Random walk with 3 values

Question

I have the following question:

Consider a one-dimensional random walk with step size $L$, where the probability of walking to the right is $1/4$, left is $1/4$, and staying where it is $1/2$. Suppose the walker starts from $x=0$.

1. Obtain the distribution of finding the walker at position $x=nL$ after $N$ steps.
2. Get the continuous probability to find the walker at an x position in the time $T=N\tau$, for $N$ too large.
3. Get the characteristic function for this continuous function from the previous item.

The first, I think that is similar to this question of StackExchange. So I found the following:

$$ P_{N}\left(x\right) =\sum_{s=0}^{N-n}\frac{N!}{\left(N-s-r\right)!s!r!}\left(\frac{1}{2}\right)^{2N-s} $$ Where $r$, $s$ is the probability to go to the right and stay, respectively. So we have some ways to end in the position $x$, and the "options" that $s$ can assume change between $s=0$ (where $n=l-r$) and $s=N-n$ (where $r=n,l=0$).

But the second question, remember me about CTRW (continuous-time random walk): $$P\left(x,T\right)=\sum_{N=0}^{\infty}P\left(N,T\right)P_{N}\left(x\right)$$

But we have a deterministic relationship. I know that in a time $t=T$ we have $N=\frac{T}{\tau}$ steps. So I can't see another consequence than just know what is the value of $N$ in every $T$. And the third I don't know how to proceed. So I would ask for some tip.

Answer 1

Disclaimer: This is only a hint / tips, not a full solution

The probability of ending at $x=nL$ after $N$ steps is given by the coefficient of $y^{n}$ from the expression $\left(\frac{1}{4}\cdot\frac{1}{y}+\frac{1}{2}+\frac{1}{4}\cdot y\right)^{N}$ or we can write the following:

$$ \begin{align} P_{N}(n)&=\left[y^{n}\right]\phantom{x} \left(\frac{1}{4}\cdot\frac{1}{y}+\frac{1}{2}+\frac{1}{4}\cdot y\right)^{N}\\ \\ &=\left[y^{n}\right]\phantom{x}\frac{1}{4^{N}}\cdot\frac{1}{y^{N}}\cdot\left(1+2y+y^{2}\right)^{N}\\ \\ &=\left[y^{n}\right]\phantom{x}\frac{1}{4^{N}}\cdot\frac{1}{y^{N}}\cdot\left(y+1\right)^{2N}\\ \\ &=\left[y^{N+n}\right]\phantom{x}\frac{1}{4^{N}}\cdot\left(y+1\right)^{2N}\\ \\ &=\frac{1}{4^{N}}\cdot\binom{2N}{N+n} \end{align} $$

Now that we have a simple expression, I believe you can finish the rest by yourself