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I came across a problem with SDE and need your help once again:

$$dX_t=tX_t \, dt+\exp \left(\frac{t^2}{2}\right)$$ and I'm supposed to solve this, in the way $X_t=f(t,W_t)$.

So I use Itô's formula: $$dX_t=\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial x}dW_t+\frac{1}{2}\frac{\partial^2 f}{\partial x^2}dt$$ and then get that: $$\frac{\partial f}{\partial t}+\frac{1}{2}\frac{\partial^2 f}{\partial x^2}=\mu(t,x)=t\cdot x \\ \frac{\partial f}{\partial x}=\sigma(t,x)=\exp \left(\frac{t^2}{2}\right).$$ From the second line, I'd get: $f=\exp\left(\frac{t^2}{2}\right)\cdot x$, but then the first line would become: $$t\cdot \exp \left(\frac{t^2}{2} \right)\cdot x + \frac{1}{2}\cdot 0$$ which is not $t\cdot x$. Where is the mistake?

all the best!


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Since no one did yet, I guess I should point out that the SDE is not written correctly. – Did May 22 '12 at 13:45
up vote 2 down vote accepted

Oh yes, I found the mistake:

The function is not $dX_t=txdt+(\cdots)$, but $dX_t=tX_tdt + (\cdots).$ So with $X_t=f(t,x)$, it becomes clear that $dX_t=t\cdot \exp\left( \frac{t^2}{2}\right)\cdot x + (\cdots) = t\cdot f(t,x)+(\cdots)= tX_t + (\cdots)$. Stupid me!

and then we get $$X_t=\exp\left( \frac{t^2}{2}\right)\cdot W_t$$ and with $X_0=1$, this is $$X_t=\exp\left( \frac{t^2}{2}\right)\cdot \frac{W_t}{W_0}$$

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