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2

I did this quickly, so hopefully I didn't mess up. $$\mathrm{d}\left(B_1^2 B_2^2 \right) = 2 B_1 B_2^2 \mathrm{d}B_1 + 2 B_1^2 B_2 \mathrm{d}B_2 + B_2^2 \mathrm{d}t + B_1^2 \mathrm{d}t$$ Since if you integrate both sides the terms with $\mathrm{d}B$ factors will be martingales what you want is $$ A_2(t) = \int_0^t \left(B_2^2(s) + ...


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I think you are on the right way. Consider $ \ \varphi _{\lambda} (s) = \lambda e ^{-\lambda(t-s)}$ , $s \in [0;t]$. Then $||\varphi _{\lambda}||_{L_1} = 1 - e^{-\lambda t}$, so $||\varphi _{\lambda}||_{L_1} \to 1$ as $\lambda \to \infty$, and at the same time for any $\varepsilon \in (0;t)$ $$ \int _0 ^{t-\varepsilon} \varphi _{\lambda} (s) ds \to 0,\ ...


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This proof is the proverbial "long and winding road". It can be made somewhat less tedious by the use of Dynkin's multiplicative system theorem, which I state and discuss in this answer. Note that the random variables $W_t : W \to \mathbb{R}$ are just the evaluation functionals on $W$: $W_t(\omega) = \omega(t)$, which are continuous with respect to the ...


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By Kolmogorov's 0-1 law, any tail measurable random variable ($\lim S_n/n$, $\limsup S_n/n$, and $\liminf S_n/n$ are all tail-measurable) is almost surely deterministic. See page 46 of Williams' Probability with Martingales if you have it. This immediately answers both parts of your question.


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Hint It follows from Itô's formula that $$\exp(\theta t) \cdot B_t = \int_0^t \exp(\theta s) \, dB_s + \theta \int_0^t \exp(\theta s) B_s \, ds,$$ i.e. $$ \int_0^t \exp(\theta s) \, dB_s = \exp(\theta t) \cdot B_t - \theta \int_0^t \exp(\theta s) B_s \, ds.$$


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Actually, there is nothing left to do. From $$dZ_t = b \exp \left( \frac{c^2}{2} t - c W_t \right) \, dt$$ it follows that $$Z_t = \underbrace{Z_0}_{0} + b \int_0^t \exp \left( \frac{c^2}{2} s - c W_s \right) \, ds.$$ Hence, $$X_t = Y_t \cdot Z_t = b \exp \left(- \frac{c^2}{2} t+c W_t \right) \cdot \int_0^t \exp \left( \frac{c^2}{2} s - c W_s \right) ...


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Suppose that $$dX_t = - \mu X_t \, dt + \sigma \, dW_t \tag{1}$$ and set $Y_t := e^{\mu t} X_t$. Itô's formula states that $$df(t,X_t)= \frac{\partial}{\partial x} f(t,X_t) \, dX_t + \frac{\partial}{\partial t} f(t,X_t) \, dt + \frac{1}{2} \frac{\partial^2}{\partial x^2} f(t,X_t) \, d\langle X \rangle_t. \tag{2}$$ Here, we have $f(t,x) = e^{\mu t} \cdot ...


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Recall that an SDE of the form $$dX_t = b(X_t) \, dt + \sigma(X_t) \, dW_t$$ has a unique solution whenenver the coefficients $b$, $\sigma$ are globally Lipschitz continuous. In particular, we see that the SDE $$dX_t = (1+X_t) \, dt+ X_t \, dW_t$$ has a unique solution. Now let $(Y_t)_t$ the (unique) solution of the SDE $$dY_t = Y_t \, dt + Y_t \, dW_t ...


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So since I was missing something and didn't get any answer, I tried to start all over from Nualart's book. The point that I missed was the invariance of the product Wiener measure to rotations. Now with this fact we have that $\mathbb{E}[\langle F,P_t y\rangle]=\mathbb{E}[\langle P_tF,u\rangle]$ \begin{align*} ...


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For $U \subseteq \mathbb{R}$, $0 \notin \bar{U}$, we set $$X^U_t := \sum_{s \leq t} \Delta X_s \cdot 1_U(\Delta X_s).$$ Then $(X_t^U)_{t \geq 0}$ is a Lévy process and, if $\int_U y \, d\nu(y)<\infty$, then $$\mathbb{E}X_t^U = t \cdot \int_U y \, d\nu(y).$$ In particular, for $U(\varepsilon) := (\varepsilon,1)$ this implies that $$X_t^{U(\varepsilon)} ...


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First question: Protter says $Z^\epsilon$ converges uniformly to $Z$ as $\epsilon\to 0$. In what sense does he mean uniformly and how is it established? $$Z^\epsilon=\sum_n Z_{T^\epsilon_n}\mathbf1_{[T^\epsilon_n,T^\epsilon_{n+1})} \\ |Z^\epsilon - Z| =\sum_n |Z_{T^\epsilon_n}-Z| \mathbf1_{[T^\epsilon_n,T^\epsilon_{n+1})}$$ now using the definition of ...


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Since the function $f$ depends on the time $t$, we have to apply the time-dependent Itô formula which states that $$f(t,B_t)-f(0,B_0) = \int_0^t \frac{\partial}{\partial x} f(s,B_s) \, dB_s + \int_0^t \left( \frac{\partial}{\partial t}f(s,B_s) + \frac{1}{2} \frac{\partial^2}{\partial x^2} f(s,B_s) \right) \, ds.$$ For $f(t,x) := t \cdot x$ we see that $$t ...


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This should be a comment, but it's too long. I'm not sure I understand what you are defining. Typically a stochastic process is something like a jointly measurable function $X : \Omega \times (a,b) \to \mathbb{R}$, where $(\Omega, \mathcal{F}, P)$ is a probability space and $(a,b)$ is some interval (which could be replaced by a closed interval, an ...



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