# Douglas Zare

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 Name Douglas Zare Member for 3 years Seen 24 mins ago Website Location Age
 3h comment Probability $k$ bins are non-empty.Just to confirm, the $k$ query bins are not uniformly chosen from the $m \choose k$ subsets, they don't have to be distinct, correct? 4h comment repeated application of binomial distribution on a set of random variablesIt doesn't look correct without some huge assumptions. If you are still confused, work on clarifying your question, and see the FAQ for other sites where clarified versions of this question might fit better. 4h comment circles required to cover the perimeter of rectangleI doubt an exact calculation is feasible. However, if the rectangle is large, then you should be able to get a good approximation by breaking the perimeter up into pieces, most of which are covered roughly independently and with probability close to $1$. Then use inclusion-exclusion. 14h revised Permutations of $(Z/pZ)^*$Added example. 15h comment Joint distribution from multiple marginalsYou can get a simple estimator from the average of $p(a=1)$, the average of $p(b=1)$, and the correlation between $\vec{p(a=1)}$ and $\vec{p(b=1)}$. However, it's not clear that this estimate is consistent as the number of samples goes to infinity unless you also increase the sample size. 15h comment Finding conditions on unspecified CDF that permit a solution to an equationFirst, it's bad to post the same question on different sites at the same time, particularly without linking in both directions. This leads to duplication of effort from the people you are asking to help you. Second, the answer on stat.stackexchange is quite incomplete. It says you can make local changes to $F′$ to get a solution without adjusting the moments by much. That's obvious. The interesting question is what conditions on the moments are sufficient. Since you accepted the trivial answer, I suppose you aren't actually interested in sufficient conditions so I won't bother to post. 19h answered Determine the probability that two random vectors over a finite field are orthogonal 1d comment Finding conditions on unspecified CDF that permit a solution to an equationThese variables must have some meaning, or else you should simplify the expressions by using $M=N−1$ instead of $N$, for example. Why not include that meaning, so the equation you are trying to solve is intuitive? 1d comment Is this cube packing possible?If you center the $9$th cube, you can rotate it $45$ degrees about an axis parallel to an edge, and then when you project perpendicular to that edge you get the $2$-dimensional picture. 2d revised Another colored balls puzzle (part II)minor clarifications 2d comment Another colored balls puzzleThis is very nice, and it can be applied to part 2: mathoverflow.net/questions/130513/… 2d comment Random walk on the hypercubeOne upper bound would be the time to reach a particular permutation by adjacent transpositions in the symmetric group. However, this value should be much lower, since you only need to hit one of $k! (n-k)!$ permutations. 2d comment Another colored balls puzzle (part II)@Jon Peterson: Yes, that's why I looked at the graph $C/\sim$, where $\sim$ identifies antipodal points of the cube, which has $2^{n-1}$ vertices. $C/\sim$ is also connected and regular. Stopping at either $00\ldots 0$ or $11 \ldots 1$ is equivalent to stopping at the original point in a random walk on $C/\sim$. 2d revised Another colored balls puzzle (part II)Rule 2 bounds 2d comment Blue and red balls puzzleWell, I don't think this is far from proving that there is a $n^{3/4}$ scaling law, but it is missing some details. May18 comment Reference request: affine transforms + circle inversion?Circle inversion doesn't preserve generalized ellipses, or conics. Curves of degree n are typically sent to curves of degree 2n. If you invert an hyperbolas about its center you get a figure-8s, a lemniscates. Ellipses can be sent to dimpled limaçons or hippopedes. Parabolas can be sent to cissoids or cardiods. If the center of inversion is on the conic, though you get a cubic curve like a crunode $y^2 = x^2(x+1)$. en.wikipedia.org/wiki/Inverse_curve xahlee.info/SpecialPlaneCurves_dir/Inversion_dir/… May17 comment computing an integral involving standard normal pdf and cdf@Mark: I don't understand what you are confused about. I translated the integral into a statement about random variables. Inside the integral is a density for $X=x, Y \lt x$. Integrating from $x=c$ to $x = \infty$ means $X \gt c, Y \lt X$. May17 comment computing an integral involving standard normal pdf and cdf@Mark: $P(Y \gt X \gt c) + P(X \gt Y \gt c) = P(X,Y \gt c) = P(X\gt c)^2$. By symmetry, $P(Y \gt X \gt c) = P(X \gt Y \gt c) = \frac{1}{2} P(X \gt c)^2$. May17 comment computing an integral involving standard normal pdf and cdfThis questions asks for the probability that a standard normal distribution is in a wedge-shaped region, or equivalently for the CDF of a two-dimensional Gaussian distribution with a general covariance matrix. Wikipedia says that there isn't an analytic expression for this, but that approximations are known. One related paper I remember is by Marsaglia, jstatsoft.org/v16/i04/paper, on the distribution of the ratio between two normal distributions, but that would be a double wedge, between two lines instead of two rays. May17 comment computing an integral involving standard normal pdf and cdf@Mark: That case is easy to compute by hand. It would be the probability that $X \gt c, X \gt Y$ where $X,Y \sim N(0,1)$. If I calculate correctly, that's $P(X \gt c) - P(Y \gt X \gt c) = P(X \gt c) - 1/2 P(X \gt c)^2$. The general case doesn't simplify like that. May17 revised Blue and red balls puzzleadded variance argument May17 answered Blue and red balls puzzle May17 comment Is there any proof that you feel you do not “understand”?That statement of the Monty Hall problem is incomplete, as is mentioned in the Wikipedia article. May17 comment Can distinct open knots correspond to the same closed knot?Even if you avoid cutting at a wild point, I think you can still tell where you cut some wild knots. The points which can't be thickened up form a closed subset of $S^1$, and when you cut at a point on the complement, you can get different topological types of closed subsets of the interval, e.g., $\lbrace 1/3 \rbrace \cup \lbrace 1/3 + 1/2^n \rbrace$ vs. $\lbrace 1/5, 1/4, 1/3 \rbrace \cup \lbrace 1/3 + 1/2^n \rbrace$. May17 comment Can distinct open knots correspond to the same closed knot?Are you restricting to tame knots? If you allow wild knots, then I think you should get something different if you cut at a wild point of a wild knot than at a normal point of a wild knot. May16 comment Blue and red balls puzzleComputing the average would be a step forward, but it may not be the end. It's reasonable to say $f(0,m) = x^m$ and ask which polynomial you get. May16 comment An interesting version of the problem “balls into bins”If you find one ball in each of the $4$ bins, this could come from $A+F$, $B+E$, or $C+D$. Is that really what you wanted to ask? May16 comment Sums of uniformly random vectors from the $n$-dimensional unit ballOk, that explains why you can't use the CLT in one direction. There may still be some applications, though. See Stam, 1982. "Limit Theorems for Uniform Distributions on Spheres in High-Dimensional Euclidean Spaces." J. Applied Probability 19, 221-228. May15 comment Sums of uniformly random vectors from the $n$-dimensional unit ballWhat is unsatisfactory about the Central Limit Theorem? May15 revised Another colored balls puzzle (part II)Added upper bound. May15 answered Another colored balls puzzle (part II) May15 comment Another colored balls puzzle (part II)The denominator is something like $(n-1)(n-2)...(n-k)$, but the polynomials in the numerator are more complicated, and the variations I tried on the coefficients didn't show up in the OEIS. May15 revised Fano plane drawings: embedding PG(2,2) into the real planeembedded image May15 comment Another colored balls puzzle (part II)If the exact value for $(n-1,1)$ is $2^{n-1}-1$, then the exact value for $(n-2,2)$ is $\frac{n}{n-1}(2^{n-1}-2)$. Similarly, one can work out more complicated expressions for the exact value for $(n-3,3)$, etc. however, I don't see the general form. Mathematica's RSolve spits out a page of things involving Hypergeometric2F1Regularized. If you can verify that starting with $a(0)=0, a(1) = 2^{n-1}-1$ and the recurrence $a(k) = 1 + \frac{k}{n} a(k-1) + \frac{n-k}{n} a(k+1)$ that $a(n-1) = 2^{n-1}-1$ then this solves the problem for $2$ colors, and provides a nice lower bound for everything. May14 comment Another colored balls puzzle (part II)Exact calculations for $n \le 100$ give a value of $2^{n-1}-1$ for the expected time to completion from $(1,n-1)$. I haven't yet proved that this always holds. May14 comment Google question: In a country in which people only want boys@Rhett Butler: Here is a difference. When you bet on roulette, you don't get paid $B/(B+G)$. If you make bet $1$ at each step, you get paid $B-G$, say. Your payoff is a martingale. You do not get paid $B/(B+G)$. $B/(B+G)$ is not a martingale. Stop pretending that recognizing the fact that $B/(B+G)$ does not have expected value $1/2$ under some stopping rules means that we have a roulette strategy which wins on average. Your statement that "my" strategy (which is not mine) must win at roulette is mathematically wrong. It does not become right by spamming it or being increasingly rude. May14 comment Generalization of hypergeometric distribution?One special case was asked here: mathoverflow.net/questions/130115/… May14 comment Google question: In a country in which people only want boys@Rhett Butler: You are confused. As has been pointed out much earlier here, the payoff in roulette is not the proportion $B/(B+G)$. So, if you think there is an application to roulette of a method for changing the expected value of $B/(B+G)$, the burden is on you to show it. Your claims that my correct statements are equivalent to claiming to have a winning roulette strategy are wrong and insulting. I tried to help you to understand some math, and you lie and ridicule me. Any more lies claiming that I am trying to invent a winning roulette strategy will be flagged as spam. May14 comment Google question: In a country in which people only want boys@Rhett Butler: Do you really think I have no idea that the sex of a child is modeled as $1/2$ independently of what came before, even though this is used in my calculations? You think that triviality is what everyone else is missing, too? How stupid. The mathematically interesting thing to me is that when the families can choose how many children to have based on the previous sexes, the proportion $B/(B+G)$ is a biased estimator of that $1/2$, as I stated in my answer, which means the expected value is not $1/2$. And $E[B/(B+G)]$ is what the OP's summation tried to calculate. May13 accepted Non-Constant-Sum Blotto Game for Only 2 Players and 2 Battlefields May13 answered Non-Constant-Sum Blotto Game for Only 2 Players and 2 Battlefields May11 awarded ● Nice Answer May11 revised Show that this ratio of factorials is always an integerAdded material from Gessel's Super Ballot Numbers; added 45 characters in body; edited body; added 2 characters in body May11 revised Show that this ratio of factorials is always an integerAdded summations. May11 answered Show that this ratio of factorials is always an integer May9 answered Modern Mathematical Achievements Accessible to Undergraduates May9 accepted Why is it hard to prove that the Euler Mascheroni constant is irrational? May7 answered Probability that one RV will exceed many others May4 awarded ● Great Answer May4 revised Why is it hard to prove that the Euler Mascheroni constant is irrational?corrected statement about periods