# 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 $$0, and $$W_t$$ is the standard Brownian motion.

My questions are:

1. How can we prove $$Y_t$$ is a semi-martingale?
2. 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?

• Hint: Apply Ito's formula to $f(Z_t)$ where $f(x) = 1/x$. Aug 2, 2021 at 15:40
• The problem with your solution is that $Y_t Z_t = 1$ does not imply that $\langle Y_t, Z_t \rangle = 0$ Aug 2, 2021 at 17:47

## 2 Answers

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 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.