3
$\begingroup$

Suppose $X$ and $Y$ are Ito processes, $X_t=x+\int^t_0Y_sdB_s$ and $Y_t=y-\int^t_0X_sdB_s,\ t\geq 0$, here $B$ is a standard Brownian motion. I need to prove that $Z_t:=X^2_t+Y^2_t=(x^2+y^2)\exp\{t\}$.

My first thought was plugging the expression for $Y_s$ in the expression of $X_t$ then solving for $X$ and vice versa. I would then have

$$X_t=x+\int_0^t\left(y-\int_0^sX_udB_u\right)dB_s.$$

Applying Ito's formula yields

$$\int_0^sX_udB_u=\frac{1}{2}(X^2_s-x^2)-s$$

and plugging back into the previous formula we have

$$dX_t=\left(y-\frac{1}{2}(X^2_t-x^2)+t\right)dB_t$$

and I don't know how to solve this.

My guess is that $X$ and $Y$ might be stochastic exponents, but I don't see how to get there. I just started learning, so maybe I'm missing something simple. Any hints are very appreciated.

$\endgroup$
3
$\begingroup$

HINT: This is about applying Ito's lemma to $f(X_t, Y_t) = X_t^2 + Y_t^2$. Rewrite your vector valued process as $$ \mathrm{d} X_t = Y_t \mathrm{d} B_t \qquad \mathrm{d} Y_t = -X_t \mathrm{d} B_t \qquad X_0 = x \qquad Y_0 = y $$ $$ \begin{eqnarray} f(X_t, Y_t) = f(X_0, Y_0) &+& \int_0^t \left( \frac{1}{2} Y_s^2 \frac{\partial^2}{\partial X_s^2} f(X_s, Y_s) + \frac{1}{2} (-X_s)^2 \frac{\partial^2}{\partial Y_s^2} f(X_s, Y_s) \right) \mathrm{d} s \\ &+& \int_0^t \left( Y_s \frac{\partial}{\partial X_s} f(X_s, Y_s) - X_s \frac{\partial}{\partial Y_s} f(X_s, Y_s) \right) \mathrm{d} B_s \end{eqnarray} $$

$\endgroup$
  • $\begingroup$ Thanks for the reply, now I see it. $\endgroup$ – nokiddn Mar 4 '12 at 20:33

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.