We’re rewarding the question askers & reputations are being recalculated! Read more.

# Tag Info

4

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

The linear/non-linear classification is concerned about the dependent variables, and its derivatives. To verify whether the equation is linear, you should be checking that the equation is linear in each of these variables, and the coefficients of these are functions of the independent variables (t and x in your example). In your example, the dependent ...

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 $$x = \ln \left( F_{t, T} / K \right) = \ln \left( S e^{r \tau} / K \right).$$ Following the remark that you cited, we then define $$e^{-r \tau} C(x, \nu, \tau) = V(S, \nu, t).$$ 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. 2 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} 2u(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)\\ &...

Only top voted, non community-wiki answers of a minimum length are eligible