For $Y_i$ martingale w.rt. $X_1,X_2,...$ show $\mathbb{E}(Y_{n+m}|X_1,...,X_n]=\mathbb{E}(Y_{n}|X_1,...,X_{n+m}]=Y_n$

Question

I tried to write

\begin{align} \mathbb{E}[Y_{n+m}| X_1 ,...,X_n = x_1,...,x_n] &= \sum_y y \mathbb{P}[Y_{n+m}=y|X_1,...,X_n=x_1,...,x_n] \\ &= \sum_{x_{n+1},...,x_{n+m-1}} \sum_y y \mathbb{P}[Y_{n+m}=y|X_1,...,X_{n+m-1}=x_1,...,x_{n+m-1}] \mathbb{P}[X_{n+1},...,X_{n+m-1} = x_{n+1},...,x_{n+m-1}] \end{align}

But I dont know if this will lead to an answer. Can anyone please show how to do this?

Thank you

Answer 1

$E(Y_n|X_1,X_2,..,X_{n+m})=Y_n$ because $Y_n$ is already measurable w.r.t. $\sigma (X_1,X_2,..,X_{n+m})$.

Note that $E(Y_{n+m}|X_1,X_2,..,X_n)=Y_n$ (for all $n$) follows from definition of a martingale when $m=1$.

Now use induction on $m$. Suppose the equaion holds for a certain $m$ and consider $E(Y_{n+m+1}|X_1,X_2,..,X_n)$. Write this as $E(E(Y_{n+m+1}|X_1,X_2,..,X_{n+1}|X_1,X_2,...,X_n))$. By induction hypothesis with $n$ changed to $n+1$ this gives $E(Y_{n+1}|X_1,X_2,...,X_n))$ which is $Y_n$.