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I'm a first year computer science graduate student. My research interests are mathematical logic, complexity theory, algorithms, spectral graph theory, lower bounds, computational economics, algorithmic game theory, and probability.


Sep
6
awarded  Popular Question
Sep
5
asked Linear programming with an independent variable
Jul
13
awarded  Yearling
Jul
12
awarded  Good Question
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Jun
10
awarded  Enthusiast
May
16
answered Statistics Review
May
15
comment Statistics Review
Not quite, no - I'll post an answer. Thanks for sharing your thoughts.
May
15
comment Statistics Review
Can you post what you've tried so far? (1) is pretty straightforward.
May
9
comment Probability of having her purse
@Winther: that's not right at all, see below. That gives probabilities of $\frac{4}{16}, \frac{3}{16}, \frac{9}{16}$ for stores 1, 2, 3, respectively.
May
9
comment Probability of having her purse
Yes, that's exactly right -- does that clarify things?
May
8
answered Probability of having her purse
Apr
20
comment Basic questions about descriptive complexity
Most of my questions were answered by these gentle slides: msor.victoria.ac.nz/foswiki/pub/Events/ALC2011/LectureSlides/…
Apr
20
asked Basic questions about descriptive complexity
Apr
6
comment Pure vs mixed strategy Nash Equilibria
@BalazsRau : Let's say you're playing a game where the players can pick between $n$ different strategies. In general, there might be mixed equilibria that only mix over $k$ of those $n$ strategies. In this case, there will be some linear combination of my $k$ strategies that will make you like some $k$ of your strategies equally well (and the other $n-k$ strategies less than that). In the pure case ($k=1$), this statement is uninteresting: it just says that you like $1$ of your strategies at least as well as all the others, which is obvious.
Apr
6
comment Pure vs mixed strategy Nash Equilibria
This is off on a bit of a tangent from your original post, though =) the answer to your original question is "a pure strategy is just a mixed strategy over exactly 1 strategy."
Apr
6
comment Pure vs mixed strategy Nash Equilibria
I agree with your NE analysis. The "equally happy with anything" bit of my post referred specifically to mixed equilibria, sorry if that was unclear. It turns out that if $A$ plays a mixed-strategy equilibrium where he has nonzero chance of playing $k$ different strategies, then there is some set of $k$ strategies for $B$ that are all equally good (and therefore $B$ can randomize among them however he likes). In the special case of $k=1$ ("pure strategy equilibrium"), that means there is (at least) $1$ strategy for $B$ that stands up to scrutiny, which is not so informative.
Apr
6
answered Pure vs mixed strategy Nash Equilibria
Mar
25
comment Questions on “simple-connectedness-like” property
One problem with your definition: some disconnected sets will not be "non-simply connected" (because, for example, there do not exist open connected subsets $U, V \subset X$ eith $U \cup V = X$) and therefore it will be defined as "simply connected." I doubt that's what you want.