

Saku

The result is correct. For simplicity, imagine that his original two cards were the top two cards in the deck, and his replacement card is the third from the top. We first learn

A = Card 2 is a Q.

Then we learn

B = Cards 1 and 3 are a Q and a non-Q, in some order.

Given A and B, we want the probability of

C = Card 1 is a Q,

which is

P(C | A and B) = P(A and B and C)/P(A and B)
= P(Q, Q, non-Q)/[P(Q, Q, non-Q) + P(non-Q, Q, Q)].

We now compute

P(Q, Q, non-Q) = (4/52)*(3/51)*(48/50), and
P(non-Q, Q, Q) = (48/52)*(4/51)*(3/50).

But both of these are just (4*3*48)/(52*51*50), so they are equal and the final answer is 1/2.

How much physics do you know or have you tried to learn?

[It is interesting to me because I know tons of people who are interested in learning physics but do not know enough math so I am curious about people who know enough math but simply aren't that interested]