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If $f$ i twice differentiable scalar function and $X_t, Y_t$ are Ito processes then Ito lemma holds. But in 90% of sources I can only find the case, when $Y_t=t$ (it is deterministic function). The Ito formula is given by:

$$df=(\mu f_x+f_t+\frac{\sigma^2}{2}f_{xx})dt+\sigma f_xdB_t.$$

But what if $Y_t$ is not deterministic function but "normal" Ito process?

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  • $\begingroup$ I don't understand ii if $X_t$ is some Ito processand $Y = f(X_t)$, then $Y$ is a usual kind of Ito process -- even in the simplest case of $f(x)=x$, $Y$ would have its own SDE, identical to $X$... $\endgroup$ – gt6989b Jul 6 '15 at 16:40
  • $\begingroup$ Perhaps it would help if you would give a specific example of a problem... Are you interested in a case where $f(x)$ would not be a single-valued function, but would satisfy some SDE of its own? $\endgroup$ – gt6989b Jul 6 '15 at 16:41
  • $\begingroup$ Let $dX_t=adt+bdB_t$ and $dY_t=\alpha dt + \beta dB_t$. What dynamics have process $Z_t=f(X_t,Y_t)$? For example $f=e^{xy^2}$ $\endgroup$ – ksl Jul 6 '15 at 16:52
  • $\begingroup$ That just becomes the vector case of Ito's formula. (You need to be careful about whether they are the same Brownian motion or different ones, however.) $\endgroup$ – Ian Jul 6 '15 at 16:55
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    $\begingroup$ en.wikipedia.org/wiki/… Look at the part after "in higher dimensions" in the middle. (And again, be careful about whether the two Brownian motions are the same, correlated but different, or independent, because it affects the entries of the matrix $G_t$). $\endgroup$ – Ian Jul 6 '15 at 16:59
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If $f: \mathbb{R} \times \mathbb{R} \longrightarrow \mathbb{R}$ then $Z_t = f(X_t,Y_t)$ is a stochastic process given by $$dZ_t = f_x dX_t + f_ydY_t + \frac{1}{2}f_{xx}d[X]_t+ \frac{1}{2}f_{yy}d[Y]_t + \frac{1}{2}f_{xy}d[X,Y]_t$$ where $d[X]_t$ is the quadratic variation of $X_t$ and $d[X,Y]_t$ is the quadratic covariation of $X_t$ and $Y_t$.

People often write informally $d[X]_t = ``(dX_t)^2"$ and $d[X,Y]_t = ``(dX_t)(dY_t)"$.

(So if $dX_t=adt+bdB_t$ and $dY_t=αdt+βdB_t$ then $d[X]_t = b^2dt$, $d[Y]_t = \beta^2 dt$ and $d[X,Y]_t = b\beta dt$)

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  • $\begingroup$ I think the $f_{xy}$ term shoud not be multiplied by $1/2$ $\endgroup$ – Matias Puig Jun 5 '17 at 17:44
  • $\begingroup$ Yes only 1/2 multiplier on the pure (non-mixed) 2nd order partials. $\endgroup$ – Charles Pehlivanian Jan 26 '18 at 20:25

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