Hypothesis Testing Two Measures Coin Tosses Martingale Consider a sequence of iid tosses of a coin with $X_i$ denoting the outcome of the $i^{th}$ toss. These random variables are defined on some $(\Omega, \mathcal{F})$ on which we have two probability measures $\mathbb{P}_{A}$ and $\mathbb{P}_{B}$. Under hypothesis $A$, $\mathbb{P}_{A}$ is the true measure, and the probability of a head on any toss is $p=a$. Under hypothesis $B$, the measure is $\mathbb{P}_{B}$ and $p=b$ for some $a,b \in (0,1)$.
Let $P_{A}(x_1, x_2, \dots, x_n)$ denote the probability of a sequence of outcomes $(x_1, x_2,\dots, x_n)$ under the hypothesis $A$, i.e.
$$P_{A}(x_1, x_2, \dots, x_n) = \mathbb{P}_{A}(X_1 = x_1, X_2 = x_2, \dots, X_n = x_n),$$
with the analogous definition for $P_B$.
I need to show that $$Z_n = \frac{P_A(X_1, X_2, \dots, X_n)}{P_B(X_1, X_2, \dots, X_n)}$$ is a martingale under $\mathbb{P}_{B}$, relative to the filtration generated by the tosses, $\mathcal{F}_n = \sigma(X_k : k \le n)$. Then I need to see what happens to the distribution of its limit (which I will know exists almost surely by the Martingale Convergence Property).
I am struggling to prove that $Z_n$ is indeed a martingale: it is clear that it is adapted, but I am not sure how to show integrability and the expectation property.
Would it also follow that $\frac{1}{Z_n}$ is a $\mathbb{P}_{A}-$martingale?
 A: I will only address the martingale part. I will indicate with $\mathcal{S}$ our discrete state space. First we check integrability:
$$\begin{aligned}E^{\mathbb{P}_B}[Z_n]&=\sum_{x_1,...,x_n\in \mathcal{S}}\frac{P_A(x_1,...,x_n)}{P_B(x_1,...,x_n)}\mathbb{P}_B(X_1=x_1,...,X_n=x_n)=\\
&=\sum_{x_1,...,x_n\in \mathcal{S}}P_A(x_1,...,x_n)=1,\,\forall n\end{aligned}$$
since probabilities sum up to one. We have
$$P'_A(x_1,...,x_n):=\frac{P_A(x_1,...,x_n)}{P_A(x_1,...,x_{n-1})}=\mathbb{P}_A(X_n=x_n|X_1=x_1,...,X_{n-1}=x_{n-1})$$
and equivalently for $P_B$. So
$$\begin{aligned}E^{\mathbb{P}_B}[Z_{n}Z_{n-1}^{-1}|\mathscr{F}_{n-1}]&=E^{\mathbb{P}_B}\bigg[\frac{P_A(X_1,...,X_n)}{P_B(X_1,...,X_n)}\frac{P_B(X_1,...,X_{n-1})}{P_A(X_1,...,X_{n-1})}\bigg|\mathscr{F}_{n-1}\bigg]=\\
&=E^{\mathbb{P}_B}\bigg[\frac{P_A'(X_1,...,X_{n})}{P_B'(X_1,...,X_{n})}\bigg|\mathscr{F}_{n-1}\bigg]\end{aligned}$$
and
$$\begin{aligned}&E^{\mathbb{P}_B}\bigg[\frac{P_A'(X_1,...,X_{n})}{P_B'(X_1,...,X_{n})}\bigg|X_1=x_1,...,X_{n-1}=x_{n-1}\bigg]=\\
&=\sum_{x_n \in \mathcal{S}}\frac{P_A'(x_1,...,x_{n-1},x_n)}{P_B'(x_1,...,x_{n-1},x_n)}\mathbb{P}_B(X_n=x_n|X_1=x_1,...,X_{n-1}=x_{n-1})=\\
&=\sum_{x_n \in \mathcal{S}}P_A'(x_1,...,x_{n-1},x_n)=1\end{aligned}$$
since again probabilities sum up to one. So we conclude.
