Today my prof gave me an equation of random walk: $$p(x_i,t+\Delta t)=\frac{1}{2}(p(x_i-\Delta t)+p(x_i+\Delta t))-p(x_i,t)$$ Using this he get$$P_t=P_{xx}$$ when $\Delta t<<1$
But how and what's actually the meaning of the probability equation? e.g. one half of particles goes left and another half goes right, while nothing left in the middle. But why divided by 2 an how does this actually make sense?
The expression in the OP is incorrect. It should be
$$p(x_i,t+\Delta t)=\frac12(p(x_i+\Delta x,t)+p(x_i-\Delta x,t)) \tag 1$$
Expanding the left-hand side of $(1)$ in a Taylor series in $\Delta t$, we obtain
$$p(x_i,t+\Delta t)=p(x_i,t)+p_t(x_i,t)\Delta t+O(\Delta t)^2 \tag 2$$
Expanding the left-hand side of $(1)$ in a Taylor series in $\Delta t$, we obtain
$$\begin{align} p(x_i+\Delta x,t)+p(x_i-\Delta x,t)&=p(x_i,t)+p_x(x_i,t)\Delta x+\frac12 p_{xx}(x_i,t)(\Delta x)^2+O(\Delta x)^3\\\\ &+p(x_i,t)-p_x(x_i,t)\Delta x+\frac12 p_{xx}(x_i,t)(\Delta x)^2+O(\Delta x)^3\\\\ &=p(x_i,t)+p_{xx}(x_i,t)(\Delta x)^2+O(\Delta x)^3 \tag 3 \end{align}$$
Equating $(2)$ and $(3)$, dividing both sides by $\Delta t$, eveals
$$p_t(x_i,t)+O(\Delta t)=\frac12 p_{xx}(x_i,t)\frac{(\Delta x)^2}{\Delta t}+\frac{O(\Delta x)^3}{\Delta t} \tag 4$$
If $\lim_{\Delta t\to 0}\frac{(\Delta x)^2}{2\Delta t}=1$, then $(4)$ becomes in the limit as $\Delta t\to 0$,
$$p_t(x_i,t)=p_{xx}(x_i,t)$$
as was to be shown!