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.


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?


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.


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.

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

1 Answer 1


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 $M$.

  • $\begingroup$ As an aside, I might not have been able to figure this out if I didn't feel compelled to clean up my notation in order to make my question readable for MSE. The moral of the story is to keep your notation clean. $\endgroup$ Oct 17, 2019 at 17:10

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