Itos Lemma problem

Can someone help me with calculus for this problem.

I have these 3 equations and with Itos Lemma I have to find $$dXt$$.

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

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:

$$dX=\frac{1}{2}cdY+\frac{1}{2}Ydc=cY(\frac{\mu}{2}-\frac{a\alpha}{2})dt+\frac{\sigma}{2}YcdB$$

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

• Hello, thank you. You explained very well ! – user41013 May 12 '19 at 20:09
• I understand correctly... this formula is from Multidimensional Ito's Lemma? – user41013 May 12 '19 at 20:31
• Can we assume that $a$ and $\alpha$ are constants? In this case $c(t) = e^{-a \alpha t}$ – Jónás Balázs May 12 '19 at 20:50
• Can we write $\frac{\partial X}{\partial t} = -1/2 e^{-a \alpha t} ( a \alpha + \mu ) Y$ and $\frac{\partial X}{\partial Y} = 1/2 e^{-a \alpha t} dY$? – Jónás Balázs May 12 '19 at 21:02