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1) Feynmann-Kac and Girsanov First you should remember that the process $X$ is independent of the measure you are considering. Now let's consider a change of measure from ${\mathbb{P}}$ to ${\mathbb{Q}}$. Let us assume $\mathbb{E}_t^{\mathbb{P}}[\tfrac{d{\mathbb{Q}}}{d{\mathbb{P}}}] = e^{\theta W_t^P - \frac{1}{2}\theta^2 t}$ for some constant $\theta$. ...


3

You must separate the mathematical theory from the financial theory. The notion of numéraire specifically pertains to the latter. [A] Mathematical perspective You reach the following PDE (regardless of how you did it) $$ u_t + \mu u_x + \frac{1}{2}\sigma^2 u_{xx} - ru = 0 $$ Feynman-Kac then tells you that the unique solution can be written as $$ u(t,x) = \...


3

As there are many comments but no answer, let me just sum up some of the comments. Solving a PDE or doing some probability? Usually when one wants to solve the SDE they get a distribution for $X$. In the real world though it is often more useful know some of the properties of this distribution, such as its expected value $\mathbb{E}(X)$, or more generally $...


2

First note that you have a typo in the definition of the moneyness. It should be \begin{equation} x = \ln \left( F_{t, T} / K \right) = \ln \left( S e^{r \tau} / K \right). \end{equation} Following the remark that you cited, we then define \begin{equation} e^{-r \tau} C(x, \nu, \tau) = V(S, \nu, t). \end{equation} Note that $C$ is a function of $\tau$ ...


2

self financing portfolios have discounted prices that are martingales. So if the products involves paying fees, these have to taken account of to get a martingale. The product is not a self financing portfolio if these are ignored.


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Let $P(t, r_t, T)$ be the bond price at time $t$, where $0 \leq t \leq T$. Then, by Ito's formula, \begin{align*} &\ P(t, r_t, T) \\ =& P(0, r_0, T) + \int_0^t\partial_s P(s, r_s, T) ds + \int_0^t\partial_r P(s, r_{s-}, T) dr_s + \frac{1}{2}\sigma^2 \int_0^t r_s\partial_{rr} P(s, r_s, T)ds\\ & \quad +\sum_{s \leq t}\big[P(s, r_s, T) - P(s, r_{s-}...


2

a few pointers: did you use appropriate boundary conditions ? how did you truncate the space domain ? in particular it is common to first do a change of variable to get a better behaved $A$ matrix with all terms roughly of the same size. For instance in the "log normal" SABR case ($\beta=1$) it would be appropriate to work in $(x, y) \in ]-\infty, +\infty[...


2

The domain would be twice differentiable in S, once differentiable in t functions of S and t defined on the Cartesian product of S in (0,infty) and t in (0,T) for finite T >0. To obtain a unique solution when boundary conditions are added, one must curtail the growth as S goes to infinity by adapting the corresponding condition for uniqueness of solutions ...


2

Since we have $\frac{\partial\mathbb{E}[u(t,x)]}{\partial t}=0$, the infinitesimal generator is: $Af(x)=\frac{1}{2}x^2\frac{\partial}{\partial x^2}$


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$$u(t,x)=\mathbb{E}\left[h(B_T)+\int_t^Tg(B_s)ds|B_t=x\right]$$ where $B$ is a brownian motion. So if you enter a contract whose underlying asset is $B$, such that you pay every day $t$, $-g(B_t)dt$ up to time $T$ where you receive $h(B_T)$, then the value of this contract is $u$ $$\partial_t u + \frac{1}{2}\partial_{xx}u + g = 0$$ there is $\frac{1}{2}$ ...


2

Assuming that you Have an (or a set of) SDE(s) describing the dynamics of an asset $X$, with $t$-value $X_t$; Define $V$ as a claim contingent on the asset $X$, with $t$-value $V_t$; Define $N$ as a claim that may but need not be contingent on the asset $X$, with $t$-value $N_t$; Define a probability measure $\mathbb{Q}^N$ associated to the asset $N$ such ...


2

Note that $r(t)=e^{\ln r(t)}$. Then \begin{align*} dr(t) &= e^{\ln r(t)} d\ln r(t) + \frac{1}{2}e^{\ln r(t)}\langle d\ln r(t), d\ln r(t)\rangle\\ &=r(t) \big[\alpha(t)(\theta (t)-\log r(t))dt+\sigma dW(t) \big] + \frac{1}{2}\sigma^2 r(t) dt\\ &=r(t)\Big[\Big(\frac{1}{2}\sigma^2 + \alpha(t)(\theta (t)-\log r(t))\Big)dt + \sigma dW(t)\Big]. \end{...


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)\\ &...


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