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6

That should not be true. The problem is that you're not considering the square of the stochastic integral, i.e. the random variable $J= (\int_0^1 {W_s} ds)^2$, which would indeed be the square of a normally distributed random variable, but the r.v. $S= (\int_0^1 {W_s}^2 ds)$. A note in the direction of finding a solution would be a discretization attempt, ...


4

The distribution of $\int_0^t W^2_s\,ds$ was found by Cameron and Martin in the 1940s. Mark Kac (1949, Transactions of the AMS) applied his method to find the Laplace transform of this integral: $$ \Bbb E\left[\exp(-u\int_0^t W^2_s\,ds)\right]={1\over\sqrt{\cosh(t\sqrt{2u})}}. $$


3

$\text{Cov} \, (X_t,X_r)=\text{Cov} \, (\int_0^ts^3W_sdW_s,\int_0^rs^3W_sdW_s)=\int_0^t s^6 W^2_s ds$ This identity does not hold true. Note that the left-hand side is a fixed real number whereas the right-hand side is a random variable. If you apply Itô's isometry correctly, you find $$\text{cov} \, (X_t,X_r) = \color{red}{\mathbb{E} \bigg( }\int_0^t ...


3

Being predictable (as in being measurable with respect to the predictable $\sigma$-algebra) always implies being progressive (as in being measurable with respect to the progressive $\sigma$-algebra), but the other implication only holds in special cases, such as when e.g. the underlying filtration is the usual completion of the filtration generated by a ...


2

I'll rewrite the proof in terms of the process $\tilde{S}_t$, rather than using stochastic integrals. Well, let me first say that the idea of Girsanov's theorem is to 'eliminate' the drift term of $\tilde{S}_t$ by changing the probability measure (to $\mathbb{Q}$). We have, $$d\tilde{S}_t = \tilde{S}_t(\mu-r)dt + \tilde{S}_t \sigma dB_t.$$ Since $\tilde{B}_t ...


2

Let $X_{t}$ and $W_{t}$ be defined as above. Let's apply Ito with $f(x,y)=xy^{2}$ to get the semimartingale decomposition of $f(X_{t},W_{t})=X_{t}W_{t}^{2}$. \begin{align*} X_{t}W_{t}^{2}&=\int_{0}^{t}W_{s}^{2}\text{sgn}(W_{s})dW_{s}+2\int_{0}^{t}X_{s}W_{s}dW_{s}\\ ...


2

Given that $\sigma(s)$ is a deterministic function, then the process $(X(t))_{t \geq 0}$, where $$X(t) = \int_{0}^{t} \sigma(s) dW_s,$$ is a Gaussian process with zero mean and covariance function $\rho(s,t) = \displaystyle \int_{0}^{\min(s,t)} \sigma(u)^2 du$. A proof of this theorem can be found in Schreve's stochastic calculus for Finance II. Hence, ...


2

Check out George Lowther's blog: https://almostsure.wordpress.com/2010/01/03/the-stochastic-integral/ Especially Lemma 3, answers your question "why cadlag?"


1

It's enough that $f$ be Lebesgue measurable and $\int_0^t [f(s)]^2\,ds<\infty$ for each $t$. Indeed, in this case your stochastic integral is well defined. Moreover, if we set $A_t:=\int_0^t[f(s)]^2\,ds$, then $Y$ is a (mean zero) continuous martingale with quadratic variation $\langle Y\rangle_t=A_t$. Let $\tau(t):=\inf\{s:A_s>t\}$. Then ...


1

This might be helpful: Theorem 29 on page 173 of Protter's "Stochastic integration and differential equations": Let $M$ be a local martingale and let $H\in \mathcal P$ be locally bounded. then the stochastic integral $H\cdot M$ is a local martingale. I'm reading that book, but haven't got that far yet, will be interested to see if someone can provide a ...


1

Your calculation is almost good. However, your application of Ito's lemma is not entirely correct. I will outline the details for you: Since $X(t) = \int_{0}^{t} \sigma(s) dW_s$, it follows that $dX(t) = \sigma(t)dW_t$. Define $Z(t) = \exp(iuX(t)) = f(t,X(t))$. Now, an easy calculation shows $$\begin{cases} \displaystyle\frac{\partial f}{\partial t} = 0, \\ ...


1

I would like to add something to the excellent answer of Siron: although it is good to know the Ito Integral of a deterministic function is Gaussian, we can solve my problem without that property. Let $\quad g \colon [0,T] \longmapsto \mathbb{R}$ a deterministic function. Then if we define: $\quad Y_t = \int_0^t g(s) \mathrm{d} W_s \iff \mathrm{d} Y_t = ...


1

To answer your first question, stochastic integrals of this type are Ito integrals. They are defined as a specialized limit of a Riemann-like sum with respect to a partition. Under appropriate conditions on processes $X_s$ and $B_s,$ we can define the integral as a limit of left-hand (non-anticipatory) sums: $$\int_0^t X_s dB_s = \lim_{n \to \infty} ...


1

We have $\mathbb{P}(B_s=0)=0$ for all $s>0$, and therefore it follows directly from Tonelli's theorem that $$\mathbb{E} \left( \int_0^t 1_{\{B_s=0\}} \, ds \right) = \int_0^t \underbrace{\mathbb{E}(1_{\{B_s=0\}})}_{=\mathbb{P}(B_s=0)} \, ds = 0.$$ Since the integrand is non-negative this implies $$\int_0^t 1_{\{B_s=0\}} \, ds = 0 \quad \text{a.s.}$$


1

For Brownian motion it is true that $B=C$. This means that if $f\in B$ there exists $g\in C$ such that $f(\omega,t)=g(\omega,t)$ for $\Bbb R\times\mu$-a.e. $(\omega,t)\in\Omega\times[0,T]$. This does not mean that $t\mapsto f(\cdot,t)$ is a predictable process. The matter is discussed in some detail in Chapter 3 of Introduction to Stochastic Integration by ...


1

Choose $$f(t, B_t) = \frac{B_t^3}{3}$$ so that we have $$ \frac{B_t^3}{3} = \int_0^t B_s^2 dB_s + \int_0^t 0 ds + \frac{1}{2}\int_0^t 2B_s ds $$ $$ \to E[\frac{B_t^3}{3}] = E[\int_0^t B_s^2 dB_s + \int_0^t 0 ds + \frac{1}{2}\int_0^t 2B_s ds] $$ $$ \to E[\frac{B_t^3}{3}] = E[\int_0^t B_s^2 dB_s] + \frac{1}{2}\int_0^t 2E[B_s] ds $$ $$ \to ...


1

Let's show that $\mathbb{E}[\int_0^t (e^{Z_s}(1-X_su_s))^{2}ds] < \infty $ By Cauchy-Schwarz inequality: $\mathbb{E}[\int_0^t (e^{Z_s}(1-X_su_s))^{2}ds] \leq \mathbb{E}[\int_0^t e^{4 Z_s}ds]^{\frac{1}{2}} \mathbb{E}[\int_0^t (1-X_su_s)^{4}ds]^{\frac{1}{2}} $ Because $u$ is bounded, it is clear that $\mathbb{E}[\int_0^t (1-X_su_s)^{4}ds] < ...


1

It depends very much on what you know about stochastic integrals. If you know Novikov's condition, then this might be your first choice. If you know local martingales, then you can use the fact that $(e^{Z_t})_t$ is a local martingale and the boundedness of $u$ to conclude that $e^{Z_t} \in L^2$. In my answer, I'll present a solution which does not require ...


1

Let's try to prove this rigorously. Lemma 1 Let $E$ be a topological space $t\ge 0$ and $t_0^{(n)},\ldots,t_n^{(n)}\ge 0$ with $$0=t_0^{(n)}<\cdots<t_n^{(n)}=t$$ for some $n\in\mathbb N$ $F:[0,t]\times H\to E$ be continuous $(X_t)_{t\ge 0}$ be a left-continuous $H$-valued stochastic process on $(\Omega,\mathcal A,\operatorname P)$ ...


1

There is exactly a definition of the term $\int_0^t\langle\Phi_s{\rm d}W_s,F_x(s,X_s)\rangle$. For $\Phi_s$ taking values in $\mathfrak L_2(Q^\frac{1}{2}U,H)$ and satisfying the condition that the integral of $\Phi_s$-'s square-norm in $\mathfrak L_2(Q^\frac{1}{2}U,H)$ is a.s. finite (just called the "Energe Condition" privately), and for $\Psi_s$ a ...


1

@webbster: I'm going to flip this around and deal instead with right-limited functions. So let $h:[0,1]\to\Bbb R$ have right limits ate each point of $[0,1)$. Extend $h$ to all of $[0,\infty)$ by setting $h(t)=h(1)$ for $t\ge 1$. It suffices to show that for a fixed $\epsilon>0$, the set $B:=\{t\in[0,1): |h(t)-h(t+)|\ge\epsilon\}$ is countable. For ...



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