# Tagged Questions

This tag is used for questions about stochastic integrals - especially for calculations . For questions related to more theoretic aspects of stochastic integrals such as its construction. Stochastic-analysis may be a more appropriate tag.

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### Time integral of Brownian motion's running maximum

Let $\mu \geq 0$ and consider $B_{\mu}(t) := B(t) + \mu t$ a one-dimensional BM with drift $\mu,$ and let $M_t := \max_{0 \leq s \leq t} B_{\mu}(t)$ be its running maximum. My question involves two ...
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### How to solve a nonlinear SDE analytically

I have a numerical solution for the following equation: x'(t)= x(t) - x^3(t) + n(t) where n(t) is a white gaussian noise with zero mean and unit variance. I am a bit confused on how to go about the ...
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### Covariance of nonlinear sde

My problem is to compute the covariance of the following Ito process $$dX_t=AX_t+\sum_{k=1}^{n}B_kX_tdW_k,$$ where $A,B_k$ are nonlinear operators defined on a complex separable Hilbert space $H$. ...
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### Convergence in $L^2$ of the stochastic integral $\int\limits^{t}_{0}\frac{B_s}{e}1_{B_s\in(-e,e)}dB_s$

Let $e\in \mathbb{R}^+$ and $B_t$ 1-dimensional Brownian motion. Consider $$X_t=\int^{t}_{0}\frac{B_s}{e}1_{B_s\in(-e,e)}dB_s.$$ How to show that $X_t \to 0$ in $L^2$ as $e\to0$? Obviously the ...
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### How to compute this integral using Ito isometry? [duplicate]

I am trying to evaluate the following integral: $E\Bigg[\Bigg(\int^{t}_{0} \frac{B_s}{e}1\big(B_s\in(-e,e)\big)\Bigg)^2\Bigg]$ I cannot figure out how to apply Ito isometry when the indicator ...
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### Construction of Ito integral

This is in regards to constructing the Ito integral, specifically the first step of approximating bounded, continuous functions by elementary functions. Let $(\Omega, \mathcal{F}, P)$ be a ...
Suppose the following integrals $$\int_t^T X_s \, ds\ \text{ and }\ \int_t^T Y_s \, ds$$ are well-defined, where $X_s$ and $Y_s$ are continuous stochastic process. Do we ...
Good day. Imagene we have a martingale $M(t)=\int_0^t f(s)dB(s)$ which satisfies Dambis-Dubins-Schwarz Theorem. At the same time $M(t)^2 - <M>(t)$ is a Martingale starting in $0$ as well. If i ...