# Proving that $Z_n=\frac{p_b(X_1,\dots,X_n)}{p_a(X_1,\dots,X_n)}$ is a martingale iff $a=c$

Problem: For a parameter $$a\in(0,1)$$ let $$p_a(x_1,\dots,x_n)$$ denote the probability that the first $$n$$ terms in the i.i.d. sequence $$X_1,X_2,\dots$$ of $$\operatorname{Bernoulli}(a)$$ random variables are exactly $$x_1,\dots,x_n$$. Let $$a\ne b$$ and $$Z_n=\frac{p_b(X_1,\dots,X_n)}{p_a(X_1,\dots,X_n)}$$ where $$X_1,X_2,\dots$$ are i.i.d. $$\operatorname{Bernoulli}(c)$$ random variables. Prove that $$Z_n$$ is a martingale if and only if $$c=a.$$

My Attempt: We first need to compute the functions $$p_a$$ and $$p_b$$, which are the joint pmf's of $$n$$ Bernoulli$$(a)$$ and $$n$$ Bernoulli$$(b)$$ random variables, respectively. But this is not hard, thanks to independence. In particular, we have $$p_a(x_1,\dots,x_n)=a^{x_1+\cdots+x_n}(1-a)^{n-(x_1+\cdots+x_n)},$$ $$p_b(x_1,\dots,x_n)=b^{x_1+\cdots+x_n}(1-b)^{n-(x_1+\cdots+x_n)}.$$ Therefore, we have $$Z_n=\frac{b^{X_1+\cdots+X_n}(1-b)^{n-(X_1+\cdots+X_n)}}{a^{X_1+\cdots+X_n}(1-a)^{n-(X_1+\cdots+X_n)}},$$ where $$X_1,\dots,X_n$$ are i.i.d. Bernoulli$$(c)$$ random variables. Now we need to show that the martingale property $$E[Z_{n+1}\mid\mathcal F_n]=Z_n$$ is satisfied if and only if $$c=a$$. Using the properties of conditional expectation and the independence of the random variables, \begin{align*} E[Z_{n+1}\mid\mathcal F_n] &=E\left[\frac{b^{X_1+\cdots+X_{n+1}}(1-b)^{n+1-(X_1+\cdots+X_{n+1})}}{a^{X_1+\cdots+X_{n+1}}(1-a)^{n+1-(X_1+\cdots+X_{n+1})}}\Bigg\vert\mathcal F_n\right]\\ &=\frac{b^{X_1+\cdots+X_n}(1-b)^{n-(X_1+\cdots+X_n)}}{a^{X_1+\cdots+X_n}(1-a)^{n-(X_1+\cdots+X_n)}}E\left[\left(\frac{b}{a}\right)^{X_{n+1}}\left(\frac{1-b}{1-a}\right)^{1-X_{n+1}}\right].\\ \end{align*} Computing the expectation above, it follows that $$Z_n$$ is a martingale if and only if we have $$\frac{(1-b)(1-c)}{1-a}+\frac{bc}a=1.$$ It follows that we must have $$a=c.$$

Do you agree with my work above?
Any help is much appreciated. Thank you for your time.

You are correct. Also, you can simplify the argument a little, and should be a little more explicit at the conclusion.

First, since $$X_1,X_2,\dots$$ are independent, we have that

$$Z_{n+1} = Z_n \frac{p_b (X_{n+1})}{p_a (X_{n+1})}.$$

Therefore $$E[Z_{n+1} | {\cal F}_n ] = Z_n E [ \frac{p_b(X_{n+1})}{p_a(X_{n+1})}].$$

The expectation on the righthand side is equal

$$(*) \quad \frac{b}{a} c + \frac{1-b}{1-a} (1-c),$$

and therefore this is a martingale if and only if $$(*)=1$$.

Fixing $$a$$ and $$b$$, $$(*)$$ is a convex combination of $$\frac{b}{a}$$ and $$\frac{1-b}{1-a}$$. As one of the two numbers is necessarily $$> 1$$ and the other is necessarily $$<1$$, there exists a unique convex combination of the two giving $$1$$. Since the choice of $$c=a$$ gives $$1$$, this is the only solution.