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 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.
