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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.

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1 Answer 1

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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 $$ \mathrm{d} X_t = Y_t \mathrm{d} B_t \qquad \mathrm{d} Y_t = -X_t \mathrm{d} B_t \qquad X_0 = x \qquad Y_0 = y $$ $$ \begin{eqnarray} f(X_t, Y_t) = f(X_0, Y_0) &+& \int_0^t \left( \frac{1}{2} Y_s^2 \frac{\partial^2}{\partial X_s^2} f(X_s, Y_s) + \frac{1}{2} (-X_s)^2 \frac{\partial^2}{\partial Y_s^2} f(X_s, Y_s) \right) \mathrm{d} s \\ &+& \int_0^t \left( Y_s \frac{\partial}{\partial X_s} f(X_s, Y_s) - X_s \frac{\partial}{\partial Y_s} f(X_s, Y_s) \right) \mathrm{d} B_s \end{eqnarray} $$

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