This week in class I get to teach one of my favorite probability arguments, which makes use of a very unusual embedding. Here's a short description (for the longer description, see the book or the original paper, where the idea is ascribed to Rubin).
The setting is balls and bins with feedback: I have two bins, and I'm repeatedly randomly throwing balls into the bins one at at time. When there are x balls in bin 1 and y balls in bin 2, the probability the ball I throw lands in bin 1 is x^p/(x^p+y^p), and the probability it lands in bin 2 is y^p/(x^p + y^p). Initially both bins start with one ball. The goal is to show that when p is greater than 1, at some point, one bin gets all the remaining balls thrown. That is, when there's positive feedback, so the more balls you have the more likely it is that you'll get the next one in a super-linear fashion, eventually the system becomes winner-take-all.
We use the following "exponential embedding". Consider the following process for bin 1. At time 0, we associated an an exponentially distributed random variable X_1 with mean 1 = 1/1^p with the bin. The "time" that bin 1 receives its next ball is X_1. Now it has two balls. We then associate an exponentially distributed random variable X_2 with mean 1/2^p with the bin. And so on.
Do the same thing with bin 2, using random variables Y_1, Y_2, ...
Now, at any point in time, due to the properties of the exponential distribution -- namely, it's memoryless, and the minimum of two exponentials with mean a_1 and a_2 will be the first with probability proportional to 1/a_1 and the second with probability 1/a_2 -- if the loads in the bins are x for bin 1 and y for bin 2, then the next ball will fall into bin 1 is x^p/(x^p+y^p), and the probability it lands in bin 2 is y^p/(x^p + y^p). That is, this exponential process is equivalent to the initial balls and bins process.
Now let X = sum X_i and Y = sum Y_i. The infinite sums converge with probability 1 and are unequal with probability 1. So suppose X < Y. Then at some finite "time" in our exponential embedding, bin 1 receives an infinite number of balls while bin 2 just has a finite number of balls, and similarly if Y < X. So eventually, one bin will be the "winner" and take all the remaining balls.
In my mind, that's a beautiful proof.