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## New answers tagged stochastic-integrals

0

Question 1: Yes, that's it - because the Brownian motion $(W_t)_{t \geq 0}$ is a martingale. Question 2: Fix $i<k$. By the definition of $I_t^n$ we have \begin{align*} I_t^n &= \sum_{j=0}^{i-1} \Delta_{t_j}(W_{t_{j+1}}-W_{t_j}) + \sum_{j=i}^{k-1} \Delta_{t_j}(W_{t_{j+1}}-W_{t_j}) + \Delta_{t_k}(W_t-W_{t_k}) \\ &=: J_1+J_2+J_3 \end{align*} ...

-1

The general idea is the following : suppose that the ODE $$\frac 12 f''(x) + {\rm sgn}(x) f'(x) = 1$$ has a solution everywhere except in 0, with $f'(X_t)\in L^2(dt)$. Then: $$df(X_t) = dt + dW_t f'(X_t) \\ E\tau = E[f(X_\tau) - f(X_0)]$$because the stochastic integral is in this case a martingale. Now, it remains to find the law of $X_\tau$.

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For 1), we have $$E[\int_0^{t(\omega)^*\wedge T}c dW_s]=cE[W_{t(\omega)^*\wedge T}]$$. $t(\omega)^*\wedge T$ is a stopping time for which is bounded by $T$. Hence by the optinoal stopping theorem $$E[W_{t(\omega)^*\wedge T}]=E[W_0]=0$$ For 2). This depends on the exact definition of your $t(\omega)^*$. For a general stopping time this can not be ...

2

An obvious lower bound is $$\int_0^{t^{\ast}(\omega) \wedge T} f(t,\omega) \, dt \geq \inf_{t \in [0,t^{\ast}(\omega) \wedge T]} f(t,\omega) \cdot (t^{\ast}(\omega) \wedge T).$$ Without further assumptions on the stopping time $t^{\ast}$ or $f$, it will be difficult to get a better lower bound. Since the considered integral is defined pathwise you are ...

0

I would only change a little detail at the end: \begin{eqnarray*} \mathbb{E}\{\int_U^T g(s)dB(s)\mid \mathcal{F}_U\} &=& \sum_k\mathbb{E}[g_k\Delta B_k\mid\mathcal{F}_U] \\ &=&\Sigma_k\mathbb{E}[g_k\Delta B_k]\\ &=&\Sigma_k \mathbb{E}[\mathbb{E}[g_k\Delta B_k\mid \mathcal{F}_k]]\\ &=&\sum_k\mathbb{E}[g_k \mathbb E[\Delta ...

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Maybe not as clean as you would hope for but perhaps you could do the following: $\mathbb{E}_{W_T}\left[ \frac{\int_0^T e^{\alpha W_t} dt}{\int_0^T e^{-\alpha W_t} dt + \int_0^T e^{\alpha W_t} dt} \right] = \mathbb{E}_{W_T}\left[ \frac{1}{\frac{\int_0^T e^{-\alpha W_t} dt}{\int_0^T e^{\alpha W_t} dt} + 1} \right]$ Then at least formally $... 0 Both are L2 and probability limits of Riemann sums (as the size of the subdivision goes to 0): 1-$\sum X_{t_{i-1}} (S_{t_i} - S_{t_{i-1}}) $for the Ito integral. Imagine one controls the$X$process who does not know the future of the process$S$. This is typically the case for mathematical finance, where$S$is the value of an asset and$X$is the number ... 1 I am not so sure about your result I would have said that it converges to 0 in probability$\forall t>0$. First observe that$\frac{W_t^2}{2\epsilon}\chi_{(-\epsilon,\epsilon)}(W_t)\le \frac{\epsilon^2}{2\epsilon}=\frac{\epsilon}{2}$so the first term goes to 0 almost surely. As for the second term using Itô's isometry we see that : $$E[(\int_0^t ... 1 If M is a martingale and X is properly integrable, then$$I_t := \int_s^t X(r) \, dM_r, \qquad t \geq s$$defines a martingale and, consequently,$$\mathbb{E} \left( \int_s^t X(r) \, dM_r \mid \mathcal{F}_s \right) = \mathbb{E}(I_t \mid \mathcal{F}_s) = I_s =0$$for any t \geq s. Now, if M is a local martingale and X locally integrable, then ... 0 Well the first moment is relatively straightforward. If the integral exists than E[\int_a^b X(t) dh(t)] = \int_a^b E[X(t)]dh(t) See for example Grigoriu 3.69. Also by for example Grigoriu 3.58 E[ \int_a^b X(t) dt \int_a^b Z(t) dt] = \int_{[a,b]^2}E[X(u)Z(v)]du dv 1 I think you are right. The stochastic Itô integral is a Gaussian random variable with mean zero, since the integrand is an L_a^2([0,T]\times \Omega) process (in fact, deterministic) The mean is 0, (it is a martingale) and the variance (second moment), by Itô's isometry, is$$\sigma^2 e^{-2\kappa t} E\left[\int_0^t e^{-2\kappa s}ds\right] = ... 1 An alternate way of showing this result is to use the (stochastic)-integration by part formula :$\int_0^tW_sds= W_t-W_0 - \int_0^tsdWs\$ ( no quadratic covariation term has s has 0 quadratic variation) The first term is normal and independent from the stochastic integral which is a Wiener integral. As Wiener integrals are normal random variables, we have ...

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