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4

We consider the case where the Novikov condition is satisfied, that is, \begin{align*} E\left[\exp\left(\frac{1}{2}\int_0^T \theta^2_s ds \right)\right] < \infty. \end{align*} Then $\{L_t \mid t \ge 0\}$ is a $(\mathscr{F}_t, \mathbb{P})$-martingale. On $\mathscr{F}_T$, we define the probability measure $Q$ by \begin{align*} ...

3

I'd like to give an alternative derivation not involving the clever (mystifying?) transformation to the heat equation and thus present a more general technique for solving constant coefficeint advection-diffusion PDEs. All we need is the Fourier transform: \begin{align*} \mathcal{F}[f] & = \int_{-\infty}^\infty e^{-i \omega y} f(y) dy, \end{align*} ...

1

By Bayes' rule for conditional expectation (or here), $$E_{\mathbb Q}[X_t | \mathscr F_u] E[L_T| \mathscr F_u] = E[X_tL_T| \mathscr F_u]$$ $$\to E_{\mathbb Q}[X_t | \mathscr F_u] L_u = E[X_tL_T| \mathscr F_u]$$ $$\to E_{\mathbb Q}[X_t | \mathscr F_u] = E[\frac{X_tL_t}{L_u}| \mathscr F_u]$$ $$= \frac{1}{L_u} E[ \frac{X_tL_t}{1} | \mathscr F_u]$$ = ...

1

Based on the form of your equation, we can consider the SDE \begin{align*} dX_t = \sigma dW_t, \end{align*} where $W$ is a standard Brownian motion. Since, for $0 \leq t \leq T$, \begin{align*} X_T = X_t + \sigma (W_T-W_t), \end{align*} based on Feynman–Kac formula, the solution is given by \begin{align*} F(t, x) &= E\left(X_T^2 \mid X_t = x\right)\\ ...

1

As you have guessed correctly, these type of questions can be answered using Ito's Lemma.We have: $$d(M_t)= d(Z_t e^{\int_0^tF(Z_u)du})=d(Z_t) e^{\int_0^tF(Z_u)du}+Z_t d(e^{\int_0^tF(Z_u)du})+d(Z_t)d(e^{\int_0^tF(Z_u)du})$$ For the first two terms on R.H.S, we have: d(Z_t) e^{\int_0^tF(Z_u)du} = (f(W_t)dW_t + ...

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