# Prove that martingales have zero covariance

• Apr 19th 2012, 09:22 AM
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.
• Apr 29th 2012, 08:52 AM
Moo
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 $E[X_i-X_{i-1}]=E[E[X_i-X_{i-1}|X_{i-1}]]=0$.