# Two methods to solve SDE yield different answers

The Questions is the following:

Consider the system of SDE: be a 1-d Brownian Motion, issued from the origin. For every $$c>0$$ and $$\alpha, \beta \in \mathbb{R}$$, consider the system of SDE: $$\left\{\begin{array}{l} d X_t^1=c X_t^2 d B_t+\frac{1}{2} X_t^1 d t \\ d X_t^2=\frac{1}{c} X_t^1 d B_t+\frac{1}{2} X_t^2 d t \\ X_0^1=\alpha, \quad X_0^2=\beta \end{array}\right.$$ Simplify the expression of $$Y_t$$. $$Y_t=\left(\frac{1}{c} X_t^1\right)^2-\left(X_t^2\right)^2$$

For both attempts, I computed:

$$d[X_s^1,X_s^1] = c^2(X^2_s)^2ds \quad \text{and} \quad d[X_s^2,X_s^2] = \frac{1}{c^2}(X^1_s)^2ds$$

My first Attempt:

Consider $$f(x,y) = (x/c)^2-y^2$$, we have $$\frac{\partial f}{\partial x} = 2x/c^2,\ \frac{\partial f}{\partial y} = -2y,\ \frac{\partial^2 f}{\partial x^2} = 2/c^2,\ \frac{\partial^2 f}{\partial y^2} = -2,\ \frac{\partial^2 f}{\partial x \partial y}= 0$$. Now by Ito's Lemma, \begin{align*} &f(X_t^1,X_t^2) - f(X_0^1,X_0^2) \\ =\ &\int_0^t \frac{\partial f}{\partial x}(X_s^1)dX_s^1+ \int_0^t \frac{\partial f}{\partial y}(X_s^2)dX_s^2 + \frac12 \left[\int_0^t \frac{\partial^2 f}{\partial x^2}(X_s^1)d[X^1,X^1]_s+ \int_0^t \frac{\partial^2 f}{\partial y^2}(X_s^2)d[X^2,X^2]_s\right] \\ =\ & \int_0^t \frac{2X_s^1}{c^2}\left(c X_s^2 d B_s+\frac{1}{2} X_s^1 d s\right) - 2\int_0^t X_s^2\left(\frac{1}{c} X_s^1 d B_s+\frac{1}{2} X_s^2 d s\right) \\ \quad &+\frac12 \int_0^t \frac{2}{c^2}c^2(X^2_s)^2ds + \frac12\int_0^t -2\frac{1}{c^2}(X^1_s)^2ds \\ =\ &\int_0^t \frac{(X_s^1)^2}{c^2} d s - \int_0^t (X_s^2)^2d s + \int_0^t(X^2_s)^2ds -\int_0^t \frac{1}{c^2}(X^1_s)^2ds \\ =\ &0 \end{align*} which of course should not be correct.

And my second attempt is the following:

Recall that $$(X_t)^2 = X_0^2 + 2\int_0^tX_{-}dX + [X,X]$$ We then have \begin{align*} Y_t &= \left(\frac1c X_t^1\right)^2 - (X_t^2)^2 \\ &= \frac{1}{c^2}(X_0^1)^2 + \frac{2}{c^2}\int_0^tX_s^1dX_s^1 + \frac{1}{c^2}[X^1,X^1]_t \\ &\quad -(X_0^2)^2 - 2\int_0^tX_s^2dX_s^2 - [X^2,X^2]_t \\ &= Y_0 + \frac{2}{c^2}\int_0^tX_s^1\left(c X_s^2 d B_s+\frac{1}{2} X_s^1 d s\right)+ \frac{1}{c^2}c^2(X^2_t)^2 \\ &\quad - 2\int_0^tX_s^2\left(\frac{1}{c} X_s^1 d B_s+\frac{1}{2} X_s^2 d s\right) - \frac{1}{c^2}(X^1_t)^2 \\ &= Y_0 + \frac{1}{c^2}\int_0^t(X_s^1)^2 d s+ (X^2_t)^2 - \int_0^t (X_s^2)^2 d s- \frac{1}{c^2}(X^1_t)^2 \\ &= Y_0 +\int_0^t Y_s d s -Y_t \\ \end{align*} What is going on?

I feel like I genuinely don't understand how to do this at all. Thank you for your help and appreciate your time in reading.

The first is correct, the second wrong. Quick check, $$[A+B,A+B]_t=t$$, where $$A_t=f(t)$$ is a pure function of time. This is in no way related to $$(A_t+B_t)^2$$.
Note that the first tells you that $$Y_t$$ is constant, which is to be expected for $$\frac1cX^1_t=\fracαc\cosh(B_t)+β\sinh(B_t),\\ X^2_t=\fracαc\sinh(B_t)+β\cosh(B_t).$$
• But in my referrence textbook the author actually first defines $[X,X]$ to be $[X,X] = X^2 - \int XdX$, how is the second one wrong? The book is "Stochastic integration and differential equations" by Philip Protter Commented Nov 17, 2022 at 22:43
• This is incompatible with $d((X_t)^2)=2X_t\,dX_t+d[X,X]_t$, as that gives $[X,X]_t=(X_t)^2-(X_0)^2-2\int_0^tX_t\,dX_t$. Commented Nov 17, 2022 at 22:54
• Why is it incompatible? From $[X,X]_t = (X_t)^2 - (X_0)^2 -2 \int_0^tX_tdX_t$, shouldn't we get $d[X,X]_t = d((X_t)^2) - d(2 \int_0^tX_tdX_t)$, and hence $d((X_t)^2)= d[X,X]_t + d(2 \int_0^tX_tdX_t) = d[X,X]_t + 2X_tdX_t$? Commented Nov 17, 2022 at 22:59