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## New answers tagged stochastic-integrals

1

Since $\cosh(x) \ge 1$ for all real $x$ you have $\frac{e^{t/2}}{\cosh X_t} \le e^{t/2}$. So on any finite time interval you have a bounded local martingale, which is a martingale. Alternatively, note that $\frac{\sinh(x)}{(\cosh(x))^2} \le \frac{1}{2}$ for all $x$. So for any $T$, $$\int_0^T E\left[\left|-\frac{e^{t/2} \sinh ... 0 I did that like that Let f(t) and g(t) be integrating factors By Ito's Formula$$d(f(t)X1(t))=f(t)dX1(t)+f'(t)X1(t)dtd(f(t)X1(t))=(f(t)X2(t)+f'(t)X1(t))dt+af(t)dW_{1}(t)$$Similarly$$d(g(t)X2(t))=(-g(t)X1(t)+g'(t)X2(t))dt+bg(t)dW_{2}(t)$$Need to find f(t)and g(t) such that f(t)X2(t)+f'(t)X1(t)=0 and -g(t)X1(t)+g'(t)X2(t) So, ... 3 Let$$ X_1(t) = A(t) \sin (t) + B(t) \cos (t)\\ X_2(t) = A(t) \cos (t) - B(t) \sin (t) then thanks to the Ito fornula: \begin{align} dX_1(t) &= dA(t) \sin (t) + A(t) \cos (t) dt + dB(t) \cos (t) - B(t)\sin(t) dt \\ &= X_2(t)dt + dA(t) \sin (t) + dB(t) \cos (t) \\ dX_2(t) &= dA(t) \cos (t) - A(t)\sin (t) dt - dB(t) \sin (t) - B(t) \cos (t) ... 0 The product rule for finite differences is \Delta(fg)=\Delta f g + \Delta g f + \Delta f \Delta g. In classical calculus, when we send \Delta t \to 0, the second term is much smaller than \Delta t, so it vanishes in the limit. In stochastic calculus, the lowest order terms in \Delta X_t and \Delta Y_t are on the order of (\Delta t)^{1/2}: namely, ... 0 Stochastic calculus is to do with mathematics that operates on stochastic processes. The best known stochastic process is the Wiener process used for modelling Brownian motion. Other key components are Ito calculus & Malliavin calculus. Stochastic calculus is used in finance where prices can be modelled to follow SDEs. In the Black-Scholes model, ... 0 I think I may have a decent answer but I'd like to have some input. As we consider Z= \mathcal{E}(W) we have that Z_t = \exp \left( W_t-\frac12 \langle W \rangle_t\right) = \exp\left( W_t - \frac12 t \right). $$Z_t is the so called exponential Brownian martingale with$$ \mathbb{E} Z_t = \mathbb{E} \left[ \exp \left( W_t - \frac12 t \right) ...

3

To use Ito's formula to calculate a stochastic integral, you want to find an Ito process whose differential is the integrand. When calculating $\int_0^t B_s dB_s$, you initially guess (by intuition from regular calculus) that the process might be $\frac{1}{2} B_t^2$. So you use Ito's formula to calculate its differential and get $d \left ( \frac{1}{2} B_t^2 ... 0 We assume that the PDF of$v(t)$, namely$f(x, t)$, always has a finite first moment (that is, the mean). Let us consider the behavior of the position from$t = 0$to$t = \Delta t$, where$\Delta t$is chosen small enough that the variation in the first moment is as small as we like. (That we can do this is established by the continuity of$f(x, t)$.) As ... 2 Obviously, $$\ln X = \int_0^T \sigma(s) \, dW_s \qquad \text{for} \quad \sigma(s) := 1_{[0,T/2]}(s) + 1_{[0,T]}(s). \tag{1}$$ Define an Itô process$(Y_t)_{t \geq 0}by $$Y_t := \int_0^t \sigma(s) \, dW_s - \frac{1}{2} \int_0^t \sigma(s)^2 \, ds. \tag{2}$$ Then, by Itô's formula, \begin{align*} \exp(Y_t)-1 &= \int_0^t \exp(Y_s) \, dY_s + ... 0 Remark: What follows is a different argument from my initial solution attempt. Here, each of the steps in the derivation hold almost surely. We proceed by a sequence of steps. Step 1: Define the process f_t:=f(W_{t/2}, W_t)=W_{t/2}+W_t and consider that, by an application of Itô's lemma, we deduce the quadratic variation for the process as\langle ... 1 For notational convenience, definef_{n,t}:=\cos(n-1/2)\pi t$and choose any finite set of$k$-distinct natural numbers$n_1, \ldots, n_k$. Then, the Itô integrals$\{\int_0^1 f_{n_1,t}~\text dW_t,\ \ldots,\ \int_0^1 f_{n_k,t}$$\text dW_t\} are jointly normal. Consequently, these intergrals are independent if any pair of integrals has zero correlation. So, ... 1 By Fubini's theorem,$$\begin{align*} \int_0^t \int_0^s (s-r)^{n-1} B_r \, dr \, ds &= \int_0^t \left(\int_r^t (s-r)^{n-1} \, ds\right) B_r \, dr\\ &= \frac{1}{n} \int_0^t (t-r)^n \, B_r \, dr. \end{align*}$$Combining this with the last equation in your question, finishes the proof. Remark: Actually, we only have to do the "manipulation" using ... 0 First of all, recall that it follows from Wald's identities that$$\mathbb{P}(X_{T}=-b) = \frac{a}{a+b} \qquad \mathbb{P}(X_{T} = a) = \frac{b}{a+b} \qquad \mathbb{E}(T)=ab, \tag{1}$$see e.g. this answer for a proof or Corollary 5.11 in Brownian Motion - An Introduction to Stochastic Processes (by René Schilling and Lothar Partzsch). Part I: Calculate ... 1 You are correct that for each finite t, the stochastic integral \int_0^t e^{-\lambda s}\,dW(s) has a normal distribution. What is being asserted is that the limit$$N = \lim_{t \to \infty} \int_0^t e^{-\lambda s} \,dW(s)$$exists almost surely and also has a normal distribution. 1 I think it takes courage to identify what you are weak at and make that public. The mindset of identify weaknesses and seeking to improve is essential. If you is able to keep this mindset awake even when struggling, progress will be but a mere trivial corollary to the rich understanding that you will gain. Here are some thoughts, based on my experience: a) ... 1 As you pointed out, we know that \langle B \rangle_t = 0 in probability for t \geq 0. In fact, \langle B \rangle_t = 0 almost surely (mind that the exceptional set may depend on t). Indeed: By definition,$$\langle B \rangle_t = \mathbb{P}-\lim_{|\Pi| \to 0} V_t(B,\Pi)$$where$$V_t(B,\Pi) := \sum_{t_k \in \Pi} (B_{t_k}-B_{t_{k-1}})^2.Here ... 0 As I commented already, this does not have a known distribution. To see this let's derive the characteristic function of X_t: \begin{align} \phi(u)&=E\exp{\Big(iuX(t)\Big)}\\ &=E\exp{\Big(iu\int_0^tW(s)^2ds\Big)}\\ &=E\exp{\Big(iu\int_0^tJ_c(s)^2ds+c\int_0^tJ_c(s)dJ_c(s)+\frac12c^2\int_0^tJ_c(s)ds \Big)}\\ \end{align} where I have used ... 1 Out of Borodin and Salminen Handbook of Brownian Motion - Facts and Formulae: E_x\left[\exp\left(-\frac{\gamma^2}{2} \int_0^t W_s^2 ds \right) \right] = \frac{1}{\sqrt{\cosh(t\gamma)}}\exp\left( -\frac{x^2\gamma \sinh(t\gamma)}{2 \cosh(t\gamma)} \right) $$You can obtain it by using Girsanov's theorem. Really nifty formula when x=0. I also recommend ... 1 Why are you trying to show convergence in L^1 of the \phi_n when what is required is to show L^2-convergence of the stochastic integrals \int^T_S \phi_n \text dB_t? In any case, in order to show \ \int^T_S \phi_n \text dB_t\ {}\stackrel{L^2}{\to}\int_{S}^{T}f\text dB_{t}\  (assuming \ \mathbb E[\int_{S}^{T}f^2\text d{t}] < \infty), Oksendal ... 2 Consider an equation like $$\tag{1}dY_t = f(Y_t) dW_t$$ where Y_t is an unknown function and W_t is a continuous, but not differentiable, function. If W_t is Brownian motion then there is a classical theory for how to understand (1). Brownian motion has finite p-variation only for certain values of p. As you may know, ... 1 Hints: If \ \Bbb E[\int_{t}^{T}|h_s|\,\text ds],\,\Bbb E[\int_{t}^{T}|M_sh_s|\,\text ds]<\infty\  then, in succession, use the Fubini-Tonelli theorem, the tower property of conditional expectation, the martingale property, and Fubini-Tonelli one more time, as follows:$$ \Bbb E \left[ M_T\int_t^T h_s \text ds \, \vert\, \mathcal F_t\right] = \int_t^T ...

1

Seems like an acceptable proof method. I don't really see how that identity simplifies anything, but it is correct as far as I can tell.

1

Use the Law of Iterated Expectation. \begin{align} \mathbb{P}(\lambda_1 X + \lambda_2 Y = a) & = \mathbb{E}[\mathbb{1}_{\lambda_1 X + \lambda_2 Y = a}] \\[2ex] & = \Bbb E[\Bbb E[\mathbf 1_{\lambda_1 X + \lambda_2 Y = a}\mid Y]] \\[2ex] & = \int_{\Bbb R}\Bbb E[\mathbb{1}_{\lambda_1 X + \lambda_2 Y = a}\mid Y](y) \, \text{d} \mathbb{P}_Y(y) ... 3 I'm not an expert on this, but I did some research and found this Topics in Gaussian rough paths theory. I suggest you read it from the begining, pay special attention on page 5 and 6. From all the books and papers I found on Google this paper gives the most comprehensible explanation why you need those iterated integrals. Thy naturaly occur when you ... 0 \begin{align}var \int_0^t W_t dt &= \int_0^t\int_0^t \min(r,s)dr ds\\ &=\frac 13 t^3\\ E \int_0^t W_t dt &= \int_0^t E W_t dt\\ &=0 \end{align} Now let's use the hint for the variance \begin{align}var \int_0^t W_t dt &= var \Big(t\int_0^tdW_{\tau}-\int_0^t \tau dW_{\tau}\Big)\\ &=var \Big(t\int_0^tdW_{\tau}\Big)+var \Big(\int_0^t ... 1 Question 1: Note that [M]^{\tau_k}_{t \geq 0} is an increasing process. Therefore,|[M]^{\tau_k}_b - [M]^{\tau_k}_{b-\frac{1}{n}}| \leq 2 [M]_b.$$Since$$\mathbb{E}([M]_b^{\tau_k}) = \mathbb{E}((M_b^{\tau_k})^2)<\infty,$$we can apply the dominated convergence theorem to conclude$$\mathbb{E}([M]_b^{\tau_k}-[M]_{b-1/n}^{\tau_k}) \to 0 \qquad ...

0

I found the solution: If $f \in C_c(E)$, then $\forall t \in \mathbb{R}: t * f \in C_c$. For $f_1, ... f_n \in C_c(E), t \in \mathbb{R^n}$ set $f = (f_1, ..., f_n)$ $$\phi_X(t \cdot f) = \mathbb{E}[\exp(i t \cdot I_f(X))] = \int_E \exp(it \cdot x) d\mathbb{P}_{I_f \circ X}$$ From the uniqueness of the characteristic function $\mathbb{P}_{I_f \circ X}$ is ...

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