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- 0 posts edited
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- 220 votes cast
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 |
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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 |