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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?

user63010
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1 Answers1

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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

chinnychinchin
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NECing
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  • Thank you Shu, but I do not understand the first thing you do. By the law of total expectation, what you have on the right hand side is just the expectation E$(Y_{n+m})$? – user63010 Mar 19 '13 at 18:56
  • You just ignore the $X_1,\dots,X_n$ part. Now compare the rest expression with the law of total expectation. In this case, $Y_{n+m}$ is $Y$ and $X_1,\dots,X_{n+m-1}$ is $X$. – NECing Mar 19 '13 at 18:58
  • The only way I can get it to work with your idea is if I use this version of the law of total expectation: E$(Y|X) =$ E$(\text{E}(Y|Z)|X)$, with $Y$ acting as $Y_{n+m}$, $Z$ as $X_1,\ldots,X_{n+m-1}$ and $X$ as $X_1,\ldots,X_n$ – user63010 Mar 20 '13 at 17:21
  • I am sorry but I can't just ignore the $X_1,\ldots,X_n$ part, although I am sure you know your stuff. – user63010 Mar 20 '13 at 17:29
  • @user63010 That version is correct. – NECing Mar 20 '13 at 17:33