let X be the step of an asymmetric random walk\begin{equation} X = \left| \begin{array}{cc} 1 & p \\ -1 & 1-p \\ \end{array} \right. \end{equation} Let $Y_n$ be the position after $n$ steps. I am trying to prove $$E(Y)=n(2p-1)$$ and $$V(Y)=4np(1-p)$$
Solution: $$Y = \Sigma^n X_i$$ so clearly $$E(Y) = nE(X) = n(2p-1)$$ Next I attempt to use $$V(Y)= E(Y^2) - E^2(Y)$$
$$Y^2=(\Sigma^n X_i)^2$$ so then $$Y^2 = \Sigma^n X_i^2 + 2\Sigma_{i \neq j}^n X_iX_j$$ and $$E(Y^2) = nE(X^2) + 2nE^2(X)$$ And $$E(X^2) =1 \ and \ E^2(X)=(2p-1)^2$$ Finally $$V(Y)=n +2n(2p-1) - n^2(2p-1)^2$$ Thus $$V(Y)\neq4np(1-p)$$ Where have i gone wrong?
The number of pairs $(i,j)$ with $i \neq j$ ($1\leq i,j \leq n$) is $\binom {n } {2}=\frac {n(n-1)} 2$. So the second term in $EY^{2}$ is $n(n-1)(2p-1)^{2}$.
With this change you will get $V(Y)=4np(1-p)$.