Consider the following problem.
A particle takes discrete steps $\sigma_1, \sigma_2, \sigma_3, \ldots, \sigma_n$ which take on values $+1$ or $-1$. However, $P(\sigma_i = +1) = p$ if $i$ is odd and $1-p$ if $i$ is even. Equivalently, $P(\sigma_i=-1)$ will be $1-p$ if $i$ is odd and $p$ if $i$ is even.
I am looking for the PMF of $M_n = \sigma_1 + \sigma_2 + \ldots + \sigma_n$. For large $n$ the solution using the central limit theorem is quite straightforward says that $M\sim N(0,n)$, a normal with zero mean and variance $n$.
However, I am stuck at trying to find a solution for arbitrary n. So far, what I have tried is: write $M_n = M_o+ M_e$. For these M's I can compute their PMF, which is equivalent to the usual random walk. So let $n_o = n - Floor(n/2)$ and $n_e = Floor(n/2)$. Then, if $q=1-p$,
\begin{equation} P_{n_o}(M_o = m_o) = \binom{n_o}{\frac{n_o+m_o}{2}} p^{\frac{n_o+m_o}{2}} q^{\frac{n_o-m_o}{2}} \end{equation}
\begin{equation} P_{n_e}(M_e = m_e) = \binom{n_e}{\frac{n_e+m_e}{2}} q^{\frac{n_e+m_e}{2}} p^{\frac{n_e-m_e}{2}} \end{equation}
But now I would have to convolve these two PMFs, something which I have no idea how to do. I am not even sure this is a good approach (it sure is a messy one). Anyone has any idea to share?