Questions tagged [stochastic-analysis]

For questions about stochastic analysis or stochastic calculus, for example the Itô integral.

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Malliavin derivative wrt time changed Brownian motion

The Malliavin derivative $D^W_\alpha$, $\alpha \in \mathbb{R}$, with respect to a standard Brownian motion $W_t$ is $$D^W_\alpha W_t = 1_{[0,t]}(\alpha).$$ What would be the Malliavin derivative ...
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Is a right-continuous function on a compact space even "uniformly right-continuous"?

Let $E$ be a normed $\mathbb R$-vector space, $I\subseteq\mathbb R$ be nonempty, $O\subseteq I$ be open and $x:I\to\mathbb R$. Assume $x$ is right-continuous on $O$ and $$x(t-):=\lim_{s\to t-}x(s)$$ ...
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How to calculate the jump of $e^{{\rm i}N_ty-\operatorname E[N_t](e^{{\rm i}y}-1)}$ for a Poisson process $N$?

Let $(N_t)_{t\ge0}$ be a (nonhomogeneous) càdlàg Poisson process on a probability space $(\Omega,\mathcal A,\operatorname P)$. Assume $\alpha(t):=\operatorname E\left[N_t\right]$ for $t\ge0$ is ...
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Stochastic integration of $s^2$ with respect to the Brownian motion

When applying integration by part formula for Itô's integrals, I get $$\int_0^T s^2dB_s=T^2B_T-\int_0^T2sB_sds$$ but how do we deal with the term $\int 2sB_sds$?
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Interchanging limit and expectation regarding integrated brownian motion

I am currently working with integrated brownian motion and have to find its mean and variance. I know they are $0$ and $t^3/3$, but I am having trouble showing this. Using the Îto formula we can ...
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Equivalent characterization of Poisson processes

Let $\alpha>0$. We usually say that an $\mathbb N_0$-valued process $(N_t)_{t\ge0}$ on a filtered probability space $(\Omega,\mathcal A,(\mathcal F_t)_{t\ge0},\operatorname P)$ is Poisson with ...
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A continuous Markov process that is not Gaussian?

Given a probability space, we say that $(X_t)_{t \geq 0}$ is Markov w.r.t its own filtration $(\mathcal F_t)$ if for all $s<t$, $$P(X_t \in \cdot | \mathcal F_s) = P(X_t \in \cdot | X_s).$$ ...
Consider the centered OU process given by $X(t)=\frac{\beta}{\sqrt{2\alpha}}e^{-\alpha t}B(e^{2\alpha t})$ where $(B(t))_t$ is a standard Brownian motion. It is obvious why this is a 0 mean process ...