Exponential Martingales

Question

This is a two-part question concerning exponential martingales.

It is stated that an application of Ito's lemma to \begin{align} \rho_t = \exp\left[-\int_{0}^{t} \lambda_s\,dW_s - \frac{1}{2}\int_{0}^{t} \lambda_s^2\,ds\right] \end{align} gives \begin{align} d\rho_t = -\rho_t\lambda_t\,dW_t \end{align}

I have tried to obtain this by treating $\lambda$ as a constant, hence \begin{align} \rho_t &= \exp\left[-\int_{0}^{t} \lambda_s\,dW_s - \frac{1}{2}\int_{0}^{t} \lambda_s^2\,ds\right]\\ &= \exp\left[ -\lambda W_t - \frac{1}{2} \lambda_t^2 t \right] \end{align} thus, \begin{align} d\rho_t &= \frac{\partial\rho_t}{\partial t} dt + \frac{\partial\rho_t}{\partial W_t} dW_t + \frac{1}{2}\frac{\partial^2\rho_t}{\partial W_t^2} (dW_t)^2\\ &= -\frac{1}{2}\lambda^2\exp(\cdots)\,dt -\lambda\exp(\cdots)\,dW_t + \frac{1}{2}\lambda^2\exp(\cdots)\,dt\\ &= \lambda\rho_t\,dW_t \end{align}

$\textbf{Question 1}:$ I can't help but feel treating $\lambda$ as a constant isn't the correct way to do this. Can anyone confirm or explain how to do it treating $\lambda$ as a function of $t$?

With regards to the second question I am reading some lecture notes which state the following. Let, \begin{align} X(t) = -\int_{0}^{t} \lambda_s\,dW_s \end{align} and define $f(t,x) = \exp x$, then since $X$ is an Ito process, \begin{align} f(t,X(t)) = 1 - \int_{0}^{t} f(s,X(s))\,\lambda(s)\,dW(s) \end{align} or in differential form, \begin{align} df(t,X(t)) = -f(t,X(t))\,dX(t) \end{align}

$\textbf{Question 2}:$ In light of question 1 I find it hard to agree the result stated. e.g. if I was to let \begin{align} f(t,x) = \exp\left[-\int_{0}^{t} \lambda_s\,dW_s \right] \end{align} then treating $\lambda$ as constant so $f(t,x) = \exp\left[-\lambda W_t\right]$ and applying Ito I obtain, \begin{align} df(t,x) &= 0\,dt -\lambda\exp(\cdots)dW_t + \frac{1}{2}\lambda^{2}\exp(\cdots)\,dt\\ &= f(t,x)\,dX + \tfrac{1}{2}\lambda^{2}\,f(t,x)\,dt \end{align} where I have an additional term and I seem to be missing a minus sign (assuming $dX(t) = - \lambda_t dW_t$).

$\textbf{Question 2}:$ I keep thinking to myself the original $X(t)$ should have been defined like $\rho_t$ in question 1 (i.e. two terms), then it would make sense. Would anyone be able to confirm if I'm making serious error or whether the original definition of $X(t)$ appears incorrect?

All help is appreciated.

Many thanks,

John

Answer 1

What follows assumes that processes involved are well-defined for our immediate purposes so that, for instance, $(W_s)_{s\geqslant 0}$ is a regular Brownian-motion adapted to some underlying filtration, $(\lambda_s)_{s\geqslant 0}$ is previsible and adapted to the same filtration, and $\int_{0}^{t}\lambda_s^2\text ds<\infty$. We move by a sequence of steps:

Step 1: Identify a relevant $C^{2}$ real-valued function defined on $\mathbb R$: $$ f(x)=e^{-x} $$

Step 2: Identify the underlying Itô Process: $$ \text dX_t=\lambda_t\text dW_t+\frac{1}{2}\lambda_t^2\text dt\tag{1} $$

Notice, this follows, by definition, from the stochastic process $X_t=\int_0^t\lambda_s\text dW_s+\frac{1}{2}\int_0^t\lambda_s^2\text ds$ (where $X_0=0$). Also, from (1), we deduce that $\text d\langle X \rangle_t=\lambda_t^2\text dt$.

Step 3: Apply Itô's lemma to $f(X_t)$: $$\text df(X_t) = \frac{\partial f}{\partial x}\Big\vert_{X_t}\text dX_t+\frac{1}{2}\frac{\partial^2 f}{\partial x^2}\Big\vert_{X_t}\text d\langle X\rangle_t = -e^{-X_t}\text dX_t+\frac{1}{2}e^{-X_t}\text d\langle X\rangle_t =-f(X_t)\lambda_t\text dW_t\,, $$

by using (1). That is,

$$\text de^{-\int_0^t\lambda_s\text dW_s-\frac{1}{2}\int_0^t\lambda_s^2\text ds} = -e^{-\int_0^t\lambda_s\text dW_s-\frac{1}{2}\int_0^t\lambda_s^2\text ds}\lambda_t\text dW_t$$

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Remarks: Had we chosen the alternative Itô process $X_t=\int_0^t\lambda_s\text dW_s$ instead (or, in sde form, $\text dX_t=\lambda_t\text dW_t$), while keeping the same function $f$, you can check that we would obtain the different (but still related) process sde

$$\text de^{-\int_0^t\lambda_s\text dW_s} = -e^{-\int_0^t\lambda_s\text dW_s}(\lambda_t\text dW_t-\frac{1}{2}\lambda^2_t\text dt).$$

The difference between the two cases is that the choice of $X_t$ as in (1) ensures that $f(X_t)$ is a martingale, while the other choice of $X_t$ does not.