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| age | 23 | |
| visits | member for | 4 months |
| seen | 8 hours ago | |
| stats | profile views | 148 |
Graduate Student in Math at Courant Institute, NYU.
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1d |
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Hyperbolic PDE classification When you say 'linearize' you mean $t \sim 0$, right? What's wrong with $\sqrt{1 + \sin^2 t} \sim 1 + \frac{1}{2} \sin^2(t) \sim 1 + \frac{t^2}{2}$? |
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May 14 |
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Integration of radial functions? Rudin gives a proof in "Principles of Mathematical Analysis", which is a great read for very many other things. The essential work involves justifying the jacobian change-of-variables formula in the case of 'polar' coordinates. He does this for $n = 2$ in some detail. |
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May 14 |
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Can a Accumulation Point be an Eigenvalue? I think the point is that $0$ must be an accumulation point of the spectrum of this compact operator. Now ask yourself: what if my compact operator were invertible? You should know that the unit ball is relatively compact iff the ambient space is finite-dimensional, by the Riesz lemma. |
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May 13 |
answered | bifurcation value |
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May 9 |
revised |
Gambling Game: Martingales added 760 characters in body |
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May 9 |
revised |
Gambling Game: Martingales added 760 characters in body |
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May 9 |
revised |
Gambling Game: Martingales added 760 characters in body |
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May 9 |
revised |
Gambling Game: Martingales added 760 characters in body |
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May 9 |
answered | Gambling Game: Martingales |
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May 9 |
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Probability using C(n,k) 1 You're missing some data: we need to know how likely it is to have a girl or a boy, and we need to know if there's any dependence between subsequent birth sexes. |
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May 9 |
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Elements in Product Sigma Algebras The moral of the comments under your question is that the measurability of sections tells you nothing about measurability in $m \otimes \text{Bor}[0,1)$. |
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May 9 |
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Elements in Product Sigma Algebras @Catfish The key is how $m \otimes \text{Bor}[0,1)$ is defined: it's the smallest sigma algebra over $X \times [0,1)$ for which all measurable rectangles $A \times B$ are measurable, when $A \in m, B \in \text{Bor}[0,1)$. In particular, finite and countable unions of rectangles are measurable in $m \otimes \text{Bor}[0,1)$. |
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May 8 |
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Inequalities of expectations There's no way the second inequality can hold. Let $X = Y = \lambda$ be a constant function. Then the inequality would read $\lambda^4 \leq \lambda \cdot \lambda^2 = \lambda^3$ which is false for all $\lambda > 1$. |
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May 8 |
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Inequalities of expectations In a certain sense, yes! See my above answer. Practically speaking, the Holder inequality one derives is usually the optimal inequality one can obtain. |
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May 8 |
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Inequalities of expectations This is also an explicit counterexample to the second inequality you posited, for all $p < 4$. |
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May 8 |
answered | Inequalities of expectations |
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May 8 |
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Elements in Product Sigma Algebras @tomasz Okay, so the empty set is not a singleton, my bad : P. That's an interesting result, though- I didn't expect the diagonal to be so poorly behaved. |
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May 8 |
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Elements in Product Sigma Algebras I think @tomasz has the following counterexample in mind: let $X = [0,1)$ with the Borel sigma-algebra and let $A$ be any non-Borel (or nonmeasurable, depending on your mood) set. Let $\Delta = \{(t,t) \mid t \in A\}$ be the diagonal of $A$ in $[0,1)^2$. Then each section is a singleton, but as one can show, $\Delta$ is not measurable in the product sigma-algebra. |
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May 8 |
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Elements in Product Sigma Algebras If a set is measurable in the product sigma algebra, then its sections are measurable. This is an essential ingredient towards proving Fubini, for e.g.. Tomasz is saying that the converse is not true, necessarily. |
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May 8 |
answered | Elements in Product Sigma Algebras |
