Douglas Zare
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Registered User
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3h |
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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? |
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4h |
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repeated application of binomial distribution on a set of random variables It 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. |
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4h |
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circles required to cover the perimeter of rectangle I 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. |
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14h |
revised |
Permutations of $(Z/pZ)^*$ Added example. |
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15h |
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Joint distribution from multiple marginals You 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. |
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15h |
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Finding conditions on unspecified CDF that permit a solution to an equation First, 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. |
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19h |
answered | Determine the probability that two random vectors over a finite field are orthogonal |
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1d |
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Finding conditions on unspecified CDF that permit a solution to an equation These 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? |
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1d |
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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. |
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2d |
revised |
Another colored balls puzzle (part II) minor clarifications |
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2d |
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Another colored balls puzzle This is very nice, and it can be applied to part 2: mathoverflow.net/questions/130513/… |
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2d |
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Random walk on the hypercube One 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. |
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2d |
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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 $C/\sim$. |
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2d |
revised |
Another colored balls puzzle (part II) Rule 2 bounds |
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2d |
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Blue and red balls puzzle Well, I don't think this is far from proving that there is a $n^{3/4}$ scaling law, but it is missing some details. |
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May 18 |
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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/… |
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May 17 |
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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 $X \gt c, Y \lt X$. |
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May 17 |
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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, $P(Y \gt X \gt c) = P(X \gt Y \gt c) = \frac{1}{2} P(X \gt c)^2$. |
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May 17 |
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computing an integral involving standard normal pdf and cdf This 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. |
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May 17 |
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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. |
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May 17 |
revised |
Blue and red balls puzzle added variance argument |
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May 17 |
answered | Blue and red balls puzzle |
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May 17 |
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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. |
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May 17 |
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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. $\lbrace 1/5, 1/4, 1/3 \rbrace \cup \lbrace 1/3 + 1/2^n \rbrace$. |
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May 17 |
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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. |
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May 16 |
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Blue and red balls puzzle Computing 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. |
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May 16 |
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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? |
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May 16 |
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Sums of uniformly random vectors from the $n$-dimensional unit ball Ok, 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. |
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May 15 |
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Sums of uniformly random vectors from the $n$-dimensional unit ball What is unsatisfactory about the Central Limit Theorem? |
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May 15 |
revised |
Another colored balls puzzle (part II) Added upper bound. |
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May 15 |
answered | Another colored balls puzzle (part II) |
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May 15 |
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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. |
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May 15 |
revised |
Fano plane drawings: embedding PG(2,2) into the real plane embedded image |
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May 15 |
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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. |
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May 14 |
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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. |
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May 14 |
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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. |
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May 14 |
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Generalization of hypergeometric distribution? One special case was asked here: mathoverflow.net/questions/130115/… |
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May 14 |
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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. |
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May 14 |
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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. |
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May 13 |
accepted | Non-Constant-Sum Blotto Game for Only 2 Players and 2 Battlefields |
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May 13 |
answered | Non-Constant-Sum Blotto Game for Only 2 Players and 2 Battlefields |
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May 11 |
awarded | ● Nice Answer |
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May 11 |
revised |
Show that this ratio of factorials is always an integer Added material from Gessel's Super Ballot Numbers; added 45 characters in body; edited body; added 2 characters in body |
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May 11 |
revised |
Show that this ratio of factorials is always an integer Added summations. |
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May 11 |
answered | Show that this ratio of factorials is always an integer |
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May 9 |
answered | Modern Mathematical Achievements Accessible to Undergraduates |
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May 9 |
accepted | Why is it hard to prove that the Euler Mascheroni constant is irrational? |
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May 7 |
answered | Probability that one RV will exceed many others |
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May 4 |
awarded | ● Great Answer |
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May 4 |
revised |
Why is it hard to prove that the Euler Mascheroni constant is irrational? corrected statement about periods |
