I asked this question in Math Exchange and obtained no 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 $0<d\leq 1$.
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: Let $u(x,t)=\text{Pr}(X(t)=x)$. By the Total Probability Theorem, $$ u(x,t+\Delta t)=\frac{d}{2} u(x-h,t)+(1-d)u(x,t)+\frac{d}{2}u(x+h,t).\quad (1) $$ This may be written as $$ \frac{u(x,t+\Delta t)-u(x,t)}{\Delta t}=\frac{d h^2}{2\Delta t}\frac{u(x+h,t)-2u(x,t)+u(x-h,t)}{h^2}. \quad (2) $$ Put $$ \frac{dh^2}{2\Delta t}=D. \quad (3) $$ 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 $$ u(x,t)=\frac{1}{\sqrt{4\pi Dt}}\text{exp}\left(-\frac{x^2}{4Dt}\right).\quad (4)$$ 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:
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?
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.
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 $(4)$.
