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Girsanov Example

Let \(u(x)\colon \mathbf{R}^d \rightarrow \mathbf{R}^d\) such that \(\sup_x |u(x)| \leq K \). Define \(X_t\) by
\[dX_t = u(X_t)dt + \sigma dB_t\]
for \(\sigma >0\) and \(X_0=x\).

For any open set \( A \subset \mathbf{R}^d\) assume that you know that \(P(B_t \in A) >0)\) show that the same holds for \(X_t\).

Hint: Start by showing that \(\mathbf{E}[ f(x+ \sigma B_t)] = \mathbf{E}[\Lambda_t f(X_t)]\) for some process \(\Lambda_t\) and any function \(f\colon \mathbf{R}^d\rightarrow \mathbf{R}\). Next show that \(\mathbf{E}[\Lambda_t^2] < \infty\)

Exit Through Boundary II

Consider the following one dimensional SDE.
\begin{align*}
dX_t&= \mu dt+ \sin( X_t )^\alpha dW(t)\\
X_0&=\frac{\pi}2
\end{align*}
Consider the equation for \(\alpha >0\) and \(\mu \in \mathbf{R}\). On what interval do you expect to find the solution at all times ? Classify the behavior at the boundaries in terms of the parameters.

For what values of \(\alpha < 0\) does it seem reasonable to define the process ? any ? justify your answer.

Shifted Brownian Motion and a PDE

Let \(f \in C_0^2(\mathbf R^n)\) and \(\alpha(x)=(\alpha_1(x),\dots,\alpha_n(x))\) with \(\alpha_i \in C_0^2(\mathbf R^n)\) be given functions and consider the partial differential equations
\begin{align*}
\frac{\partial u}{\partial t} &= \sum_{i=1}^n \alpha_i(x)
\frac{\partial u}{\partial x_i} + \frac{1}{2} \frac{\partial^2
u}{\partial x_i^2} \ \text{ for } t >0 \text{ and }x\in \mathbf R^n \\
u(0,x)&=f(x) \ \text{ for } \ x \in \mathbf R^n
\end{align*}
Use the Girsonov theorem to show that the unique bounded solution \(u(t,x)\) of this equation can be expressed by
\begin{align*}
u(t,x) = \mathbf E_x \left[ \exp\left(\int_0^t \alpha(B(s))\cdot dB(s) –
\frac{1}{2}\int_0^t |\alpha(B(s))|^2 ds \right)f(B(t))\right]
\end{align*}

where \(\mathbf E_x\) is the with respect to \(\mathbf P_x\) when the Brownian Motion starts at \(x\). (Note there maybe a sign error in the above exponential term. Use what ever sign is right.) For the remainder, assume that \(\alpha\)
is a fixed constant \(\alpha_0\). Now using what you know about the distribution of \(B_t\) write the solution to the above equation as an integral kernel integrated against \(f(x)\). (In other words, write \(u(t,x)\) so that your your friends who don’t know any probability might understand it. ie \(u(t,x)=\int K(x,y,t)f(y)dy\) for some \(K(x,y,t)\))

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