4 votes
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Question about Ito formula and BSDE

Note that your BSDE implies the following dynamics for $y$: $$dy_s = g_0(s) ds - z_s dB_s$$ with terminal condition $y_T = \xi$. This implies that the dynamics for $y_s^2$ are given by: $$d(y_s^2) = ...
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3 votes
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Solution of a Stochastic Differential Equation

Writing \begin{align} E_t&:=\mathcal{E}(X)_t\,,\\ F_t&:=y_0 + \int_0^t \mathcal{E}(X_s)^{-1}dY_s - \int_0^t \mathcal{E}(X)_s^{-1}d\langle X,Y\rangle_s \end{align} Applying Ito to $Z_t=E_tF_t$ ...
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3 votes
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Exponential submartingale inequality

Under some mild regularity conditions, the stochastic exponential process $$X_t = \mathcal{E}(\epsilon M)_t = \exp\left( \epsilon M_t - \frac{1}{2} \epsilon^2 \langle M \rangle_t \right)$$ is a non-...
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2 votes
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Constructing the solution to a linear backward SDE

Your application of Itô's lemma is correct; however, you may wish to consider a different process to solve this BSDE. To that end, set $$M_t = X_t Y_t - \int_0^t X_s c_s ds$$ By Itô's lemma, the ...
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2 votes

Deterministic function that gives Snell envelope

Notice that the Snell envelope can be constructed as $U_T=Z_T$ and for $0\leqslant t\leqslant T-1$, $$\tag{*} U_t=\max\left\{Z_t,\mathbb E\left[U_{t+1}\mid\mathcal F_t\right]\right\}, $$ where $\...
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2 votes
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Does a bounded martingale imply a bounded stochastic integral?

I'm not quite sure that $Y_t$ is bounded, but I think you can show it's a true martingale by using the BDG inequalities. Recall one condition that ensures $Y$ is a martingale is $\mathbb{E}[\langle Y,...
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  • 8,467
1 vote

Dynkin's martingale formula

I do not know if I can tag this as an answer but. The first formula is the expectation of the second formula (which really is the original form). By this, I mean Dynkin's martingale formula yields ...
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1 vote
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Using the BDG inequality to show a process is uniformly integrable

Assume that $g=0$. The quadratic variation of $y$ is given by: $$\langle y\rangle_t = \int_t^T z_s^2 ds \leq \int_0^T z_s^2 ds $$ The BDG inequality tells you that, for some $C>0$, we have: $$\...
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1 vote
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Using BDG inequality to show the solution to a BSDE belongs to $S^2_{\mathscr{F}}$

By an application of the BDG inequality or Doob's maximal inequality, one can show that $$\mathbb{E}\left[ \sup_{0 \leq t \leq T} |y_t|^2 \right] < \infty$$ This was shown in your other recent ...
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1 vote

On continuous modifications being indistinguishable for random fields

Right-continuity with regard to every variable $t_1,\ldots,t_d$ is sufficient to conclude. Indeed by your assumption, the event $A : = \bigcap_{t \in \mathbf{Q}^d}[X_t=Y_t]$ is almost sure. Assume ...
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1 vote
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How to show that $(X,B)$ and $(Y,W)$ satisfy the same SDE if their joint law is equal?

Suppose $\Big(X,B,\Omega,\mathcal F, \big(\mathcal F_t\big)_{t\geq0}, \mathbb P\Big) \stackrel{\mathcal{Law}}{=} \Big(Y,W,\Theta,\mathcal G, \big(\mathcal G_t\big)_{t\geq0}, \mathbb Q\Big)$ Since $X$ ...
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  • 3,154
1 vote

Malliavin derivative of stopped Brownian motion

Stopped Brownian motion is not Malliavin differentiable because if it was it would imply that $T$ is constant (see footnote pg.4 Locally Lipschitz BSDE driven by a continuous martingale path-...
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