4
votes
Accepted
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:
$$d(y_s^2) = ...
- 10.2k
3
votes
Accepted
Solution of a Stochastic Differential Equation
Writing
\begin{align}
E_t&:=\mathcal{E}(X)_t\,,\\
F_t&:=y_0 + \int_0^t \mathcal{E}(X_s)^{-1}dY_s - \int_0^t \mathcal{E}(X)_s^{-1}d\langle X,Y\rangle_s
\end{align}
Applying Ito to $Z_t=E_tF_t$ ...
- 4,304
3
votes
Accepted
Exponential submartingale inequality
Under some mild regularity conditions, the stochastic exponential process $$X_t = \mathcal{E}(\epsilon M)_t = \exp\left( \epsilon M_t - \frac{1}{2} \epsilon^2 \langle M \rangle_t \right)$$
is a non-...
- 10.2k
2
votes
Accepted
Constructing the solution to a linear backward SDE
Your application of Itô's lemma is correct; however, you may wish to consider a different process to solve this BSDE.
To that end, set
$$M_t = X_t Y_t - \int_0^t X_s c_s ds$$
By Itô's lemma, the ...
- 10.2k
2
votes
Deterministic function that gives Snell envelope
Notice that the Snell envelope can be constructed as $U_T=Z_T$ and for $0\leqslant t\leqslant T-1$,
$$\tag{*}
U_t=\max\left\{Z_t,\mathbb E\left[U_{t+1}\mid\mathcal F_t\right]\right\},
$$
where $\...
- 158k
2
votes
Accepted
Does a bounded martingale imply a bounded stochastic integral?
I'm not quite sure that $Y_t$ is bounded, but I think you can show it's a true martingale by using the BDG inequalities.
Recall one condition that ensures $Y$ is a martingale is $\mathbb{E}[\langle Y,...
- 8,467
1
vote
Dynkin's martingale formula
I do not know if I can tag this as an answer but. The first formula is the expectation of the second formula (which really is the original form). By this, I mean Dynkin's martingale formula yields ...
1
vote
Accepted
Using the BDG inequality to show a process is uniformly integrable
Assume that $g=0$. The quadratic variation of $y$ is given by:
$$\langle y\rangle_t = \int_t^T z_s^2 ds \leq \int_0^T z_s^2 ds $$
The BDG inequality tells you that, for some $C>0$, we have:
$$\...
- 10.2k
1
vote
Accepted
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 ...
- 10.2k
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 ...
- 2,202
1
vote
Accepted
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 $X$ ...
- 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-...
- 136
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