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4

Feynman–Kac Theorem: Assume that $F$ is a solution to the boundary value problem \begin{align} &F_t+\mu(t,x)F_x+\frac{1}{2}\sigma^2(t,x)F_{xx}-rF=0\\ &F(T,x)=\Phi(x), \end{align} Assume furthermore that the process $e^{-r_s}\sigma(s,X_s)F_s$ is in $\mathcal L^2$ where \begin{align} dX_s=\mu(s,x)ds+\sigma(s,x)dW_s, \end{align} then $F$ has the ...


4

let $Y_t=(X_t)^\alpha$,then $$dY_t=\alpha Y_tdt+dW_t^P$$ we define $Q$ measure by $$\frac{dQ}{dP}=exp\left(-\alpha\int_{0}^{T}Y_t\,dW_t^p-\frac{1}{2}\alpha^2\int_{0}^{T}Y_t^2 dt\right)$$ this shows that $$W_t^Q=W_t^P+\alpha\,\int_{0}^{t}Y_s\,ds$$ is standard wiener process under $Q$ measure, thus we have $$dW_t^P=dW_t^Q-\alpha\,Y_t dt$$ and $$dY_t=\alpha ...


2

Let $\{P_t \mid t \geq 0\}$ be a compound Poisson process, where \begin{align*} P_t = \sum_{i=1}^{N_t} (V_i -1), \end{align*} and $N_t$ is a Poisson process with intensity $\lambda$ and jump times $\tau_i$, $i = 1, \ldots, \infty$. Let $Y_i=\ln V_i$ and $f(x)$ be the density function. Then \begin{align*} P_t - \lambda t E(V_1) &= P_t - \lambda t ...


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Similar question has been discussed previously; see Why does the short rate in the Hull White model follow a normal distribution?. Basically, the probabilistic limit of normal random variables is still normal. Then, as $$\sum_{[t_{i-1},t_{i}]\in\pi_{n}}f(t_{i-1})(W_{t_{i}}-W_{t_{i-1}})$$ is normal, the limit $$\int_{0}^{t}f(\tau)dW_{\tau},$$ in probability, ...



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