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Math Help - Prove that martingales have zero covariance

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    Prove that martingales have zero covariance

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

    So using the covariance formula, I know that 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  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 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 E[X_m|X_n]=X_n for any m>n, it would be useful to condition by the X with the lowest index.
    For example, E[X_iX_j]=E[E[X_iX_j|X_j]]=E[X_jE[X_i|X_j]]=E[X_j^2].

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

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