# Prove that martingales have zero covariance

• Apr 19th 2012, 08:22 AM
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.
• Apr 29th 2012, 07:52 AM
Moo
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\$.