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visits | member for | 5 years, 6 months |
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If I like your problem, I'll think of it in honest. However, if the question is badly posed or the information is incomplete, I'll vote to close it when in a bad mood and waste several hours of your and my time asking for clarification, pointing out trivial counterexamples, etc., when in a good mood. In both cases, the most likely outcome is that I'll finally switch my attention to something else. ;)
Apr 14 |
awarded | Great Question |
Mar 28 |
awarded | Nice Answer |
Mar 18 |
comment |
Parodies of abstruse mathematical writing
It is just a grammatically correct combination of buzzwords. The only beauty of it is the total lack of meaning achievable only by true randomness: no matter what nonsense a human tries to write, some logic will still be there (though some writings of our administrators come amazingly close to the example in the post). What it is certainly lacking is a sparkle of weird wit and a completely unexpected angle from which the things are viewed, which are the most valuable things in parodies (IMHO, at least), and which make a really good parody something much more than just a parody. |
Mar 16 |
awarded | Stellar Question |
Mar 14 |
awarded | Nice Answer |
Mar 12 |
awarded | Nice Answer |
Feb 24 |
awarded | Nice Answer |
Feb 24 |
awarded | Popular Question |
Feb 23 |
answered | Translates of null sets |
Feb 14 |
comment |
Solving a doubly exponential generating function
Not only "algebraic". Any "explicit formula" in any minimally decent sense of this word defines a function that can be analytically continued along almost every path on the complex plane while $F$ has the unit circle as its natural boundary. |
Feb 13 |
comment |
Inserting an open and simply-connected set between a compact set and an open set
If the complement of $K$ is connected, just consider the set of points of the complement to which you can drag an open disk of fixed very small radius from infinity without touching $K$. It is a closed set and its complement $V$ is open, contains $K$, and is simply connected just because to drag the center of the disk inside a loop, you need to drag it to the boundary of the loop first (though you will need some kind of induction argument to show it with full rigor and completely from scratch). |
Feb 13 |
comment |
Must the Lebesgue measure of a $\rho$ - neighbourhood of an $(n-2)$ - dimensional set be at least $c\rho^2$?
Be a little bit careful here: the dimension is too crude a measurement to draw such a conclusion because you may have some extra sub-power decay it does not catch (consider a compact set of dimension $1$ but zero measure on the line and put it in $\mathbb R^3$). However, if the $n-2$-dimensional Hausdorff measure is positive, then it is true and follows straight from the definitions. |
Jan 13 |
awarded | Guru |
Dec 30 |
comment |
What is the reason that $\sigma$-algebra replaced $\sigma$-ring in introductory measure theory?
Oh, my! So it is "sigma-ring", not "semiring"? I was totally perplexed with that typo in the post. Then there is not much difference, really. It is just convenient to assume that the whole space is always measurable and to be able to pass to the complement freely, but otherwise it is more a question about terminology than about substance unless I misunderstand something. |
Dec 29 |
comment |
A version of Wald identity
@HMPanzo Indeed. Thank you! It doesn't look like their approach is any easier than mine (or even substantially different)... :-) |
Dec 29 |
revised |
A version of Wald identity
deleted 5 characters in body |
Dec 29 |
comment |
A version of Wald identity
@user57639 OK, see if it works now. Either I am missing some obvious approach, or it can make one of the trickiest problems about the Brownian motion on the probability exam. Where did you get this question from? |
Dec 29 |
answered | A version of Wald identity |
Dec 28 |
comment |
A version of Wald identity
@Nate Elderedge Ah, indeed. OK, let me try to answer properly then :-) |
Dec 28 |
comment |
A version of Wald identity
The (now deleted) example of Bjorn shows that at most one of those is true for the hitting time of $1$, in which case $EW_T=EW_T^2=1$, i.e., neither $0$, nor $\infty$ ;-) |