I would like to discuss a slightly different kind of random walk on $Z$ in which we include the probability of being "stuck" on the same place.
Let us denote as:
$$p/3 \; \text{the prob. of right step (which is +1 on Z)} $$
$$p/3 \; \text{for left step (which is -1 on Z)} $$
$$ 1-\frac{2}{3}p \; \text{for being in the same place.} $$
Here $p \in (0,1) $
How I get from this to the diffusion equation?
I denote as $n_1, n_2, n_3$ respectively the amount-of-steps random variables, respectively to the left,to right and on the same place. Obviously $(1) \; N=n_1+n_2+n_3$ is the total number of steps.
As expectation values I get: $$ <n_1> = <n_2> = N\frac{p}{3}$$
While variances: $$ V[n_1]=V[n_2]=N\frac{p}{3}(1-\frac{p}{3})$$
I define the distance random variable $d_n=n_1 -n_2 $ as usual but Now I cannot write $n_2(n_1,N)$ as I do in the simpler case of bernoulli process; this is because of (1). How can I proceed?
Let $P_N(x)$ denote the probability of being at $x$ after $N$ steps. assuming $P_0(0)=1$. Then:
$$P_{N+1}(x)= \left(1-\frac{2}{3}p\right)P_N(x)+\frac{p}{3}(P_N(x-1)+P_N(x+1)),$$
which is obtained by conditioning on the walk's possible value at step $N$.
If you want the continuous limit, write the above as:
$$\frac{P_{N+1}(x)-P_N(x)}{N}=\frac{p}{3}\frac{P_{N}(x-1)-2P_N(x)-P_N(x+1)}{N}$$
so that after rescaling your random walk in $x,N$ appropriately, you get:
$$\partial_t P_t(x)=\frac{p}{3}\partial_{xx}P_t(x)$$