# Tag Info

3

Applying Itô's formula for the Itô process $$X_t := \int_0^t f(s) \, dW_s$$ and $f(x) := x^4$ gives $$X_t^4 = 4 \int_0^t X_s f(s) \, dW_s + \frac{4 \cdot 3}{2} \int_0^t X_s^2 f(s)^2 \, ds.$$ Taking expecation yields \begin{align*} \mathbb{E}(X_t^4) &= 6 \int_0^t \mathbb{E}(X_s^2 f(s)^2) \, ds \\ &\leq 6 \mathbb{E} \left(\sup_{s \leq t} |X_s|^2 ... 2 Here's a different argument. DefineX_t = \int_0^t f(s,\omega) dW_s \\ Y_t = X_t^4 \\ Z_t = X_t^2.$$Apply the Ito formula to Y_t and Z_t:$$Y_t = 4 \int_0^t f(s,\omega) X_s^3 dW_s + 6 \int_0^t f(s,\omega)^2 X_s^2 ds \\ Z_t = 2 \int_0^t f(s,\omega) X_s dW_s + \int_0^t f(s,\omega)^2 ds$$Now substitute and take expectations, thereby canceling the ... 2 Basically, you can simply apply Itô's formula:$$\begin{align*} f(W_t^1, W_t^2)-f(0,0) &= \int_0^t f_x(W_s^1,W_s^2) \, dW_s^1 + \int_0^t f_y(W_s^1,W_s^2) \, dW_s^1+ \\ &\quad \frac{1}{2} \bigg( \int_0^t f_{xx} (W_s^1,W_s^2) \, \underbrace{dW_s^1 dW_s^1}_{ds} + 2 \int_0^t f_{xy}(W_s^1,W_s^2) \, \underbrace{dW_s^1 dW_s^2}_{\varrho \, ds} \\ &\quad ...

1

For #1: Let's call your isomorphism $T$, the isometry from $\mathcal{H}_K$ to $L^2(P)$ which maps $K_t$ to $B_t$. Yes, $T$ really is a stochastic integral: $Tf$ is the Itô integral, not of $f$, but of $f'$. So $$Tf = \int_0^1 f'(t)\,dB_t$$ or in other words, for deterministic $g \in L^2([0,1])$, $$\int_0^1 g(t)\,dB_t = T\left(\int_0^\cdot ... 1 I like "Brownian Motion Calculus" by Ubbo F. Wiersema (2008). I found it more approachable than other books I've seen. But it depends on your level of mathematical sophistication (I'm a mathematical hick). 1 Let H:x\mapsto F(\min(x,\tau)), then \mathrm dH(x)=\mathbf 1_{x\leqslant\tau}\,\mathrm dF(x) hence (3) reads$$g(t)=h(t)+\int_0^\infty g(t-x)\,\mathrm dH(x).$$It follows that$$\tilde{g}(s)=\frac{\tilde{h}(s)}{1-\tilde{H}(s)},$$where$$\tilde{H}(s)=\int_0^\infty\mathrm e^{-sx}\,\mathrm dH(x)=\int_0^\tau\mathrm e^{-sx}\,\mathrm dF(x).$$1 When you are not sure, there is not harm in treating entries as separate variables to apply Ito's formula. The only thing you have to take care of in that case are quadratic covariations. 1 Two ways to fix this: Any martingale with respect to a nice filtration (i.e. a filtration satisfying the usual conditions) has a càdlàg modification. This means that we can assume without loss of generality that the process has càdlàg sample paths. In the case of Brownian motion, the completion of the canonical filtration satisfies indeed the usual ... 1 The answer is certainly: yes. By definition of a martingale, \mathsf E M_t = \mathsf E M_0 (just using the tower property) and since M_0 = 0 in your case, all follows. Note that Ito integral may not be a martingale unless some integrability conditions are satisfied, in general it may be only a local martingale - so you need to see which integrability ... 1 It follows from Itô's formula that the solution to the SDE$$dM_t = M_t \sigma_t dW_t$$equals$$M_t = M_0 \exp \left( \int_0^t \sigma_s \, dW_s - \frac{1}{2} \int_0^t \sigma_s^2 \, ds \right).$$By assumption, M_0 \geq 0. Hence,$$\begin{align*} \sqrt{M_T} &= \sqrt{M_0} \exp \left( \frac{1}{2} \int_0^T \sigma_s \, dW_s - \frac{1}{4} \int_0^T ...

1

Suppose that $(X_t,Y_t)_t$ is a solution to the given system. Then in particular, we can write $$X_t-X_0 = - \underbrace{\int_0^t Y_s \, dB_s}_{\text{local martingale}} - \frac{1}{2} \underbrace{\int_0^t X_s \, ds}_{\text{bounded variation, continuous}}.$$ Applying the optional stopping theorem yields that (t,\omega) \mapsto \left( \int_0^{t \wedge ...

Only top voted, non community-wiki answers of a minimum length are eligible