bio | website | dcsc.tudelft.nl/~itkachev |
---|---|---|
location | Leiden, Netherlands | |
age | 26 | |
visits | member for | 3 years, 8 months |
seen | Aug 17 at 14:25 | |
stats | profile views | 937 |
I am a PhD student at TU Delft, working in applied probability and stochastic optimal control. My current focus is on approximate model-checking of stochastic systems via bisimulations (a part of computer science). I am interested in a wide field of applications, in particular in some areas of finance, such as risk theory.
Aug 3 |
comment |
Sufficiency of stationary policy for negative stochastic dynamic programming
@MichaelFanZhang: it may become unbounded ($-\infty$), but that's in fact important to negative dynamic programming. Also, since $g$ is sign-semidefinite, the sum is always negative so its expectation is well-defined, even though it may be $-\infty$ as well. Did I answer your question? |
Jul 22 |
comment |
Quotients of standard Borel spaces
@Burak: I see your point, indeed some Borel selection would be needed. |
Jul 22 |
awarded | Inquisitive |
Jul 21 |
comment |
Uniformization/measurable selection theorems
I am also aware of the updated version of Wagner's paper published in 1979, where he claims to add more results know in the Russian literature by that time, however I do not have an access to that. |
Jul 21 |
asked | Uniformization/measurable selection theorems |
Jul 20 |
revised |
Existence of an universally measurable pullback
deleted 175 characters in body |
Jul 20 |
accepted | Coupling of non-probability/sub-probability measures |
Jul 20 |
accepted | Particular neighborhoods of analytical sets |
Jul 20 |
asked | Stronger version of linearity for functions of measures |
Jul 20 |
comment |
Quotients of standard Borel spaces
@Burak: for some reason I was not notified, so I just came across your comment. Do you mean here, that if $E$ is smooth then $X/E$ is at least analytic, and if $X/E$ is a standard Borel space, then $E$ is smooth? Also, isn't the surjectivity in the 2nd sense (meeting each $F$ class) equivalent to bireducibility by definition? |
Jul 20 |
comment |
Inverse of a Borel surjection
I was exactly trying to use a Borel set which does not have a Borel uniformization, and relate it to a $\mathrm{Gr}(f)^{−1}$ via an isomorphism, however that did not work. Your idea of using a projection is really nice, thanks. Also, there is a survey of measurable selection theorems (Wagner 1977-79). Do you know if there was any progress after that, maybe another survey paper? Please tell me in case I should make it a separate question. |
Jul 20 |
comment |
Quotients of standard Borel spaces
@JoelDavidHamkins: my hypothesis that existence of a surjective $g$ implies existence of a Borel selector is not correct. If such a surjection $g$ exists, and there is a Borel selector $h:X \to X$ for $\sim_g$, then $h$ is $g$-measurable, and hence it factors through $g$ as $h = s\circ g$ where $s:Z\to X$ is Borel. The fact that $s\circ g$ is a Borel selector for $\sim_g$ is equivalent to $g\circ s = \mathrm{id}_Z$, and according to your answer such $s$ may fail to exist. |
Jul 20 |
revised |
Quotients of standard Borel spaces
added 57 characters in body |
Jul 20 |
accepted | Inverse of a Borel surjection |
Jul 20 |
comment |
Inverse of a Borel surjection
@Burak: nice, I haven't heard of that selection theorem before. |
Jul 20 |
comment |
Inverse of a Borel surjection
@JoelDavidHamkins: thanks, fixed that. |
Jul 20 |
revised |
Inverse of a Borel surjection
edited body |
Jul 19 |
asked | Inverse of a Borel surjection |
Jul 18 |
comment |
Quotients of standard Borel spaces
@JoelDavidHamkins: indeed, that's what I mean. I've updated this phrase again to avoid confusion. |
Jul 18 |
revised |
Quotients of standard Borel spaces
added 43 characters in body |