2
$\begingroup$

I have proven that Martingales have orthogonal increments.

From this I need to show that $\operatorname{Cov}[M(t),M(s)]$ relies only on $\min\{s,t\}$.

I used the expected value definition of Covariance and eliminated one piece using the orthogonal increments bit. Also, I introduced conditioning on $F_s$.

$\operatorname{Cov}[M(t),M(s)]= E[(M(t)-E[M(s)])(M(s)-E[M(s)])]$

$= .... $

$= E[E[((M(t)+E[M(t)])E[M(s)]|F_s)]]$

From here I am lost.

Should I....

$ = E[E[M(t)E[M(s)]|F_s] +E[E[E[M(t)]*E[M(s)]|F_s]]$

$ = E[E[M(t)M(s)|F_S] + E[E[E[M^2(s)]|F_s]]$

I know $E[M(t)] = E[M(s)]$. Thats how I got the second piece of the second line.

Is this correct? Where should I go from there?

$\endgroup$
0
$\begingroup$

You are on the right track. Assuming $s<t$, by conditioning, $$ E[M(s)M(t)]=E[M(s)E[M(t)|\mathcal F_s]=E[M(s)^2]. $$ So, $$ cov[M(s),M(t)]=E[M(s)M(t)]-E[M(s)]E[M(t)]=E[M(s)^2]-(E[M(s)])^2=var[M(s)], $$ this depends on $t$ only through the fact that $t\in(s,\infty)$.

$\endgroup$
0
$\begingroup$

What is the mathematical meaning of 'rely on'? Typically, to compute covariances of martingales you can do the following. Suppose $s<t$ \begin{align} \mbox{Cov}[M(t),M(s)] &= \mbox{Cov}[M(t)-M(s)+M(s),M(s)] \\ &=\mbox{Cov}[M(t)-M(s),M(s)] + \mbox{Var}[M(s)] \\ &= \mathbb{E}[M(s)(M(t)-M(s))] - \mathbb{E}[M(t)-M(s)]\mathbb{E}[M(s)] + \mbox{Var}[M(s)] \\ \end{align} Using the orthogonal increment property we have that $\mathbb{E}[M(s)(M(t)-M(s))] = 0$ and hence for $s<t$ $$\mbox{Cov}[M(t),M(s)] = \mbox{Var}[M(s)].$$ The proof goes similar for $t<s$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.