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Math Help - basic question on martingales definition

  1. #1
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    basic question on martingales definition

    In many books, a martingale (roughly speaking) is defined as a process X1,X2,... satisfying:

    <br />
\mathbb{E}[X_{t_{n+1}} | X_{t_n},\ldots,X_{t_0}] = X_{t_n}<br />

    which is equivalent to

    <br />
\mathbb{E}[X_{t_{n+1}} - X_{t_n} | X_{t_n},\ldots,X_{t_0}] = 0<br />

    While the latter is understandable to me, I interpret the former in this sense: "expectation is equal to a random variable", which is absurd to me.

    Now, I am sure this is a stupid question but: how should I interpret the former?
    Maybe that X_tn is no more a random variable but a number (because at time tn is known)? If so, why don't textbook put that symbol in lower case as usual... ?
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  2. #2
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    Quote Originally Posted by tossifan View Post
    In many books, a martingale (roughly speaking) is defined as a process X1,X2,... satisfying:

    <br />
\mathbb{E}[X_{t_{n+1}} | X_{t_n},\ldots,X_{t_0}] = X_{t_n}<br />

    which is equivalent to

    <br />
\mathbb{E}[X_{t_{n+1}} - X_{t_n} | X_{t_n},\ldots,X_{t_0}] = 0<br />

    While the latter is understandable to me, I interpret the former in this sense: "expectation is equal to a random variable", which is absurd to me.

    Now, I am sure this is a stupid question but: how should I interpret the former?
    Maybe that X_tn is no more a random variable but a number (because at time tn is known)? If so, why don't textbook put that symbol in lower case as usual... ?
    Hi,
    you should read again the chapter on conditional expectations in your textbook: a conditional expectation is indeed a random variable. The conditional expectation E[X|Y,Z] is intuitively "the random variable depending on Y and Z only that best approximates X"; formal definition isn't straightforward, cf. textbooks or probably wikipedia.
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  3. #3
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by tossifan View Post
    In many books, a martingale (roughly speaking) is defined as a process X1,X2,... satisfying:

    <br />
\mathbb{E}[X_{t_{n+1}} | X_{t_n},\ldots,X_{t_0}] = X_{t_n}<br />

    which is equivalent to

    <br />
\mathbb{E}[X_{t_{n+1}} - X_{t_n} | X_{t_n},\ldots,X_{t_0}] = 0<br />

    While the latter is understandable to me, I interpret the former in this sense: "expectation is equal to a random variable", which is absurd to me.

    Now, I am sure this is a stupid question but: how should I interpret the former?
    Maybe that X_tn is no more a random variable but a number (because at time tn is known)? If so, why don't textbook put that symbol in lower case as usual... ?
    They are the same since

    <br />
\mathbb{E}[X_{t_n} | X_{t_n},\ldots,X_{t_0}] = X_{t_n} <br />

    because you know  X_{t_n}

    <br />
\mathbb{E}[X_{t_{n+1}} - X_{t_n} | X_{t_n},\ldots,X_{t_0}]

     = <br />
\mathbb{E}[X_{t_{n+1}} | X_{t_n},\ldots,X_{t_0}] -<br />
\mathbb{E}[ X_{t_n} | X_{t_n},\ldots,X_{t_0}] = X_{t_n}-X_{t_n}=0<br />
    Last edited by matheagle; February 27th 2010 at 04:12 PM.
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  4. #4
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    Quote Originally Posted by matheagle View Post
    They are the same since

    <br />
\mathbb{E}[X_{t_n} | X_{t_n},\ldots,X_{t_0}] = X_{t_n} <br />

    because you know  X_{t_n}

    <br />
\mathbb{E}[X_{t_{n+1}} - X_{t_n} | X_{t_n},\ldots,X_{t_0}]

     = <br />
\mathbb{E}[X_{t_{n+1}} | X_{t_n},\ldots,X_{t_0}] -<br />
\mathbb{E}[ X_{t_n} | X_{t_n},\ldots,X_{t_0}] = X_{t_n}-X_{t_n}=0<br />
    So  X_{t_n} would be a constant at time t_n ...

    But this seems to disagree with what Laurent wrote!
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  5. #5
    MHF Contributor

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    Quote Originally Posted by tossifan View Post
    So  X_{t_n} would be a constant at time t_n ...
    Why would it be? In the equality E[X_{t_n}|X_{t_1},\ldots,X_{t_n}]=X_{t_n}, both hand sides are random variables. By the way, this equality relates to the intuition I gave: the left-hand side is the best approximation of X_{t_n} in terms of X_{t_1},\ldots,X_{t_n}, so in particular in terms of itself; therefore it is X_{t_n} itself (the best approximation is the variable itself...).
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