Cfrgtkky

Find a number of paths in random walk from $S_0=0$ to $S_{8n}=0$ satisfying the following terms:

(a) $S_k le -2$ for $2 le k le 4n-2$

(b) $S_k > 0$ for $4n le k le 8n$

Theorem: Let $S_0=a$ and $S_n=b$. $N_n(a,b)$ is the number of paths such that $S_0=a$ and $S_n=b$. If $b>0$ then the number of paths satisfying the following terms: 1) $S_0=0$ and $S_n=b$, 2) there doesn't exist $k in {{1,...,n-1}}$ that $S_k=0$, is equal to $frac bn N_n(0,b)$,

where $N_n(a,b) = binom{n}{1/2(n+b-a)}$.

I'm going to use homogeneity in time and in space (random walk properties).

(I don't want to use Catalan numbers in this exercise or any other "proven formulas".)

(a) $S_0=0, S_1=-1, S_2=-2,..., S_{4n-2}=-2$ for $S_k le -2$

then $S_2=0$ and $S_{4n-2}=0$ for $S_k le 0$

and $S_2=0$ ; $S_{4n-2}=0$ for $S_k ge 0$ (symmetry)

so $S_2=0$ ; $S_{4n-1}=1$ for $S_k > 0, ;;3 le k le 4n-1$

and from that $frac bnN_n(0,b)=frac{1}{4n-3} binom{4n-3}{1/2(4n-3+1-0)}=frac{1}{4n-3} binom{4n-3}{2n-1}$

(b)$S_{4n}=0, S_{8n-1}=1$, so $frac bnN_n(0,b)=frac{1}{4n-1} binom{4n-1}{2n}$

Is it a proper usage of the theorem mentioned above? I know how to use it when $S_k> or < ...$, but I'm not sure if I can "transform it" like that when $S_k le or ge ...$.

Please, correct me where I'm wrong or just tell me any tips. Will be grateful for any help.