SDE for $Y_t = \frac{1}{Z_t}$ where $Z_t =\exp \left(\int_0^t X_u \, dW_u-\frac{1}{2}\int_0^t X^2_u \, du \right)$ The problem is the following (Problem 3.10 from Karatzas/Shreve):

Let $Z_t = \exp \left(\int_0^t X_u \, dW_u-\frac{1}{2}\int_0^t X^2_u \, du \right)$. Define $Y_t=1/Z_t$. Show the following stochastic differential equation holds
\begin{align*}
dY_t=Y_tX_t^2\,dt-Y_tX_t\,dW_t, \quad Y_0=1
\end{align*}
where we have assumed $\int_0^t X_u^2 \, du<\infty$ almost surely for all $0<t<\infty$, and $W_t$ is the standard Brownian motion.

My questions are:

*

*How can we prove $Y_t$ is a semi-martingale?

*For the differential equation, this is my try (which I assume $Y_t$ is a semi-martingale

\begin{align*}
Y_tZ_t=1 \ \Longrightarrow Z_t\,dY_t + Y_t\,dZ_t + d\langle Y_t, Z_t \rangle = 0
\end{align*}
Since $Y_tZ_t=1$, we have $\langle Y_t, Z_t\rangle = 0$. Also, from the textbook, it is shown $dZ_t= Z_tX_t \,dW_t$
So from the above we have
\begin{align*}
Y_tZ_t \,dY_t = -Y_t^2 \, dZ_t \ \Longleftrightarrow\ dY_t = -Y_t^2\,dZ_t = -Y_tX_t\,dW_t
\end{align*}
since $dZ_t = Z_tX_t\,dW_t$ and $Z_tY_t=1$. However, it seems the $dt$ term is missing.... I can't really see what is wrong in the calculation though. Does anyone have any comments?
 A: The issue with your solution is that $Y_t Z_t = 1$ does not imply that $\langle Y_t, Z_t \rangle = 0$ for the covariation. See the answer by @ChristopherK for a computation of $\langle Y_t, Z_t \rangle$. Here's a way to solve your original problem.
Let $$R_t = \underbrace{\int_0^t X_u dW_u}_{(\star )} - \underbrace{\frac{1}{2}\int_0^t X_u^2 du}_{(\star \star)}$$ Note that with the assumption that $\int_0^t X_u^2 du < \infty$, the process $R_t$ is well-defined. Moreover, $R_t$ is a semi-martingale, because $(\star)$ is a local martingale and $(\star \star)$ is a process of bounded variation.
Let $g(x) = e^{-x}$ and observe that $Y_t = g(R_t)$. Itô's lemma implies that the class of semi-martingales is stable under the application of $C^2$ maps, which $g$ clearly is. That solves (1).
To solve (2), observe that $g_x(x) = - e^{-x}, g_{xx}(x) = e^x$. Further observe that the dynamics of $R_t$ can be written as $$dR_t = X_t dW_t - \frac{1}{2} X_t^2 dt$$ By Itô's lemma:
$$\begin{align*}
dY_t &= g_x(R_t)(dR_t) + \frac{1}{2} g_{xx}(R_t)(dR_t)^2 \\
&= - \exp (-R_t) \left( X_t dW_t - \frac{1}{2}X_t^2 dt\right) + \frac{1}{2} \exp(-R_t) X_t^2 dt \\
&=\exp (-R_t) X_t^2 dt - \exp (-R_t) X_t dW_t \\
&=Y_t X_t^2 dt - Y_tX_tdW_t
\end{align*}$$
As required.
As an additional technical point, setting $Y_t = \frac{1}{Z_t}$ is a perfectly valid thing to do you because $P \left(\inf_{0 \leq s \leq t} Z_s > 0\right) = 1 $.
A: A note on the calculation, based on the comment: $d[Y,Z]_{t} = -X_{t}^{2}\, dt$.
Letting $$U_{t} = \int_{0}^{t} X_{u}\, dW_{u} - \frac{1}{2}\int_{0}^{t} X_{u}^{2}\, du$$ and applying Itô's lemma for $f(u) = e^{u}$,
\begin{align*}
dZ_{t} = d(f(U_{t})) &= f'(U_{t})\, dU_{t} + \frac{1}{2}f''(U_{t})\, d[U,U]_{t} \\
&= e^{U_{t}}\, (X_{t}\, dW_{t} - \frac{1}{2}X_{t}^{2}\, dt) + \frac{1}{2}e^{U_{t}}X_{t}^{2}d[W,W]_{t} \\
&= Z_{t}X_{t}\, dW_{t},
\end{align*}
as claimed.
Doing same for $Y_{t} = Z_{t}^{-1} = e^{-U_{t}}$ and $g(u) = e^{-u}$,
\begin{align*}
dY_{t} = d(g(U_{t})) &= g'(U_{t})\, dU_{t} + \frac{1}{2}g''(U_{t})\, d[U,U]_{t}  \\
&= -e^{-U_{t}}\, (X_{t}\, dW_{t} - \frac{1}{2}X_{t}^{2}\, dt) + \frac{1}{2}e^{-U_{t}}X_{t}^{2}d[W,W]_{t} \\
&= -Y_{t}X_{t}\, dW_{t} + Y_{t}X_{t}^{2}\, dt.
\end{align*}
Then you can calculate the quadratic variation
\begin{align*}
d[Y,Z]_{t} = -Y_{t}Z_{t}X_{t}^{2} d[W,W]_{t} = -X_{t}^{2}\, dt.
\end{align*}
It follows by the product rule for Itô calculus that
\begin{align*}
d(Y_{t}Z_{t}) &= Y_{t}\, dZ_{t} + Z_{t}\, dY_{t} + d[Y,Z]_{t} \\
&= Y_{t}Z_{t}X_{t}\, dW_{t} + Z_{t}\, (-Y_{t}X_{t}\, dW_{t} + Y_{t}X_{t}^{2}\, dt) - X_{t}^{2}\, dt \\
&= 0, 
\end{align*}
as indeed the derivative of a constant should be.
