= Solution
Suppose under the physical measure $P$ the up indicators are independent with probability $p\in(0,1)$. If $N_k$ is the number of up moves through date $k$, a particular history has probability $p^{N_k}(1-p)^{k-N_k}$ under $P$ and $q^{N_k}(1-q)^{k-N_k}$ under $Q$. Thus the <binomial-market probability density>, the <Radon-Nikodym derivative> on $\mathcal F_k$, is
$$
\boxed{L_k=\left.\frac{dQ}{dP}\right|_{\mathcal F_k}=\left(\frac qp\right)^{N_k}\left(\frac{1-q}{1-p}\right)^{k-N_k}.}
$$
It is strictly positive. Its next multiplier has conditional expectation $p(q/p)+(1-p)((1-q)/(1-p))=1$, so $L_k$ is a mean-one <martingale> and $L_k=\mathbb E_P[L_n\mid\mathcal F_k]$. These facts verify normalization and equivalence of the two probability measures.
The <state-price density> for terminal cash is $\zeta_n=L_n/B_n=b^{-n}L_n$. In particular $V_0=\mathbb E_P[\zeta_nH]$. At an intermediate date the conditional change-of-measure identity gives
$$
V_k=\frac{B_k}{L_k}\mathbb E_P\left[\frac{L_nH}{B_n}\ \middle|\ \mathcal F_k\right].
$$
If physical up probabilities depend on the past, replace the constant-probability expression by the product over periods of $q/p_k$ on an up move and $(1-q)/(1-p_k)$ on a down move. Each multiplier still has conditional mean one, provided $0<p_k<1$, and the pricing measure retains conditional up probability $q$.
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