Assume an asymmetric random walk.
I am a little bit confused by one exercise. Does anyone know my mistake? We have to proofe: $\mathbb{P}(S_{2n+1}=2k+1|S_0=0)= \binom{2n+1}{n+k+1}p^{n+k+1}q^{n-k}$
My solution: x denote the number of ups and 2n+1-x denotes number of downs. Therefore $2n+1=2k+1$ We know: $2k+1=x-(2n+1-x)=2x-2n+1$,
We compute x by: $2k+1=2x-2n+1$ via algebra $x=k+n$ Substitude in "Down": $2n+1-(k+n)$ results in $n-k+1$ Therefore my solution would be $\mathbb{P}(S_{2n+1}=2k+1|S_0=0)= \binom{2n+1}{n+k}p^{n+k}q^{n-k+1}$
Does anyone know the mistake? Thanks for help.