By definition of a martingale $\{Y_n\}$ with respect to to a sequence $\{X_n\}$,
$\text{E}(Y_{n+1}|X_1,\ldots,X_n) = Y_n$.
According to my teacher, it is also true that $\text{E}(Y_{n+a}|X_1,\ldots,X_n) = Y_n$, for any integer $a \geq 1$. (1)
This was not so clear to me, so I started to think about it and came up with:
$\text{E}(Y_{n+a}|X_1,\ldots,X_n) = \{\text{by definition of }Y_{n+a}\} = \text{E}( \text{E}(Y_{n+a+1}|X_1, \ldots, X_{n+a})|X_1,\ldots,X_k) = (\text{by the law of total expectation}) = \text{E}(Y_{n+a+1})$, which just made me more confused.
Any ideas on how to show that (1) is true?
Best Answer
Here's a way to understand it.
First, use the law of total expectation $\mathbb{E}(Y)=\mathbb{E}(\mathbb{E}(Y|X))$. Specifically we will apply the form $\mathbb{E}(Y|X)=\mathbb{E}(\mathbb{E}(Y|X,Z)|X)$.
The sequence of X can be thought of as two discrete sequences; match X to $X_1$ thru $X_n$, and Z to $X_{n+1}$ thru $X_{n+m-1}$. This allows us to eliminate the Z sequence from the expression. $$\mathbb{E}(Y_{n+m}|X_1,\dots,X_n)=\mathbb{E}\big[\mathbb{E}(Y_{n+m}|X_1,\dots,X_{n+m-1})|X_1,\dots,X_n\big]$$ Now, note that the expectation inside is just the definition of $Y_{n+m-1}$. Thus, you get $$\mathbb{E}(Y_{n+m}|X_1,\dots,X_n)=\mathbb{E}(Y_{n+m-1}|X_1,\dots,X_n)$$ You can keep repeating this until you reach $Y_{n+1}$, which means $$\mathbb{E}(Y_{n+m}|X_1,\dots,X_n)=\mathbb{E}(Y_{n+1}|X_1,\dots,X_n)=Y_n.$$
Hope it helps