Let $\{X_i \mid i \in \mathbb{N} \}$ be a sequence of independent and identically distributed random variables with probability mass function given by:
$$ P(X_i = x)= \begin{cases} p, & \text{if } x = -1\\ 1-p, & \text{if } x = 1 \\ 0, & \text{otherwise} \end{cases} $$
For $t>0$, define $S_t = \displaystyle \sum_{i=1}^t X_i$ and $S_0 = 0$. I would like to find $\operatorname{Var} (S_t \mid S_s)$, where $0\le s <t$. Is there some property of the conditional variance that I am missing here? Any suggestion will be greatly appreciated.
First note that $\operatorname{Var}(X_i) = 4p(1-p).$
Since these are independent, you have $\operatorname{Var} S_t = t \operatorname{Var}(X_i) = 4tp(1-p).$
\begin{align} \operatorname{Var}(S_t\mid S_s) & = \operatorname{Var}\left( S_s + X_{s+1}+X_{s+2} + X_{s+3} + \cdots + X_t \mid S_s \right) \\[10pt] & = \operatorname{Var}\left(X_{s+1}+X_{s+2} + X_{s+3} + \cdots + X_t \right). \end{align} This last equality holds because when you condition on $S_s$ you treat $S_s$ as a constant. Adding a constant to a random variable does not change its variance. And we don't need to say $\operatorname{Var}(X_{s+1}+X_{s+2} + X_{s+3} + \cdots + X_t\mid S_s)$ because $X_{s+1}, X_{s+2}, X_{s+3}, \ldots, X_t$ are independent of $S_s$.
So you get $(t-s)4p(1-p).$