# Ito's product rule: Three processes

I stumbled upon the following problem, I want to compute the stochastic differential of the following 3 processes:

$dX_t = \mu_{X,t}dt + \sigma_{X,t}dW_t$

$dY_t = \mu_{Y,t}dt + \sigma_{Y,t}dW_t$

$dZ_t = \mu_{Z,t}dt + \sigma_{Z,t}dW_t$

That is, I want to calculate $d(X_tY_tZ_t)$ using Ito's product rule

Now from the aforementioned product rule I know that $d(X_tY_t) = X_tdY_t + Y_tdX_t + \sigma_{X,t}\sigma_{Y,t}dt$. I tried to write $P_t = X_tY_t$ and substitute to calculate $d(P_tZ_t)$.

I obtained the process $d(X_tY_tZ_t) = X_tY_tdZ_t + X_tZ_tdY_t + Y_tZ_tdX_t + Z_t\sigma_{X,t}\sigma_{Y,t}dt + \sigma_{P,t}\sigma_{Z,t}dt$ but do not know how to work out the last volatility product. The correct process is apparently equal to

$d(X_tY_tZ_t) = X_tY_tdZ_t + X_tZ_tdY_t + Y_tZ_tdX_t + Z_t\sigma_{X,t}\sigma_{Y,t}dt + X_t\sigma_{Y,t}\sigma_{Z,t}dt+Y_t\sigma_{X,t}\sigma_{Z,t}dt$

I was hoping if somebody could help me arriving at the above equation.

Moreover, I am also curious to find out how to calculate the SDE $d(X_t/Y_t)$, given that the processes $X_t$ and $Y_t$ are the same as given above. If somebody knows how to tackle this one, I would be very grateful for your help.

Again, Ito's product rule; $d(X_tY_t) = X_tdY_t + Y_tdX_t + dX_tdY_t$.
Substituting in for $P_t = X_tY_t$, we can write the SDE of the 3 processes as:
\begin{align} d(P_tZ_t) &= P_tdZ_t + Z_tdP_t + dP_tdZ_t\\ \text{(substituting back in ${P_t}$)}&= X_tY_tdZ_t + Z_td(X_tY_t) +d(X_tY_t)dZ_t\\ &= \underbrace{X_tY_tdZ_t + X_tZ_tdY_t + Y_tZ_tdX_t}_{C_t} + Z_tdX_tdY_t + d(X_tY_t)dZ_t\\ &= C_t + Z_tdX_tdY_t + X_tdZ_tdY_t + Y_tdZ_tdX_t + dX_tdY_tdZ_t\\ &=C_t + \sigma_{Y,t}\sigma_{Z,t}X_tdt + \sigma_{X,t}\sigma_{Z,t}Y_tdt + \sigma_{X,t}\sigma_{Y,t}Z_tdt \end{align}
Note: Neglecting all terms smaller than $dt$. However, I am still having trouble with the quotient $d(X_t/Y_t)$. Any help regarding this problem would be much appreciated.
Set $$f(x,y,z)=xyz$$ , we have $$\frac{\partial f}{\partial x}=yz\,\,,\,\,\,\frac{\partial f}{\partial y}=xz\,\,\,,\,\,\,\frac{\partial f}{\partial z}= \frac{\partial (xyz)}{\partial z}=xy$$ $$\frac{\partial^2 f}{\partial x^2}=\frac{\partial^2 f}{\partial y^2}=\frac{\partial^2 f}{\partial z^2}=0$$ $$\frac{\partial^2 f}{\partial x\partial y}=\frac{\partial^2 f}{\partial y\partial x}=z\quad ,\quad\frac{\partial^2 f}{\partial x\partial z}=\frac{\partial^2 f}{\partial z\partial x}=y\quad ,\quad\frac{\partial^2 f}{\partial y\partial z}=\frac{\partial^2 f}{\partial z\partial y}=x$$ Set $$X_1=X$$ , $$X_2=Y$$ and $$X_3=Z$$. By application of Ito's lemma, we have $$df(X_1,X_2,X_3)=\sum_{i=1}^{3}\frac{\partial f}{\partial X_i}dX_i+\frac{1}{2}\sum_{i=1}^{3}\sum_{j=1}^{3}\frac{\partial^2 f}{\partial X_i\partial X_j}dX_idX_j$$ therefore $$d(X_tY_tZ_t)=Y_tZ_t\,dX_t+X_tZ_t\,dY_t+X_tZ_t\,dY_t+Z_t\,dX_tdY_t+Y_t\,dX_tdZ_t+X_t\,dY_tdZ_t$$ Note $$dX_tdY_t=\sigma_{X,t}\sigma_{Y,t}dt$$ $$dX_tdZ_t=\sigma_{X,t}\sigma_{Z,t}dt$$ $$dZ_tdY_t=\sigma_{Y,t}\sigma_{Z,t}dt$$ Now Set $$g(x,y)=\frac{x}{y}$$ and repeat this procedure: $$\frac{\partial g}{\partial x}=\frac{1}{y}\,\,,\,\,\,\frac{\partial g}{\partial y}=-\frac{x}{y^2}$$ $$\frac{\partial^2 g}{\partial x^2}=0\quad,\quad\frac{\partial^2 g}{\partial y^2}=\frac{2x}{y^3}\quad,\quad \frac{\partial^2 g}{\partial x\partial y}=\frac{\partial^2 g}{\partial y\partial x}=-\frac{1}{y^2}$$ Set $$X_1=X$$ and $$X_2=Y$$. By application of Ito's lemma, we have $$dg(X_1,X_2)=\sum_{i=1}^{2}\frac{\partial g}{\partial X_i}dX_i+\frac{1}{2}\sum_{i=1}^{2}\sum_{j=1}^{2}\frac{\partial^2 g}{\partial X_i\partial X_j}dX_idX_j$$ therefore $$d\left(\frac{X_t}{Y_t}\right)=\frac{1}{Y_t}dX_t-\frac{X_t}{Y_t^2}dY_t+\frac{X_t}{Y_t^3}dY_tdY_t-\frac{1}{Y_t^2}dX_tdY_t$$ Note $$dX_tdY_t=\sigma_{X,t}\sigma_{Y,t}dt$$ $$dY_tdY_t=\sigma_{Y,t}\sigma_{Y,t}dt$$