UMD Probability Qualifying Exams/Aug2008Probability

Solution
Notice that since $$f,g$$ are measurable functions, then $$Y_n$$ is composed of linear combinations of $$\mathcal F_n$$-measurable functions and hence $$Y_n$$ is $$\mathcal{F}_n$$-adapted. Furthermore, for any $$n$$, $$Y_n$$ is finite everywhere, hence is $$L^1$$.

Therefore, we only need to check the conditional martingale property, i.e. we want to show $$Y_n=E(Y_{n+1}|\mathcal{F}_n$$.

That is, we want

$$\begin{align} f(X_n)-f(X_0)-\sum_{i=1}^{n-1}g(X_i)&=E[f(X_{n+1})-f(X_0)-\sum_{i=1}^{n}g(X_i)|\mathcal{F}_n]\\ f(X_n)&=E[f(X_{n+1})|\mathcal{F}_n]-g(X_n) \end{align}$$

Therefore, if $$Y_n$$ is to be a martingale, we must have

$$g(X_n)=E[f(X_{n+1})|\mathcal{F}_n]-f(X_n)$$.

Since $$\Chi=\{1,2\}$$, we can compute the right hand side without too much work.

$$g(1)=E[f(X_{n+1})|X_n=1]-f(1)=(1\cdot 1/3 + 4\cdot 2/3 )-1 =2$$

$$g(2)=E[f(X_{n+1})|X_n=2]-f(2)=(1\cdot 1/2 + 4\cdot 1/2)=-\frac{3}{2}$$

This explicitly defines the function $$g$$ and verifies that $$Y_n$$ is a martingale.