# Itô's lemma and uniqueness of the solution to the stochastic exponential SDE

Background

I am reading a proof that applies the Itô product rule and Itô's lemma for some calculations. However, I am not able to reproduce one of these calculations, so I'd appreciate it if someone could help me figure out what I'm doing wrong.

Details

Fix a càdlàg semimartingale $$X$$ with $$X_0 = 0$$. We define the stochastic exponential process $$\mathcal E(X)$$ of $$X$$ by $$\mathcal E(X)_t = \exp\left( X_t - \frac 12 \left\langle X^c \right\rangle_t\right) \prod_{s\le t} (1+\Delta X_s) e^{-\Delta X_s}, \tag{SE}\label{SEsol}$$ where $$X^c$$ is the continuous martingale part of $$X$$; $$\langle\cdot\rangle$$ denotes the (predictable) quadratic variation; and, $$\Delta X_t = X_t - X_{t-}$$, where $$X_{t-} = \lim_{s \uparrow t} X_s$$. We know that $$\mathcal E(X)$$ solves the SDE $$\mathrm d Z_t = Z_{t-} \mathrm d X_t; \quad Z_0 = 1. \tag{SE-SDE}\label{SE}$$

I am reading a proof that $$\mathcal E(X)$$ is the unique solution of \eqref{SE}. To establish this, the proof defines the process $$Y$$ by $$Y_t = \color{blue}{\exp\left(-X_t + \frac 12 \langle X^c \rangle_t\right)} Z_t =: \color{blue}{U_t} Z_t,$$ where $$Z$$ is a solution to \eqref{SE}. The proof then applies Itô's product rule and Itô's lemma to calculate that $$\mathrm d Y_t = Y_{t-} \left(\left(e^{-\Delta X_t} -1 -\Delta X_t\right)(1+\Delta X_t) - (\Delta X_t)^2\right), \tag{1} \label{1}$$ and argues that $$Y$$ uniquely solves the SDE defined by \eqref{1}, which establishes the uniqueness of the solution to \eqref{SE}.

Unfortunately, I am not able to reproduce this calculation, as the expression I find for $$\mathrm d Y_t$$ is $$\mathrm d Y_t = Y_{t-} \left(\mathrm d \langle X^c \rangle_t + \left(e^{-\Delta X_t} -1 -\Delta X_t\right)(1+\Delta X_t) - (\Delta X_t)^2\right). \tag{2}\label{2}$$

However, given the form of \eqref{SEsol}, I expect $$Y$$ to be a pure jump process, so my guess would be that I've made a mistake in my derivation.

Could anyone help me figure out what I'm doing wrong, if I am in fact mistaken?

Calculations

The calculation starts by using Itô's lemma to calculate that $$\mathrm d U_t = U_{t-} \left( -\mathrm d X_t + \mathrm d \langle X^c \rangle_t + e^{-\Delta X_t} - 1 -\Delta X_t \right).$$

I am able to follow this step, and use it to calculate that $$\Delta U_t = U_{t-} \left( -\Delta X_t + e^{-\Delta X_t} -1 -\Delta X_t \right). \tag{3}\label{3}$$

Next, we calculate using the product rule that $$\mathrm d Y_t = Z_{t-}\mathrm d U_t + U_{t-} \mathrm dZ_t + \Delta U_t \Delta Z_t \\ = Y_{t-} \left( -\mathrm d X_t + \mathrm d \langle X^c \rangle_t + e^{-\Delta X_t} -1 -\Delta X_t \right) + Y_{t-} \mathrm d X_t + Y_{t-} \left( \color{red}{-\mathrm d \langle X^c \rangle_t} + \left( - \Delta X_t + e^{-\Delta X_t} -1 -\Delta X_t \right) \Delta X_t \right) \tag{4}\label{4}.$$

It is easy to see that \eqref{4} should simplify to \eqref{1}. However, I don't see where the $$\color{red}{-\mathrm d \langle X^c \rangle_t}$$ highlighted in red comes from. My calculations do not have this term, thus explaining the discrepancy between \eqref{1} and \eqref{2}.

My understanding is that the third term in \eqref{4} is given by $$\Delta U_t \Delta Z_t = Z_{t-} \Delta U_t \Delta X_t,$$ where $$\Delta U_t$$ is given by \eqref{3}. If this is correct, then the red $$\color{red}{-\mathrm d \langle X^c \rangle_t}$$ should not appear in \eqref{4}.

What am I am not getting here? My guess is that my mistake lies in \eqref{3}, or in viewing the last term in \eqref{4} as the the quadratic co-variation term, but I don't see what the exact error is.

Update

I now know that my mistake is in the application of the product rule, which should read $$\mathrm d Y_t = Z_{t-} \mathrm d U_t + U_{t-} \mathrm d Z_t + \mathrm d [U,Z]_t,$$ and this looks like it should lead exactly to \eqref{1}. I'm going to work on this and update my question or post an answer later on.

• Unrelated to the problem: That has to be the most beautifully formatted question I have ever seen on this site. Apr 7, 2021 at 8:48

As I noted in my question, the product rule in the first line of $$(4)$$ should read \begin{align*} \mathrm d Y_t &= Z_{t-} \mathrm dU_t + U_{t-} \mathrm dZ_t + \mathrm d [U,Z]_t \\ &= Z_{t-} \mathrm dU_t + U_{t-} \mathrm dZ_t + \mathrm d \langle U^c,Z^c\rangle_t + \Delta U_t \Delta Z_t. \end{align*}
The $$\color{red}{-\mathrm d \langle X^c \rangle_t}$$ then shows up in the last line of $$(4)$$ because $$\mathrm d \langle U^c,Z^c \rangle_t = -Y_{t-} \mathrm d \langle X^c \rangle_t$$.
To see this, we use the facts that $$\mathrm d U^c_t = -U_{t-} \mathrm d X_t^c$$ and $$\mathrm d Z_t ^c = Z_{t-} \mathrm d X^c_t$$, along with the identity $$\langle H\bullet M,N \rangle = H\bullet \langle M,N\rangle$$, where $$H\bullet M$$ is the integral of the process $$H$$ with respect to $$M$$.