Random Systems

Feb 23, 2020

(6.1)

(a)

$(6.25) \implies exp \Big(\sum_{n=1}^\infty \frac{(ik)^n}{n!}C_n) = \sum_{n=1}^\infty \frac{(ik)^n}{n!} \langle x^n \rangle \\$

Expanding the exponential…

$1 + \sum_{n=1}^\infty \frac{(ik)^n}{n!}C_n + \frac{1}{2}\big(\sum_{n=1}^\infty \frac{(ik)^n}{n!}C_n\big)^2 + \frac{1}{6}\big(\sum_{n=1}^\infty \frac{(ik)^n}{n!}C_n\big)^3 + ... = 1 + ik \langle x \rangle - \frac{k^2}{2} \langle x^2 \rangle - \frac{ik^3}{6} \langle x^3 \rangle + ...$

Regroup the left side by orders of $k$

$1 + ikC_1 - k^2 (\frac{1}{2}C_2 +\frac{1}{2}{C_1}^2) - ik^3 (\frac{1}{6}C_3 + \frac{1}{2}C_1C_2 + \frac{1}{6}C_1^3) + ... = 1 + ik \langle x \rangle - \frac{k^2}{2} \langle x^2 \rangle - \frac{ik^3}{6} \langle x^3 \rangle + ...$

For each order of k, we assume the term is equal and get:

$C_1 = \langle x \rangle \\ \frac{C_2 + C_1^2}{2} = \frac{\langle x^2 \rangle}{2} \implies C_2 = \langle x^2 \rangle - {\langle x \rangle}^2 \\ \frac{C_3 + 3C_1C_2 + C_1^3}{6} = \frac{\langle x^3 \rangle}{6} \implies C_3 = \langle x^3 \rangle - 3 \langle x \rangle (\langle x^2 \rangle - {\langle x \rangle}^2) - {\langle x \rangle}^3 \implies \\ C_3 = \langle x^3 \rangle - 3 \langle x \rangle \langle x^2 \rangle + 2 {\langle x \rangle}^3$

(b)

First, let’s calculate $\langle x \rangle, \langle x^2 \rangle, \langle x^3 \rangle$ for a gaussian distribution with mean $\mu$ and variance $\sigma^2$ .

$\langle x \rangle = \mu$ $\sigma^2 = \langle x^2 \rangle - \langle x \rangle^2 \implies \langle x^2 \rangle = \sigma^2 + \mu^2$

For $\langle x^3 \rangle$ , we are going to define $z \sim \mathcal{N}(0, 1)$ and notice that since these are both normal distributions:

$x \sim \sigma z + \mu$

Therefore:

$\langle x^3 \rangle = \langle (\sigma z + \mu)^3 \rangle = \\ \langle \sigma^3 z^3 + 2 \mu \sigma^2 z^2 + 2 \sigma \mu^2 z + \mu^3 \rangle = \\ \sigma^3 \langle z^3 \rangle + 3 \mu \sigma^2 \langle z^2 \rangle + 3 \sigma \mu^2 \langle z \rangle + \mu^3$

$z$ is distributed with mean 0 and variance 1, so: $\langle z \rangle = 0, \langle z^2 \rangle = 1$ , and since $sign(z) \sim sign(z^3) \implies \langle z \rangle = \langle z^3 \rangle = 0$ (can also be proved by solving $\int z^3 e^{-\frac{z^2}{2}}dz$ by substitution $u=\frac{z^2}{2}$ , or by a symmetry argument). We put all of that in the above equation and get

$\langle x^3 \rangle = 3 \mu \sigma^2 + \mu^3$

Now for the cumulants, insert the above values and get:

$C_1 = \mu \\ C_2 = \sigma^2 + \mu^2 - \mu^2 = \sigma^2 \\ C_3 = 3 \mu \sigma^2 + \mu^3 - 3 \mu ( \sigma^2 + \mu^2) + 2 \mu^3 = 0$

(6.2)

(a)

(b)

Update after class:

We can express $|x1| = |e^{-\frac{(y_1 + y2)^2}{2}|$ so we get a gaussian! (c)

(6.3)

(a)

(b)

$2^M - 1 > 10^9 \cdot 60 \cdot 60 \cdot 24 \cdot 365 \cdot 10^{10} = 3.1104 \cdot 10^{25} \implies \\ M > \frac{log(10^{26})}{log(2)} = 26 \frac{log(10)}{log(2)} \implies \\ M > 87$

(6.4)

(a)

We have equation

$\frac{\partial p}{\partial t} = D \frac{\partial^2 p}{\partial x^2}$

Keeping t constant, we apply two-sided Fourier transform on both sides and get:

$\frac{\partial P(k, t)}{\partial t} = -k^2DP(k, t) \implies \frac{\partial P(k, t)}{\partial t} + k^2DP(k, t) = 0$

We now got an easier, differential equation. Plug in the Ansatz $P = Ae^{-\lambda t}$ and get:

$-\lambda A e^{-\lambda t} + k^2 D Ae^{-\lambda t} = 0 \implies \\ Ae^{-\lambda t} (k^2D - \lambda) = 0 \implies \\ \lambda = k^2D \implies \\ P(k, t) = Ae^{-k^2D t}$

We know that $p(0, 0) = 1, x \ne 0 : p(x, 0) = 0 \implies P(0, 0) = 1$ thus:

$P(0, 0) = Ae^{-k^2D 0} = 1 \implies A = 1 \\ P(k, t) = e^{-k^2D t}$

According to a Fourier Transform Table, we have an identity:

$\mathcal{F}\Big(\frac{1}{\sigma \sqrt{2\pi}} e^{\frac{-x^2}{2\sigma^2}}\Big) = e^{\frac{-\sigma^2k^2}{2}}$

So, $\sigma = \sqrt{2Dt}$ , and we get:

$p(x, t) = \frac{1}{\sqrt{4\pi D t }} e^{\frac{-x^2}{4Dt}}$

(b)

We got in (a) that

$\sigma^2 = 2Dt$

(c)

TBD

(d)

For the random walk, the Diffusion coefficient is $\frac{1}{2}$ (since $\langle \delta^2 \rangle = 1$ and $\rho = 1$ )

Code:

(e)

If we wanted to ask how many trajectories at any specific time t are within the error bars, this would be a Gaussian so we want to know what $P(- 1.5\sigma < X < 1.5\sigma)$ .

Using a probability table:

We get $0.9332 - (1 - 0.9332) = 0.8664 \implies 86\%$

But, the number of trajectories which are within the whole error bars is a different question, and is no longer Gaussian. It seems that when t approaches infinity it will go to zero. Need to formalize.