Can someone help me with calculus for this problem.

I have these 3 equations and with Ito`s Lemma I have to find $dXt$.

\begin{cases} dY= μYdt+σYdB \\ X=\frac{1}{2}cY\\ dc =-aαcdt\end{cases}


1 Answer 1


write down Ito's lemma for the function X:

$$dX=\frac{\partial X}{\partial Y}dY+\frac{1}{2}\frac{\partial^2 X}{\partial Y^2}(dY)^2+\frac{\partial X}{\partial c}dc+\frac{1}{2}\frac{\partial^2 X}{\partial c^2}(dc)^2+\frac{\partial^2 X}{\partial Y \partial c}dYdc+\frac{\partial^2 X}{\partial c \partial Y}dcdY$$

Using the following:

$\frac{\partial X}{\partial Y}=\frac{1}{2}c$, $\frac{\partial^2 X}{\partial Y^2}=0$

$\frac{\partial X}{\partial c}=\frac{1}{2}Y$, $\frac{\partial^2 X}{\partial c^2}=0$

$\frac{\partial^2 X}{\partial Y \partial c}=\frac{\partial^2 X}{\partial c \partial Y}=0$

Inserting these 4 expressions into the above Ito formula, one gets to:


where the initial expressions for $dY$ and $dc$ have been substituted back in the last step. The solutions for $Y$ and $c$, are trivial: They are the solution of the SDE for a GBM, and an exponential decay, respectively

  • $\begingroup$ Hello, thank you. You explained very well ! $\endgroup$
    – user41013
    May 12, 2019 at 20:09
  • $\begingroup$ I understand correctly... this formula is from Multidimensional Ito's Lemma? $\endgroup$
    – user41013
    May 12, 2019 at 20:31
  • $\begingroup$ Can we assume that $a$ and $\alpha$ are constants? In this case $c(t) = e^{-a \alpha t}$ $\endgroup$ May 12, 2019 at 20:50
  • $\begingroup$ Can we write $\frac{\partial X}{\partial t} = -1/2 e^{-a \alpha t} ( a \alpha + \mu ) Y$ and $\frac{\partial X}{\partial Y} = 1/2 e^{-a \alpha t} dY$? $\endgroup$ May 12, 2019 at 21:02

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