3
$\begingroup$

Suppose I have a general SDE

$dx_{t} = \mu dt + \sigma dz_{t}$

Then I can put $E[]$ on both sides to get

$E[dx_{t}] = E[\mu dt] + E[\sigma dz_{t}]$

Now comes the question: I've seen some formulas where

$E[dx_{t}]$ becomes $dE[x_{t}]$

Is it ok to swap $E[.]$ with $d[.]$?

Thank you.

$\endgroup$
1
$\begingroup$

Remember that $dx_t = \mu_t dt + \sigma_t dz_t$ is just a shorter notation for $$ x_t = x_0 + \int_0^t \mu_s ds + \int_0^t \sigma_s dz_s $$

Now, under mild hyopthesis on $\sigma$ the stochastic integral is a martingale so $E[\int_0^t \sigma_s dz_s] = E[\int_0^0 \sigma_s dz_s] = 0$. We are left with $$ E[x_t] = x_0 + E[\int_0^t \mu_s ds] = x_0 + \int_0^t E[\mu_s] ds $$ by Fubini's theorem.

So $dE[x_t] = E[\mu_t] dt$. This justifies writing $dE[x_t] = E[dx_t]$.

$\endgroup$

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.