Stephan Sturm
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 Mar 26 awarded Informed Feb 12 comment Stochastic calculus in $L^1$ I just wonder what you intend to achieve with your question? Feb 12 comment Stochastic calculus in $L^1$ If you do not require that the expectations are finite but allow an equality infinities, this follows also directly from localization. Feb 11 comment Stochastic calculus in $L^1$ Localization is a standard practice written down in nearly every stochastic analysis textbook. As limits in the definition of stochastic integrals are taken in probability, the generic setting for Ito's formula is neither $L^2$ nor $L^1$ but $L^0$. Feb 1 answered Malliavin derivative under change of measure Jan 24 answered Definition: Grigelionis Process?ch Jan 23 comment Malliavin derivative under change of measure What do you mean by a "stochastic process driven by $\tilde{B}_t$"? An Ito process with coefficients progressive wrt to the filtration generated by $\tilde{B}_t$? Why do you write that $W$ is correlated with $B$, not $\tilde{B}_t$? Could you define $F$ explicitly? Nov 19 comment Monte Carlo Simulation - efficient simulation of tail outcomes Sorry, but this is really not the site for layman's term questions, it is a site for questions people have that do math for a living. Moreover a quick google search should be enough to give you some ideas. Nov 19 comment Monte Carlo Simulation - efficient simulation of tail outcomes "Importance Sampling" is the keyword you are looking for. For a start you might look in Paul Glasserman's book on Monte Carlo Methods for Financial Engineering. Nov 14 comment European call option pricing under mean reverting stock return Yes, exactly. That's what I meant. Nov 13 comment European call option pricing under mean reverting stock return Yes, it is even exactly the same. But this is not really a research question. All what you have to do is to check that the integrability conditions in Girsanov's theorem are justified and this should be quite straight forward. Nov 12 comment Expectation, exponential of an additive functional of Brownian motion Why would you expect this? Just to get the left hand finite you need much stronger assumptions than integrability of $b$. Just consider the case $b(x) = e^x$. Oct 30 comment Between arithmetic and geometric Brownian motions: when are negative values possible? For SDEs Bernt Oksendal's "Stochastic Differential Equations: An Introduction with Applications" might be what you are looking for (about SPDEs I do not know). Oct 29 comment Between arithmetic and geometric Brownian motions: when are negative values possible? homepage.alice.de/murusov/papers/mu-mart.pdf Oct 29 comment Between arithmetic and geometric Brownian motions: when are negative values possible? You can write $S_t$, using stochastic calculus, as exponential $S_t = \mathcal{E}\bigl(\int_0^{\cdot} \frac{1}{S_s^{1-\beta}} \, ds\bigr)_t$. If $\beta = 1$, the exponent is just a Brownian motion which takes almost surely finite values. However, for $\beta < 1$ you have as integrand something what explodes when $S$ approaches zero, so you lose integrability. For $\beta>1$, clearly the exponent goes to zero when $S$ approaches zero, this is not a problem for the integrability. For more details you might have a look on Example 3.2 of the following paper by Mijatovic and Urusov: Oct 29 comment Between arithmetic and geometric Brownian motions: when are negative values possible? @8one6 The new formulation of the question changes nothing on the details, except that now the process is absorbed at zero as soon as it hits zero. This cannot happen a.s. if $\beta=1$, but can happen with positive probability for $0<\beta<1$. Oct 29 awarded Yearling Oct 28 answered Between arithmetic and geometric Brownian motions: when are negative values possible? Oct 28 comment Poisson kernel, follow-up question, follows that process $\left\{e^{i\theta X_t - \theta Y_t}\right\}$ is a martingale? In all your recent question you never specify the exact conditions. But assuming $X, Y$ independent Brownian motions and absorbing boundary conditions, the answer is yes, by exactly the argument laid out here: mathoverflow.net/q/221826 Oct 26 revised Poisson kernel, $E^{(x, y)}\text{exp}\{i\theta X_t - \theta Y_t\} = e^{i\theta x - \theta y}$ deleted 96 characters in body