I'm not sure what proof you have in mind, but I don't think using the word "converge" is correct.
Near as I can tell, your algorithm is x[n+1]= x[n] + sign(sample[n]-x[n]).
If you are applying the c.l.t. to sign(sample[n]-x[n]), all this says is that your algorithm will output a gaussian random variable with mean=median(samples), but sigma=sqrt(n). But maybe that isn't the proof you were thinking of?
[edit: also, the clt doesn't apply to sign(sample[n]-x[n]), since that isn't an i.i.d. sequence of random variables (it's not identical).
One possible proof I thought of: consider probability vectors of the form p[n] = [P(x[n] = 1), P(x[n] = 2), ...]. You can write the evolution of the probability vectors as matrix problem, p[n+1]=A p[n]. I'll bet you can break A = Projection onto median + remainder and show the remainder has norm less than 1. Just a guess, but I think it might work.]
Reminds me of a similar comment in the margin of a old book: http://en.wikipedia.org/wiki/Fermats_Last_Theorem. :)
But it does sound both plausible and brilliant. How do you prove it (in broad terms)?
Assume V_(n-1) = median(X) for the data stream X_n. Now we have
E[V_n | V_(n-1)] = V_(n-1) + E[signum(V_(n-1) - X_n)]
Proving it'll converge to that value would be harder, though. It'll probably depend on the data being much larger than the step size (which is 1).
"In a normal distribution the probability of a value higher or lower than mean is 50%. So if you find mean, the average of your +1/-1 noise is zero -> if you're on mean you'll remain there.
If you're not on mean you have a (50 + error)% chance of moving towards mean and (50 - error)% chance of moving away from mean on the next random value -> the trend is towards mean."
... I'm not following. I can see my reply a few posts down. I accidentally hit submit twice, deleted one but the second one is still there.
Edit: The same technique can be found in some 1-bit ADCs, I believe it's delta-sigma ADC.
And if you don't know they're integers, you have to find a scaling rule for your increment that assures convergence. This is the same problem as bucket size selection for the histogram.
The idea is to maintain a floating median. Start with an arbitrary number, say 0. If the incoming number is greater than the estimate, num += 1, else num -= 1.
It is easy to prove that for a large enough stream, assuming the stream is drawn from iid samples, this would converge to the median, by central limit theorem.
Using two variables instead of one, you could converge faster. So instead of incrementing/decrementing by one, you store how big your leap is, and have a schedule to change it.