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I asked this question in Math Exchange and MathOverflow and obtained no answer. This question may lack of mathematical rigorous, but I would like to understand why this type of reasoning is sometimes presented. Maybe you, as physicists, could provide me some insight into the correct answer.

Let $X(t)$ be a stochastic process in time such that $X(0)=0$ and, at each increment of time $\Delta t$, it can move $h$ units in space or not move: $$ X(t+\Delta t)=\begin{cases} X(t)+h &\text{ with probability $d/2$}, \\ X(t) &\text{ with probability $1-d$}, \\ X(t)-h &\text{ with probability $d/2$}, \end{cases} $$ where $0<d\leq 1$.

With this information, in my mathematical physics course, I was told and proved the following claim: the net expected displacement is not proportional to the elapsed time but to its square root.

Proof presented by the professor: Let $u(x,t)=\text{Pr}(X(t)=x)$. By the Total Probability Theorem, $$ u(x,t+\Delta t)=\frac{d}{2} u(x-h,t)+(1-d)u(x,t)+\frac{d}{2}u(x+h,t).\tag{1} $$ This may be written as $$ \frac{u(x,t+\Delta t)-u(x,t)}{\Delta t}=\frac{d h^2}{2\Delta t}\frac{u(x+h,t)-2u(x,t)+u(x-h,t)}{h^2}. \tag{2} $$ Put $$ \frac{dh^2}{2\Delta t}=D. \tag{3} $$ Letting $h\rightarrow0$, we arrive at $u_t=Du_{xx}$, with initial condition $u(x,0)=\delta(x)$. Taking Fourier transforms in the PDE (recall $\hat{\delta}=1$), one obtains $$ u(x,t)=\frac{1}{\sqrt{4\pi Dt}}\text{exp}\left(-\frac{x^2}{4Dt}\right).\tag{4}$$ Using this density function, we may compute $$ \overline{x^2}=\int_\mathbb{R} x^2 u(x,t) dx=2Dt.$$ Thus, the square of the displacement is proportional to $t$, as wanted.

Questions:

  1. The relation $(3)$ between space and time that one imposes so that $(2)$ has a limit is precisely what we want to prove!!! I mean, what happens here? Is this correct, or is there another more formal proof and this proof is just intuitive?

  2. The step in which $u(x,t)$ becomes a density function from a mass probability function is not clear. A possible solution could be as follows: we may extend $u(x,t)=u_h(x,t)$ to a step function $\tilde{u}_h(x,t)$ in $\mathbb{R}$ being constant at each interval $](2k-1)/2\cdot h,(2k+1)/2\cdot h[$, and define $\bar{u}_h(x,t)=\tilde{u}_h(x,t)/h$. Then $\bar{u}_h(x,t)$ is a density and satisfies $(2)$. Of course, when $h\rightarrow0$, we need to assume that $\bar{u}_h(x,t)$ tends to a density function, say $p(x,t)$, and that the double limit in (2) concerning $\bar{u}_h(x,t)$ and its partial derivatives converges to the partial derivatives of $p$, so that $p_t=Dp_{xx}$. I do not see this final fact.

  3. When solving $u_t=Du_{xx}$, $u(x,0)=\delta(x)$, I understand the formal procedure of taking Fourier transforms. How can one show that $(4)$ is the unique solution we are looking for? I read in the book Partial Differential Equation in Action, by Sandro Salsa, that $(4)$ is the unique density function solving $u_t=Du_{xx}$ being radial and self-similar. Imposing the radial condition seems clear, as the particle moves right and left equally. I assume self-similarity comes from the fact that, intuitively, $u(x,t)$ should have a bell shape, as being near $0$ is more probable, so at every $t$, $u(x,t)$ has a similar shape. Is this correct? Thus, the condition $u(x,0)=\delta(x)$ is not used, right? In fact, in the book, the delta function is introduced after obtaining the function given by $(4)$.

Some examples of this reasoning: In the book by Sandro Salsa I mentioned, to motivate the definition of Brownian motion; in http://nebula.physics.uakron.edu/dept/faculty/jutta/modeling/diff_eqn.pdf; and in Chapter 2 in https://arxiv.org/pdf/1504.08292.pdf (this chapter deals with the fractional laplacian from a probabilistic point of view).

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    $\begingroup$ I don't feel like solving the complete problem now but may come back to it. The first question is rather simple. Each increment is an iid $r$ with $\langle r \rangle=0, \langle r^2 \rangle = h^2 d$. Thus in time $T$ we have moved $X(T) = \sum_{i=1}^{T/\Delta t} r_i$ and we get $Var(X) = \frac{T}{\Delta t} Var(r) =\frac{T}{\Delta t} h^2 d$. Thus the variance scales as $T$. ** The net expected displace is $\langle X(T) \rangle=0$, so I presume you mean $\sqrt{\langle X(T)^2 \rangle} \propto \sqrt{T}$. $\endgroup$ – Borun Chowdhury Jan 16 '18 at 17:05
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    $\begingroup$ Seems to me that (3) is a definition, so it's not clear to me what you mean there in Q1. $\endgroup$ – Kyle Kanos Jan 16 '18 at 17:23
  • $\begingroup$ @KyleKanos Yes, for me it seems a definition as well. So I do not understand why the condition is put in the statement of the "theorem". Maybe it comes from a lack of rigour. $\endgroup$ – user39756 Jan 16 '18 at 18:19
  • $\begingroup$ @BorunChowdhury I think your comment is precisely the mathematical justification of question 1. See page 37 in this notes: http://ft-sipil.unila.ac.id/dbooks/AN%20INTRODUCTION%20TO%20STOCHASTIC%20DIFFERENTIAL%20EQUATIONS%20VERSION%201.2.pdf. Moreover, using the Central Limit Theorem, one arrives at (4). On the other hand, by net expected displacement I refered to $E[|X(t)|]$ or $\sqrt{E[X(t)^2]}$. It remains for me to understand questions 2 and 3. $\endgroup$ – user39756 Jan 16 '18 at 18:32

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