Itô's lemma on function composition

Question

I have a question on how to apply Itô's lemma on a function composition:

Let us consider a stochastic process $$ dX_t = a(X_t,t)dt+ b(X_t,t)dW_t, $$ where $W_t$ denotes the standard Brownian motion.

If we know the function $Y=Y(X_t,t)$, we can easily calculate $dY$ by applying Itô's lemma as \begin{align*} dY &= \left(\frac{\partial Y}{\partial t} + a\frac{\partial Y}{\partial X_t} + \frac{1}{2}b^2 \frac{\partial^2 Y}{\partial X_t^2} \right)dt + b\frac{\partial Y}{\partial X_t}dW_t\\ &=: \alpha(X_t,t)dt+\beta(X_t,t)dW_t. \end{align*}

Let us then consider a function $Z(Y,t)=Z(Y(X_t,t),t)$. If I wanted to calculate $dZ$, I would apply Itô's lemma again as $$ dZ = \left(\frac{\partial Z}{\partial t} + \alpha\frac{\partial Z}{\partial Y} + \frac{1}{2}\beta^2 \frac{\partial^2 Z}{\partial Y^2} \right)dt + \beta\frac{\partial Z}{\partial Y}dW_t. \tag{$\ast$} $$

My question is: In equation ($\ast$), is the partial derivative $\frac{\partial Z}{\partial t}$ to be considered as $\frac{\partial Z(Y,t)}{\partial t}$ or $\frac{\partial Z(Y(X,t),t)}{\partial t}=\frac{\partial Z(X,t)}{\partial t}$?

Answer 1

I think the first one is right.

We write $Y_t= f(X_t,t)$, where $f( \, \cdot \, , \, \cdot \, ) $ is twice continuously differentiable on $\mathbb{R}\times [0, +\infty)$; and similarly, write $Z= g(Y_t, t)$, where $g( \, \cdot \, , \, \cdot \, )\in C^2(\mathbb{R}\times [0, +\infty))$. Thus, $$dY_t= \frac{\partial f}{\partial t}(X_t,t) dt +\frac{\partial f}{\partial x}(X_t,t) dX_t +\frac{1}{2} \frac{\partial^2 f}{\partial x^2}(X_t, t)(dX_t)^2.$$ If $X_t$ is given by $dX_t = u(t) dt +v(t) dW_t$, where $u\in \mathscr{L}^1$, $v\in \mathscr{L}^2$ ($\mathscr{L}^2$ is known as the $``\text{Ito space''}$, and $\mathscr{L}^1$ is defined similarly to $\mathscr{L}^2$). Using the rules: $$ dt\cdot dt= dB_t\cdot dt= dt\cdot dB_t =0, \ \ dB_t\cdot dB_t =dt,$$ then $$dY_t= \left[ \frac{\partial f}{\partial t}(X_t, t) + u(t)\frac{\partial f}{\partial x} (X_t,t)+\frac{1}{2}v(t)^2 \frac{\partial^2 f}{\partial x^2}(X_t,t)\right]dt+ v(t)\frac{\partial f}{\partial x}(X_t,t) dW_t, $$ just as what you present in the question. It is known as $``\text{1D-Ito formula''}$.

Assume $dY_t= u_1(t)dt +v_1(t)dW_t$, where $u_1(t),v_1(t)$ are given above, i.e., \begin{align*} &u_1(t)= \frac{\partial f}{\partial t}(X_t, t) + u(t)\frac{\partial f}{\partial x} (X_t,t)+\frac{1}{2}v(t)^2 \frac{\partial^2 f}{\partial x^2}(X_t,t),\\ &v_1(t)= v(t)\frac{\partial f}{\partial x}(X_t,t). \end{align*} We calculate $dZ$: \begin{align*} dZ_t &= \frac{\partial g}{\partial t}(Y_t,t) dt +\frac{\partial g}{\partial x}(Y_t,t) dX_t +\frac{1}{2} \frac{\partial^2 g}{\partial x^2}(Y_t, t)(dY_t)^2\\ &= \left[ \frac{\partial g}{\partial t}(Y_t, t) + u_1(t)\frac{\partial g}{\partial x} (Y_t,t)+\frac{1}{2}v_1(t)^2 \frac{\partial^2 g}{\partial x^2}(Y_t,t)\right]dt+ v_1(t)\frac{\partial g}{\partial x}(Y_t,t) dW_t. \end{align*}

Now you can see, it is a problem of notation. In your eq. $(*)$, you write $\frac{\partial Z}{\partial t}$, but I write $\frac{\partial g}{\partial t}$ instead. You know, $f,g$ are given in advance. I think it is your improper notation that misleads you.