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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.

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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.

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