- Suppose you have a diffusion of 100 points with the following iteration:
$$(x_{n+1},y_{n+1}) \sim \mathcal{N}\left((x_n,y_n), \frac{x_n^2 + y_n^2}{2} I_{2 \times2}\right)$$
This will make a high variance for the points far from the origin and a low variance for the points close to the origin. ( The above equation is just a try to imitate this idea. We can use different equations.)
I think most of the points will come close to the origin after some time.
Question
- Do you also think that?
- Is there any theorem through which we can prove the above?
- Correct me if I was wrong.
Your sequence of points will diverge to infinity.
We can write the updates of the Markov chain $\left\{\left(x_n, y_n\right)\right\}_n$ as $$ \left(x_{n+1}, y_{n+1}\right) = \left(x_n, y_n\right) + \sqrt{\frac{x^2_n + y_n^2}{2}} Z_{n+1},$$ where $\left\{Z_n\right\}_n$ is an independent and identically distributed (i.i.d) sequence of standard two-dimensional Gaussian random variables. Define by $r_n^2 = x_n^2 + y_n^2$ the squared distance from the origin. The sequence $\left\{r_n^2\right\}_n$ satisfies the following recurrence $$r_{n+1}^2 = \left(1+ \sqrt{2}\left\langle \frac{\left(x_n, y_n\right)}{\sqrt{x^2_n + y_n^2}}, Z_{n+1}\right\rangle + \frac{1}{2}\left|Z_{n+1}\right|^2\right)r_n^2. $$ This sequence is equal in law to the sequence $\left\{\tilde{r}_n\right\}$ given by $\tilde{r}_0 = r_0$ and $$\tilde{r}_n = \left(1+ \sqrt{2}\left\langle \left(1,0\right), Z_{n+1}\right\rangle + \frac{1}{2}\left|Z_{n+1}\right|^2\right) \tilde{r}_n^2, $$ since a standard Gaussian is invariant by rotation. Furthermore $\left\{\tilde{r}_n\right\}$ is a Markov chain. We can then consider the logarithm of this chain, $q_n = \log\left(\tilde{r}_n^2\right)$, which satisfies $$q_{n+1} = q_n + \log\left(1 + \sqrt{2}\left\langle \left(1, 0\right), Z_{n+1}\right\rangle + \frac{1}{2}\left|Z_{n+1}\right|^2\right).$$ $q_n - q_0$ is the sum of i.i.d random variables $$ q_n - q_0 = \sum_{k = 1}^n \log\left(1 + \sqrt{2}\left\langle \left(1, 0\right), Z_{k}\right\rangle + \frac{1}{2}\left|Z_{k}\right|^2\right).$$ We can check that $ \log\left(1 + \sqrt{2}\left\langle \left(1, 0\right), Z_{1}\right\rangle + \frac{1}{2}\left|Z_{1}\right|^2\right)$ is square-integrable and numerically estimate its mean, $\mu$, and variance, $\sigma^2$, $$ \mu = \mathbb{E}\left[\log\left(1 + \sqrt{2}\left\langle \left(1, 0\right), Z_{k}\right\rangle + \frac{1}{2}\left|Z_{k}\right|^2\right)\right] \approx 0.219, $$ $$ \sigma^2 = \mathrm{Var}\left(\log\left(1 + \sqrt{2}\left\langle \left(1, 0\right), Z_{k}\right\rangle + \frac{1}{2}\left|Z_{k}\right|^2\right)\right) \approx 1.376.$$ So by the strong law of large numbers, $$\frac{q_n - q_0}{n} = \frac{1}{n}\sum_{k = 1}^n \log\left(1 + \sqrt{2}\left\langle \left(1, 0\right), Z_{k}\right\rangle + \frac{1}{2}\left|Z_{k}\right|^2\right) \to \mu >0 \quad \text{almost surely,}$$ which in turn implies that $q_n \to \infty$ almost surely. Consequently, $r_n$ also converges to $\infty$ almost surely and the chain $\left\{\left(x_n, y_n\right)\right\}_n$ diverges.