## Sunday, March 16, 2008

### Mathematical odds and sods for SDP

Hello all, hope you had a good spring break. Scribe notes for the last two lectures are up. Sorry for the delay; my fault, not the scribes'.

I thought I would record here a few small mathematical facts that came up in our lectures on SDP algorithms for Max-Cut and Coloring.

The GW constant: Recall the GW constant,

$\alpha := min_{-1 \leq \rho \leq 1} [\arccos(\rho)/\pi]/[1/2 - \rho/2]$.

Using calculus and some trigonometric substitutions, you can determine that $\alpha = 2/(\pi sin \theta)$, where $\theta$ is the positive solution of $\theta = \tan(\theta/2)$. As far as I know, it is not known if $\alpha$ is irrational, although presumably it is transcendental.

Gaussians: Remember that a real random variable X has the "standard Gaussian (normal) distribution" if its pdf is $\phi(x) := (1/\sqrt{2\pi}) \exp(-x^2/2)$. The annoying thing about this pdf is that it does not have a closed-form integrals although several closely related expressions do. For example, $x\phi(x)$ has $-\phi(x)$ as an antiderivative. It's easy to check that E[$X^k$] = 0 whenever $k$ is an odd positive integer, and with enough integration-by-parts effort you can check that E[$X^k]= (k-1)(k-3) \cdot \cdot \cdot 3 \cdot 1$ whenever $k$ is an even positive integer. In particular, X has mean 0 and variance 1.

In class we were particularly interested in the "tail", $\mathcal{N}(t) :=$ Pr[X $\geq t$] $= \int_{t}^\infty \phi(x) dx$, where $t$ is positive. Here is how to accurately estimate this quantity:

For an upper bound, just insert a factor of $x/t$ into the integrand, which is always at least $1$ on the range. Then you can immediately integrate and get the upper bound

$\mathcal{N}(t) \leq (1/t) \phi(t)$.

For a lower bound, you need a cleverer trick. Insert a factor of $1 - 3/x^4$ into the integral (!), which is always less than $1$. You can check that the resulting quantity, $(1-3/x^4)\phi(x)$ has an explicit antiderivative: $(1/x^3 - 1/x)\phi(x)$. Thus you can conclude

$\mathcal{N}(t) \geq (1/t - 1/t^3) \phi(t)$.

(I think this trick is due to Feller.) Thus we have excellent bounds for $\mathcal{N}(t)$ for large $t$: in particular,

$\mathcal{N}(t) \sim (1/t) \phi(t)$

as mentioned in class.

How to draw a Gaussian random variable: In practice, just use "randn" in Matlab :) In theory, suppose you have access to an oracle which gives you a uniform real from the interval [0,1]. Then the easiest thing is to draw two random numbers S and $\theta$ from [0,1] and then set g = $\sqrt{\ln(1/s)} \cos(2\pi \theta)$, h = $\sqrt{\ln(1/s)} \sin(2\pi \theta)$. Then g and h are two independent Gaussians! (This is the "Box-Muller Transformation".) It is a nice exercise to see why this is true.

Of course, in practice in theory (if you will), the usual model of randomness is to assume you have random bits. Also, of course, you can't compute ln, cos, sin, $\pi$, etc. to arbitrary precision. So in reality, you need to make some approximations.

Just as with the issue of solving SDPs exactly, these approximations introduce additive errors of $\epsilon$, with running time overhead of polylog($1/\epsilon$). And just as with the SDP-solving issue, we will try to never speak of such annoying details again :)