1
$\begingroup$

I am coming across a paper ( Proposition $1.1$ from http://www.sciencedirect.com/science/article/pii/0304414987901840 ) that claims the following fact which I don't understand why:

On a probability space $(\Omega, \mathcal{F} , \mathbb{P})$ with a Brownian motion $(B_t)_{t \in [0,T]}$, let $(C_t)_{t \in [0,T]}$ be a bounded process and $(Y_t)_{t \in [0,T]}$ be a process defined by $$ Y_t = Y_0 + B_t + \int_0^t C_s \,ds,\quad 0\leq t \leq T,$$ where $Y_0$ is a random variable independent of $(B_t)_{t \in [0,T]}$. Then it claims that $(Y_t)$ has a density function for every $t \in [0,T]$, by the Girsanov's theorem.

Let $\mathbb{Q}$ be an equivalent measure to $\mathbb{P}$ defined by $$ \frac{d\mathbb{Q}}{d\mathbb{P}} = \exp \bigg\{ -\int_0^T C_s \, dB_s - \frac{1}{2} \int_0^T C^2_s \,ds \bigg\}. $$ Then, by the Girsanov's theorem, $\{Y_t - Y_0 \}_{t \in [0,T]}$ is a $\mathbb{Q}$- Brownian motion. This means that $$ \mathbb{P} (Y_t \leq y) = \mathbb{E}^{\mathbb{Q}} \bigg[ \frac{d\mathbb{P}}{d\mathbb{Q}} \mathbf{1}_{ \{ Y_t \leq y \} } \bigg] = \mathbb{E}^{\mathbb{Q}} \bigg[ \exp \bigg\{ \int_0^T C_s \, dB_s + \frac{1}{2} \int_0^T C^2_s \,ds \bigg\} \mathbf{1}_{(- \infty ,y]} (Y_t) \bigg]. $$ How can we proceed from here? I don't know the distribution of $\exp \bigg\{ \int_0^T C_s \, dB_s + \frac{1}{2} \int_0^T C^2_s \,ds \bigg\}$, nor do I know anything about the distribution of $Y_0$. I am very confused. Any ideas?

$\endgroup$
2
$\begingroup$

Short answer: it follows directly from the Radon-Nikodym theorem.

Longer answer: Let $\mu$ denote the law of $Y(t)$ under $\mathbb P_{y_0}$ and let $\nu$ denote the law of $Y(t)$ under $\mathbb Q_{y_0}$ for some initial condition $y_0$. From your application of Girsanov theorem, $\nu(A) = 0$ implies $\mu(A) = 0$. Therefore by the Radon-Nikodym theorem, $\mu$ admits a density $\rho(y)$ with respect to $\nu$. Of course, $\nu$ is just the Gaussian distribution centered at $y_0$ with variance $t$, with density function $f(y)$ with respect to Lebesgue measure say. Now $\rho(y) f(y)$ is the density of $\nu$ with respect to Lebesgue measure. If $y_0$ is not fixed but has a distribution $\gamma$, an analogous argument holds, basically replacing $y_0$ above by $\gamma$, and noting that $$ \nu(dy) = \frac 1 { \sqrt {2 \pi}} \left\{ \int_{\mathbb R} \exp(-(y-\eta)^2/(2 t) ) \gamma(d\eta)\right\} \ d y.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.