Random Walk

$$X_n = \sum_{k=1}^{n} \xi_k, \quad X_0 = 0.$$$$W(m,n) = {N \choose (N+m)/2} \left( \frac{1}{2} \right)^N$$$$\mathbb{E}[X_N] = 0, \quad \sigma^2_{X_N} = N$$

Diffusion Coefficient

$$D = \frac{\langle (X_N - X_0)^2 \rangle}{2N}$$$$D = \lim_{t \rightarrow \infty} \frac{\langle (X_t - X_0)^2 \rangle}{2dt}$$

where $d$ is the space dimension.

Intuitively, the diffusion coefficient tells us the rate at which variance changes [1]. For the random walk, $D = 1/2$.

Continuum limit of the random walk

$$N,m \rightarrow \infty, \quad l,\tau \rightarrow 0, \quad N\tau = t, \quad ml=x$$$$\begin{align} D &= \lim_{t \rightarrow \infty} \frac{\langle (X_t - X_0)^2 \rangle}{2t} \\ &= \lim_{t \rightarrow \infty} \frac{\langle (X_{N\tau} - X_0)^2 \rangle}{2N\tau} \\ &= \lim_{t \rightarrow \infty} \frac{\langle X_{N\tau}^2 \rangle}{2N\tau} \\ &= \lim_{t \rightarrow \infty} \frac{N l^2}{2N\tau} \\ &= \frac{l^2}{2\tau} \\ \end{align}$$

⚠️ the book mentions “fixing” the diffusion constant, is that different from the computation above?

$$\begin{align} W(m,N) &= \frac{N!}{\left(\frac{N+m}{2}\right)! \left(\frac{N-m}{2}\right)!} \left(\frac{1}{2}\right)^N \\ \log W(m,n) &= \log \left( \frac{N!}{\left(\frac{N+m}{2}\right)! \left(\frac{N-m}{2}\right)!} \left(\frac{1}{2}\right)^N \right) \\ &\approx \log N! - N\log 2 - \left(\log\left(\frac{N + m}{2}\right)! + \log\left(\frac{N-m}{2}\right)! \right) \\ &\approx \left(N+\frac{1}{2}\right)\log N - N + \frac{1}{2}\log 2\pi - N\log 2 \\ & \quad \quad - \left(\log\left(\frac{N + m}{2}\right)! + \log\left(\frac{N-m}{2}\right)! \right) \\ &\approx \left(N+\frac{1}{2}\right)\log N - N + \frac{1}{2}\log 2\pi - N\log 2 \\ & \quad \quad - \frac{1}{2} \left( N + m + 1 \right) \log \left(\frac{N+m}{2}\right) + \frac{N+m}{2} - \frac{1}{2}\log 2\pi \\ & \quad \quad - \frac{1}{2} \left( N - m + 1 \right) \log \left(\frac{N-m}{2}\right) + \frac{N-m}{2} - \frac{1}{2}\log 2\pi \\ &\approx \left(N+\frac{1}{2}\right)\log N \\ & \quad \quad - \frac{1}{2} \left( N + m + 1 \right) \log \left( \frac{N}{2}\left(1+\frac{m}{N}\right)\right) \\ & \quad \quad - \frac{1}{2} \left( N - m + 1 \right) \log \left(\frac{N}{2}\left( 1 - \frac{m}{N} \right)\right) \\ & \quad \quad -\frac{1}{2}\log 2\pi - N\log 2 \end{align}$$$$\begin{align} \log W(m,N) &\approx \left(N+\frac{1}{2}\right) \log N - (N+1) \log\left(\frac{N}{2}\right) \\ & \quad \quad - \frac{1}{2} \left( N + m + 1 \right) \log \left(1+\frac{m}{N}\right) \\ & \quad \quad - \frac{1}{2} \left( N - m + 1 \right) \log \left( 1 - \frac{m}{N} \right)\\ & \quad \quad -\frac{1}{2}\log 2\pi - N\log 2 \\ &\approx -\frac{1}{2} \log N + \log(2) - \frac{1}{2}\log 2\pi \\ & \quad \quad - \frac{1}{2} \left( N + m + 1 \right) \log \left(1+\frac{m}{N}\right) \\ & \quad \quad - \frac{1}{2} \left( N - m + 1 \right) \log \left( 1 - \frac{m}{N} \right) \\ &\approx -\frac{1}{2} \log N + \log(2) - \frac{1}{2}\log 2\pi \\ & \quad \quad - \frac{1}{2} \left( N + m + 1 \right) \left( \frac{m}{N} - \frac{1}{2}\left(\frac{m}{N}\right)^2\right) \\ & \quad \quad - \frac{1}{2} \left( N - m + 1 \right) \left( -\frac{m}{N} - \frac{1}{2}\left(\frac{m}{N}\right)^2\right) \\ &\approx -\frac{1}{2} \log N + \log(2) - \frac{1}{2}\log 2\pi - \frac{m^2}{2N} \\ \end{align}$$$$\implies W(m,N) \approx \left( \frac{2}{\pi N} \right)^{1/2}\exp\left(-\frac{m^2}{2N}\right) $$$$\begin{align} W(x,t)\Delta x &\approx \int_{x-\Delta/2}^{x+\Delta/2} W(y, t) dy \\ &\approx \sum_{\substack{k = \{m, m \pm 2, m \pm 4, \ldots\} \\ kl \in (x-\Delta x/2, x+\Delta x/2)}} W(k,N) \\ &\approx W(m,N)\frac{\Delta x}{2m} \end{align}$$$$W(x,t) = \frac{1}{\sqrt{4\pi D t} } \exp\left(-\frac{x^2}{4Dt}\right)$$$$\begin{cases} \frac{\partial W(x,t)}{\partial t} = D \frac{\partial^2 W(x,t)}{\partial t^2} \\ W(x,0) = \delta(x) \end{cases}$$

I wonder how it satisfies the boundary condition of maintaining a constant area under the graph for all time $t$.

Arcsine Law

$$P_{2k,2n} = u_{2k}u_{2n-2k}$$$$\mathbb{P}\left(\frac{1}{2} < \frac{\gamma (2n)}{2n} \leq x \right) = \sum_{k,1/2<2k/2n\leq x} P_{2k,2n}$$$$u_{2k} \sim \frac{1}{\sqrt{\pi k}}, \quad P_{2k,2n} \sim \frac{1}{\pi \sqrt{k(n-k)}}$$$$\begin{align} \mathbb{P}\left(\frac{1}{2} < \frac{\gamma (2n)}{2n} \leq x \right) &= \sum_{k,1/2<2k/2n\leq x} P_{2k,2n} \\ &= \sum_{k,1/2<2k/2n\leq x} \frac{1}{\pi n \sqrt{(k/n)(1-k/n)}} \\ &\rightarrow \frac{1}{\pi} \int_\frac{1}{2}^x \frac{dt}{\sqrt{t(1-t)}} \end{align}$$

Which leads us to

$$ \mathbb{P}\left(\frac{\gamma (2n)}{2n} \leq x \right) = \frac{2}{\pi} \arcsin \sqrt{x} $$

The consequence of this theorem is that it is most likely for a radnom walk to spend either almost all of its time on the positive side, or for it to spend almost no time on the positive side.

We can do this because there was no reason to limit the lower bound to $1/2$ rather than $0$ in the derivation

References

  1. Helena (https://physics.stackexchange.com/users/12948/helena), What is the physical meaning of diffusion coefficient?, URL (version: 2014-12-03): https://physics.stackexchange.com/q/52977https://physics.stackexchange.com/a/52977/250863
  2. Ackelsberg, Ethan (https://math.osu.edu/sites/math.osu.edu/files/What_is_2018_Arcsine_Law.pdf), What is the Arcsine Law?, URL (version: 2024-08-11): https://math.osu.edu/sites/math.osu.edu/files/What_is_2018_Arcsine_Law.pdf