If $X_t$ is an Ito process, such that:

$dX_t = \mu(t, X_t)dt + \sigma(t, Xt)dW_t$ where $W_t$ is a standard brownian motion.

Then we can say that:

$E(dX_t) = \mu(t, X_t)dt$ and that $Var(dX_t) = \sigma^2(t, Xt)dt$

Is this equivalent to saying that (I removed the differential operator):

$E(X_t) = \mu(t, X_t)\times t$ and that $Var(X_t) = \sigma^2(t, Xt)\times t$


This is wrong! Notice that $dX_t=\mu(t,X_t)dt + \sigma(t,X_t)dW$ is a shorthand for $$\int_0^tdX_s = \int_0^t \mu(s,X_s)ds + \int_0^t\sigma(s,X_s)dW_s$$ Integrating: $$X_t-X_0 = \int_0^t \mu(s,X_s)ds + \int_0^t\sigma(s,X_s)dW_s \text{ (eq.1)} $$ If we take expectations, remembering that $\mathbb{E}[\int_0^t\sigma(s,X_s)dW_s]=0$, we have $$\mathbb{E}[X_t]=X_0 + \int_0^t \mu(s,X_s)ds$$ To have your result, we need $X_0=0$ and constant drift $\mu(t,X_t)=\mu$. You compute the second moment of (eq.1) to check what happens to the variance

  • $\begingroup$ That's right. I should have thought about it. Thank you! $\endgroup$ – MarinD Dec 7 '15 at 7:50

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