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# 数学代写|随机微积分代写STOCHASTIC CALCULUS代考|GRA6550 Quadratic Variation of a Square Integrable Martingale

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## 数学代写|随机微积分代写STOCHASTIC CALCULUS代考|Quadratic Variation of a Square Integrable Martingale

The next lemma connects the quadratic variation map $\Psi$ and r.c.l.l. martingales.
Lemma 5.16 Let $\left(N_t, \mathcal{F}t\right)$ be an r.c.l.l. martingale such that $\mathrm{E}\left(N_t^2\right)<\infty$ for all $t>0$. Suppose there is a constant $C<\infty$ such that with $$\tau=\inf \left{t>0:\left|N_t\right| \geq C \text { or }\left|N{t-}\right| \geq C\right}$$
one has
$$N_t=N_{t \wedge \tau} .$$
Let
$$A_t(\omega)=\Psi(N .(\omega))(t) .$$
Then $\left(A_t\right)$ is an $\left(\mathcal{F}t\right)$ adapted r.c.l.l. increasing process such that $X_t:=N_t^2-A_t$ is also a martingale. Proof Let $\Psi_n(\gamma)$ and $t_i^n(\gamma)$ be as in the previous section. \begin{aligned} A_t^n(\omega) & =\Psi_n(N .(\omega))(t) \ \sigma_i^n(\omega) & =t_i^n(N .(\omega)) \ Y_t^n(\omega) & =N_t^2(\omega)-N_0^2(\omega)-A_t^n(\omega) \end{aligned} It is easy to see that for each $n,\left{\sigma_i^n: i \geq 1\right}$ are stopping times (see Theorem 2.46) and that $$A_t^n=\sum{i=0}^{\infty}\left(N_{\sigma_{i+1}^n \wedge t}-N_{\sigma_i^n \wedge t}\right)^2 .$$

## 数学代写|随机微积分代写STOCHASTIC CALCULUS代考|Square Integrable Martingales Are Stochastic Integrators

The main aim of this section is to show that square integrable martingales are stochastic integrators.

The treatment is essentially classical, as in Kunita-Watanabe [46], but with an exception. The role of $\langle M, M\rangle$-the predictable quadratic variation in the KunitaWatanabe treatment-is here played by the quadratic variation $[M, M]$.

Recall that $\mathbb{M}^2$ denotes the class of r.c.l.l. martingales $M$ such that $\mathrm{E}\left[M_t^2\right]<\infty$ for all $t<\infty$ with $M_0=0$.

Lemma 5.27 Let $M, N \in \mathbb{M}^2$ and $f, g \in \mathbb{S}$. Let $X=J_M(f)$ and $Y=J_N(g)$. Let $Z_t=X_t Y_t-\int_0^t f_s g_s d[M, N]s^\psi$. Then $X, Y, Z$ are martingales. Proof The proof is almost the same as proof of Lemma 3.10, and it uses $M_t N_t-$ $[M, N]_t^\psi$ is a martingale along with Theorem $2.59$, Corollary $2.60$ and Theorem $2.61$. Corollary 5.28 Let $M \in \mathbb{M}^2$ and $f \in \mathbb{S}$. Then $Y_t=\int_0^t f d M$ and $Z_t=\left(Y_t\right)^2-$ $\int_0^t f_s^2 d[M, M]_s^\psi$ are martingales and $$\mathrm{E}\left[\sup {0 \leq t \leq T}\left|\int_0^t f d M\right|^2\right] \leq 4 \mathrm{E}\left[\int_0^T f_s^2 d[M, M]_s^\nu\right] .$$
Proof Lemma $5.27$ gives $Y, Z$ are martingales. The estimate (5.4.1) now follows from Doob’s inequality.

Theorem 5.29 Let $M \in \mathbb{M}^2$. Then $M$ is a stochastic integrator. Further, for $f \in$ $\mathbb{B}(\widetilde{\Omega}, \mathcal{P})$, the processes $Y_t=\int_0^t f d M$ and $Z_t=Y_t^2-\int_0^t f_s^2 d[M, M]s^*$ are martingales, $[Y, Y]_t^\psi=\int_0^t f_s^2 d[M, M]_s^\psi$ and $$\mathrm{E}\left[\sup {0 \leq t \leq T}\left|\int_0^t f d M\right|^2\right] \leq 4 \mathrm{E}\left[\int_0^T f_s^2 d[M, M]s^\psi\right], \quad \forall T<\infty .$$ Proof Fix $T<\infty$. Suffices to prove the result for the case when $M_t=M{t \wedge T}$. The rest follows by localization. See Theorem $4.49$. Recall that $\widetilde{\Omega}=[0, \infty) \times \Omega$ and $\mathcal{P}$ is the predictable $\sigma$-field on $\widetilde{\Omega}$. Let $\mu$ be the measure on $(\widetilde{\Omega}, \mathcal{P})$ defined for $A \in \mathcal{P}$
$$\mu(A)=\int\left[\int_0^T 1_A(\omega, s) d[M, M]s^\psi(\omega)\right] d \mathrm{P}(\omega) .$$ Note that $$\mu(\widetilde{\Omega})=\mathrm{E}\left[[M, M]_T^{\nsim}\right]=\mathrm{E}\left[\left|M_T\right|^2\right]<\infty$$ and for $f \in \mathbb{B}(\tilde{\Omega}, \mathcal{P})$ the norm on $\mathbb{L}^2(\tilde{\Omega}, \mathcal{P}, \mu)$ is given by $$|f|{2, \mu}=\sqrt{\mathrm{E}\left[\int_0^T f_s^2 d[M, M]_s^\psi\right]}$$

## 数学代写|随机微积分代写STOCHASTIC CALCULUS代考|Quadratic Variation of a Square Integrable Martingale

\left 缺少或无法识别的分隔符

$$N_t=N_{t \wedge \tau} .$$

$$A_t(\omega)=\Psi(N .(\omega))(t) .$$

$$A_t^n(\omega)=\Psi_n(N .(\omega))(t) \sigma_i^n(\omega) \quad=t_i^n(N .(\omega)) Y_t^n(\omega)=N_t^2(\omega)-N_0^2(\omega)-A_t^n(\omega)$$

$$A_t^n=\sum i=0^{\infty}\left(N_{\sigma_{i+1}^n \wedge t}-N_{\sigma_2^n \wedge t}\right)^2 .$$

## 数学代写|随机微积分代写STOCHASTIC CALCULUS代考|Square Integrable Martingales Are Stochastic Integrators

$$\mu(A)=\int\left[\int_0^T 1_A(\omega, s) d[M, M] s^\psi(\omega)\right] d \mathrm{P}(\omega) .$$

$$\mu(\widetilde{\Omega})=\mathrm{E}\left[[M, M]_T^{\infty}\right]=\mathrm{E}\left[\left|M_T\right|^2\right]<\infty$$

$$|f| 2, \mu=\sqrt{\mathrm{E}\left[\int_0^T f_s^2 d[M, M]_s^\psi\right]}$$

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