# Tag Info

## Hot answers tagged stochastic-integrals

3

just to offer two cents on this (health warning is that this is from an ex-trader rather than quant, and based on personal experience through self-study rather than eg as part of a formal PhD programme): Williams really is fantastic, learned the basics of measure theoretic probability from that as an undergrad, and it's stood the test of time and is still a ...

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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 ...

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 ... 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$p. As you may know, ... 1 Seems like an acceptable proof method. I don't really see how that identity simplifies anything, but it is correct as far as I can tell. 1 By Fubini's theorem, \begin{align*} \int_0^t \int_0^s (s-r)^{n-1} B_r \, dr \, ds &= \int_0^t \left(\int_r^t (s-r)^{n-1} \, ds\right) B_r \, dr\\ &= \frac{1}{n} \int_0^t (t-r)^n \, B_r \, dr. \end{align*} Combining this with the last equation in your question, finishes the proof. Remark: Actually, we only have to do the "manipulation" using ... 1 Why are you trying to show convergence inL^1$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$), Oksendal ... 1 I think it takes courage to identify what you are weak at and make that public. The mindset of identify weaknesses and seeking to improve is essential. If you is able to keep this mindset awake even when struggling, progress will be but a mere trivial corollary to the rich understanding that you will gain. Here are some thoughts, based on my experience: a) ... 1 L'hopital? The limit is a$\frac{0}{0}$type, so if we let$f(t) = \int_0^t W_udu$$$\lim_{t\to0} \dfrac{f(t)}{t} = \lim_{t\to0} \dfrac{f'(t)}{1}$$ using fundamental calculus theorem $$f(t) = \int_0^tW_udu \implies f'(t)=W_t$$ $$\therefore \lim_{t\to0} \dfrac{f'(t)}{1} = \lim_{t\to0}W_t = W_0$$ 1 The fundamental theorem of calculus states that $$\lim_{t \to 0} \frac{1}{t} \int_0^t f(u) \, du = f(0)$$ for any continuous function$f$. Applying this to$f(t) := W_t$yields the result. 1 Consider the Langevin system of equations: $$\begin{cases} X_t = \varepsilon^{-1} U_t \,dt\\ U_t = - \varepsilon^{-2}\mu U_t\,dt + \varepsilon^{-1} \sigma \,dW_t \end{cases}$$ You will recognize that the process$U$is the same as your process$X$. In Pardoux and Veretennikov's article On Poisson equation and diffusion approximation 2, it ... 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 $$\mathbb{E}([M]_b^{\tau_k}-[M]_{b-1/n}^{\tau_k}) \to 0 \qquad ... 1 Out of Borodin and Salminen Handbook of Brownian Motion - Facts and Formulae:$$ 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) $$You can obtain it by using Girsanov's theorem. Really nifty formula when x=0. I also recommend ... 1 You are correct that for each finite t, the stochastic integral \int_0^t e^{-\lambda s}\,dW(s) has a normal distribution. What is being asserted is that the limit$$N = \lim_{t \to \infty} \int_0^t e^{-\lambda s} \,dW(s)$$exists almost surely and also has a normal distribution. 1 For notational convenience, define f_{n,t}:=\cos(n-1/2)\pi t and choose any finite set of k-distinct natural numbers n_1, \ldots, n_k. Then, the Itô integrals \{\int_0^1 f_{n_1,t}~\text dW_t,\ \ldots,\ \int_0^1 f_{n_k,t}$$\text dW_t\}$ are jointly normal. Consequently, these intergrals are independent if any pair of integrals has zero correlation. So, ...

1

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 ...

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) ... 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 ...

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