Finding a PDE with Random Walks I have a question about Random Walks, but am confused by the question. So, the question is:

Given the Probabilistic Equation:
  $$p(x,t+\Delta t)=\frac{p(x+\Delta x,t)+p(x-\Delta x,t)}{2}$$
  Find the PDE for the 2 cases:$$$$
  a) If $\Delta x\rightarrow 0$, then $$\frac{(\Delta x)^2}{\Delta t}=4$$
  b) If $\Delta x\rightarrow 0$, then $$\frac{(\Delta x)^3}{\Delta t}=4$$

Now I solved $(a)$, by subtracting $p(x,t)$ from both sides, dividing by $\Delta t$, and multiplying the right hand side by $1=\frac{(\Delta x)^2}{(\Delta x)^2}$ to get:
$$\frac{p(x,t+\Delta t)-p(x,t)}{\Delta t} = \frac{(\Delta x)^2}{2\Delta t}\left(\frac{p(x+\Delta x,t)+p(x-\Delta x,t)-2p(x,t)}{(\Delta x)^2}\right)$$
As we take $\Delta t\rightarrow 0, \>\>\Delta x\rightarrow 0$, we have that the above becomes:
$$p_t = 2p_{xx}$$
But I have no idea how to do $(b)$. I tried using the third order approximation of the derivative, which can be done to the right hand side, but it doesn't add up on the left hand side. If anyone could give me a hint about this question it would be appreciated. Thanks!
 A: Taylor expand both sides:
$$p(x,t+\Delta t) = p + p_t \Delta t + O(\Delta t^2),$$
and
$$p(x\pm \Delta x,t) = p \pm p_x \Delta x + \frac{1}{2} p_{xx} \Delta x^2 \pm \frac{1}{6} p_{xxx} \Delta x^3 + O(\Delta x^4).$$
Therefore
$$\frac{1}{2}(p(x+\Delta x,t) + p(x-\Delta x,t)) =p + \frac{1}{2} p_{xx} \Delta x^2 + O(\Delta x^4).$$
Now equate the largest order terms on both sides. Which term is largest (besides the first term $p$) depends on whether you are scaling like (1) or (2).
Edit: Here are some more details. Equating the Taylor expansions for each side of your equation above we get that
$$p_t \Delta t + O(\Delta t^2) = \frac{1}{2} p_{xx} \Delta x^2 + O(\Delta x^4)$$
as $\Delta t,\Delta x \to 0$. In the first case $\Delta t = \frac{1}{4}\Delta x^2$ and you get 
$$p_t = 2 p_{xx} + O(\Delta x^4).$$
Sending $\Delta x \to 0$ you get your heat equation $p_t = 2p_{xx}$. In the second case $\Delta t = \frac{1}{4}\Delta x^3$ and we get
$$p_t \Delta x = 2p_{xx} + O(\Delta x^4).$$
When you send $\Delta x \to 0$ you get $p_{xx}=0$.
