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金融代写|随机偏微分方程代写Stochastic Differential Equation代考|MMA630 WEAK CONVERGENCE OF STOCHASTIC INTEGRALS

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金融代写|随机偏微分方程代写Stochastic Differential Equation代考|WEAK CONVERGENCE OF STOCHASTIC INTEGRALS

In order to talk of the convergence of a sequence of stochastic integrals $f \cdot M^{n}$, we must be able to define the integrand $f$ for each of the $M^{n}$. We can do this by defining all of the $\mathrm{M}^{\mathrm{n}}$ on the same probability space. The most convenient spate for this is the skorokhod space $\mathrm{D}^{\text {. }}$ Thus we will define all our martingale measures canonically on $\underline{D}$, so that once we define $\mathrm{f}(\mathrm{x}, \mathrm{t}, \mathrm{w})$ on $\mathbf{R}^{\mathrm{d}} \times \mathbf{R}^{+} \times \underline{D}$, we can define all the $f \cdot M^{n}$
The stochastic integral is not in general a continuous function on $\mathrm{D}$, so that it is not always true that $M^{n} \Rightarrow M$ implies that $f \cdot M^{n} \Rightarrow f \cdot M$, even for classical martingales. Two examples, both of real valued martingales, will illustrate some of the pitfalls.

金融代写|随机偏微分方程代写Stochastic Differential Equation代考|AN APPLICATION

In many applications – the neurophysiological example of Chapter 3, for instance – the driving noise is basically impulsive, of a poisson type, but the impulses are so small and so closely spaced that, after centering, they look very much like a white noise. The following results show that for some purposes at least, one can approximate the impulsive model by a continuous model driven by a white noise. One might think of this as a diffusion approximation.

Let us return to the setting of Chapter 5 . Let $D$ be a bounded domain in $\mathbf{R}^{\mathrm{d}}$ with a smooth boundary, and consider the initial-boundary value problem (5.3) with two changes: we will allow an initial value given by a measure on $R^{d}$, and we will replace the martingale measure $M$ by a Poisson point process $\Pi$.

Let $\Pi^{\mathrm{n}}$ be a sequence of time-homogeneous Poisson point processes on $\mathrm{D}$ with characteristic measures $\mu_{n}$, (Recall that this means that $\Pi^{n}$ is a random $\sigma$-finite signed measure on $D \times[0, \infty)$ which is a sum of point masses. If $A C D$ is Borel and $K \subset R$ is compact with $0 \notin K$, let $N_{t}^{n}(A \times K)$ be the counting process: $N_{t}^{n}(A \times K)$ is the number of points in $A \times[0, t]$ whose masses are in $K$. Then $\left{N_{t}^{n}(A \times K), t \geq 0\right}$ is a
Poisson process with parameter $\mu_{n}(A \times K)$, and if $A_{1} \times K_{1} \cap A_{2} \times K_{2}=\phi$, then $N^{n}\left(A_{1} \times K_{1}\right)$ and $N^{n}\left(A_{2} \times K_{2}\right)$ are independent.) Let
$$m_{n}(A)=\int_{D \times \mathbf{R}} r \mu_{n}(d x d r)$$
and
$$\sigma_{n}(A)=\int_{D \times R} r^{2} \mu_{n}(d x d r)$$
be the mean and intensity measures, respectively, of $\Pi^{\mathrm{n}}$. For $\delta>0$, let
$$Q_{n}(\delta)=\int_{n} x^{2+\delta_{n}}(d x d y) .$$

随机偏微分方程代写

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