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Thread: Prove that martingales have zero covariance

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    Prove that martingales have zero covariance

    Let $\displaystyle X_n $ be a martingale. Show that $\displaystyle Cov ( X_i-X_{i-1}, X_j-X_{j-1})=0 $

    So using the covariance formula, I know that $\displaystyle Cov ( X_i-X_{i-1}, X_j-X_{j-1})=E[(X_i-X_{i-1})(X_j-X_{j-1})]-E[X_i-X_{i-1}]E[X_j-X_{j-1}] $

    Now I know that I must use the property of martingale in which $\displaystyle E[X_{i+1}|X_i] = X_i $, but how should I mix that property in this problem? Should I take the double expectation on it? Thanks.
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    Re: Prove that martingales have zero covariance

    Hello,

    WLOG, we can suppose that i>j. Expand everything in $\displaystyle E[(X_i-X_{i-1})(X_j-X_{j-1})]=E[X_iX_j]-E[X_iX_{j-1}]-E[X_jX_{i-1}]+E[X_{i-1}X_{j-1}]$
    Then yes you have to take the double expectation for each term.

    Just note that according to the martingale property $\displaystyle E[X_m|X_n]=X_n$ for any $\displaystyle m>n$, it would be useful to condition by the X with the lowest index.
    For example, $\displaystyle E[X_iX_j]=E[E[X_iX_j|X_j]]=E[X_jE[X_i|X_j]]=E[X_j^2]$.

    For $\displaystyle E[X_i-X_{i-1}]E[X_j-X_{j-1}]$, note that with the martingale property, $\displaystyle E[X_i|X_{i-1}]=X_{i-1}\Rightarrow E[X_i-X_{i-1}|X_{i-1}]=0$

    So that $\displaystyle E[X_i-X_{i-1}]=E[E[X_i-X_{i-1}|X_{i-1}]]=0$.
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