# Girsanov theorem and the density of a process

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?

## 1 Answer

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 $$\nu(dy) = \frac 1 { \sqrt {2 \pi}} \left\{ \int_{\mathbb R} \exp(-(y-\eta)^2/(2 t) ) \gamma(d\eta)\right\} \ d y.$$