9 votes

Are differentials on their own in stochastic calculus just an abuse of notation?

Differentials in stochastic calculus have very precise interpretations. The stochastic differential equation in differential form $$dX_t = \mu (t, X_t) dt + \sigma (t, X_t) dB_t$$ translates precisely ...
Jose Avilez's user avatar
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5 votes
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Is the stochastic integral process $\left(\int_{0}^{t}e^{-\lambda\left(t-s\right)}\mathrm{d}B_s\right)_{t\geq 0}$ a martingale?

The criterion you mentioned is certainly a sufficient condition, but not a necessary condition. So, let's go back to the basic. Let $M_t = \int_{0}^{t} e^{-\lambda (t-s)} \, \mathrm{d}B_s$. Then for $...
Sangchul Lee's user avatar
4 votes
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Computing the quadratic covariation $\langle B,B^2\rangle_t$ of a Brownian motion and its square

See first this discussion on how to obtain the quadratic variation of two Itô processes. By Itô's formula we obtain that $B_{t}^{2}$ follows the dynamics $$ B_{t}^{2} = \int_{0}^{t}ds + \int_{0}^{t}...
minginator's user avatar
3 votes
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Canonical lift of a smooth path as a rough path

Surely if $X_t$ is nice and smooth, then $\alpha > \frac{1}{2}$, but then $2\alpha > 1$, and wouldn't this degenerate into saying that $\mathbb{X}$ is constant? From the known result on Holder ...
Thomas Kojar's user avatar
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3 votes
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What is the solution to the SDE $X^x_t = B_t + \int_0^t \frac{x − X^x_s}{1-s} ds$

Setting $Z_t$ as what you did in your work, and then writing (2) in Differential form gives us that $$dX^x_t = xdt + dB_t - \frac{B_t}{(1-t)}dt + Z_tdt \tag{4}\label{eq4} $$ And (2) also gives us that ...
vinayak's user avatar
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2 votes
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Understanding stochastic integration with semimartingale as integrator: The locally bounded variation part.

Indeed, for each $\omega$ you get a different measure $\mu_g$. So the integral can be defined path-by-path: i.e. for a fixed $\omega$ $$\left( \int_a^b f(s) dA_s \right)(\omega) = \int_a^b f(\omega, s)...
Jose Avilez's user avatar
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2 votes

Weak Uniqueness of solution of SDE

We will add a few more details, let me know if you need more. Let $X_t$ and $Y_t$ be the strong solutions constructed from $\tilde{B}_t$ and $\hat{B}_t$, respectively, as above. Here they simply ...
Thomas Kojar's user avatar
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2 votes

Is the stochastic integral process $\left(\int_{0}^{t}e^{-\lambda\left(t-s\right)}\mathrm{d}B_s\right)_{t\geq 0}$ a martingale?

Define $A_t := \int_0^t e^{\lambda s} dB_s$. You are asking if $X_t := e^{-\lambda t} A_t$ is a martingale. By Ito's formula, we have \begin{align*} dX_t &= e^{-\lambda t} dA_t -\lambda e^{-\...
user6247850's user avatar
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1 vote

Book recommendation for stochastic integral wrt local martingales

I would suggest, that you consider "Brownian Motion, Martingales, and Stochastic Calculus " by Le Gall. He does not use the Reisz representation theorem to define the stochastic integral, ...
user123234's user avatar
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1 vote

Example of a process that yields a non-martingale

Easy example: $C_k = 1_{X_k > X_{k-1}}$. This is adapted because $X$ is adapted, but $\sum_{k=1}^n C_k (X_k-X_{k-1}) = \sum_{k=1}^n (X_k-X_{k-1})^+$ is non-decreasing and hence cannot be a ...
user6247850's user avatar
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1 vote
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Converse of a martingale transform theorem

Under the stated conditions, it is not true as you can easily construct a counterexample since you basically left the process $X$ to be unrestricted. For example, assume for simplicity that $C_{n}(\...
minginator's user avatar
1 vote

Using Ito's Lemma to take a stochastic integral

Please avoid to ask too many questions at the same time in future posts. Let's tackle the side questions first. First side question. I mean, why not ? If you have a stochastic variable $X_t$ itself ...
Abezhiko's user avatar
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1 vote
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Stochastic integration: Computing $\mathbb{E}[\exp(- \lambda \int_0^{t∧T_\epsilon}\frac{ds}{B_s^2}) ]$

The trick to being able to use the Optional Stopping Theorem here is to realize that, in the equation $\alpha (\alpha - 1) = 2\lambda$, we can always choose $\alpha < 0$. This implies $Y_t = (B_{t ...
user6247850's user avatar
  • 13.5k
1 vote

Verifying Ito isometry for simple stochastic processes

A simple process $\Phi = (\Phi_t)_{t\geq 0}\in\mathcal{E}$ is a process of the form $$\Phi_t = \sum_{i=0}^{n-1}U_i1_{(t_i,t_{i+1}]}(t)$$ where $0=t_0<\dots<t_n<\infty$ and $U_i\in b\mathcal{F}...
Wilfred Montoya's user avatar

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