Michael Hardy

Minneapolis

After doing nearly all the coursework for a Ph.D. in math, I then did all the coursework for a Ph.D. in statistics and completed that degree.
6h
answered How to conceptualize "dividing out" a number (e.g. in permutations, Bayes' Theorem)?
7h
reviewed Approve suggested edit on Can't Finish Double Integral in Polar or Cartesian
7h
revised Olympiad Problem on Modular Arithmetic
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7h
revised $C(M)=\{A\in M_n(\mathbb{C}) \mid AM=MA\}$ is a subspace of dimension at least $n$.
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7h
revised Showing that $1/x$ is NOT Lebesgue Integrable on $(0,1]$
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12h
comment Probability question from GRE subject test
One can write $\Pr(X>3 \wedge Y>3)$ or $\Pr(X>3)\Pr(Y>3)$ but it makes no sense to write $\Pr(X>3)\wedge\Pr(Y>3)$. Those probabilities are numbers. The "$\wedge$" operation is not applied to numbers. ${}\qquad{}$
12h
revised Project a signal $S(t) = \sum_0^{\infty}A(k)e^{if(k)t}$ to 3d domain $ \psi_{n+1}(t) = \psi_n(t) + \hat{v}_nA_ne^{\hat{w}_ntf_n} $
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12h
revised Can someone present a visualization of the partitioning of a $L^p$ space into equivalent classes?
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14h
answered Distribution of a product of Multinomials
15h
comment Distribution of a product of Multinomials
This is easy in the case $n=1$. ${}\qquad{}$
15h
comment Why doesn't the "naive" scalar product for $SO(n)$ yield something invariant?
@JakobH : If you just google "latex symbols" you can find things like this. (What we use here is not LaTeX but MathJax. But it's way of coding mathematical notation is largely the same.) ${}\qquad{}$
16h
comment calculate-binomio-newton
@Zach466920 : Your first phrase above lost me. Do you mean the poster is self-centered, or that I am, or that the others who answer are, or what? ${}\qquad{}$
16h
revised Polarity on a Hyperboloid of one sheet
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16h
revised Distribution of a product of Multinomials
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16h
comment calculate-binomio-newton
Three answers have appeared (including mine) but so far I'm the only one who's up-voted the question. This often gets neglected. ${}\qquad{}$
16h
answered calculate-binomio-newton
16h
comment Conditional expectation of a set of Gaussian variables
You wrote "For a pair of Gaussian random vectors", but this actually works only if the whole tuple $[x,y]$ including all components of $x$ and all components of $y$ together is jointly Gaussian, which means it is so distributed that every linear combination of the components is Gaussian (where the coefficients in the linear combination are constant, i.e. not random). ${}\qquad{}$
16h
revised Conditional expectation of a set of Gaussian variables
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16h
comment Why doesn't the "naive" scalar product for $SO(n)$ yield something invariant?
A couple of typesetting oddities: In $b^2|0>+c^2|0>$, the software treats the plus sign as a unary rather than binary operation symbol, so you don't see the same spacing between the plus sign and $c^2$ that you see if you write $5+c^2$. That is not surprising since if you write something like $5>+c^2$ it is indeed a unary thing. But then in $c^2|-1\rangle$ it treats the minus sign as a binary operator, which is incorrect. So I coded it as c^2|{-1}\rangle, so that you see $c^2|{-1}\rangle$, with spacing appropriate to a unary rather than binary use of the minus sign. ${}\qquad{}$
16h
comment Why doesn't the "naive" scalar product for $SO(n)$ yield something invariant?
I found $\displaystyle\vphantom{\frac\int{\displaystyle\int}} v^T v = a^2|1>^T|1>+b^2|0>^T|0>+c^2|-1>^T|-1> = a^2|-1>|1>+b^2|0>+c^2|1>|-1> = a^2|0>+b^2|0>+c^2|0> $ and changed it to $\displaystyle\vphantom{\frac\int{\displaystyle\int}} v^T v = a^2|1\rangle^T|1\rangle+b^2|0\rangle^T|0\rangle+c^2|{-1} \rangle^T|{- 1}\rangle = a^2|{-1}\rangle|1\rangle+b^2 |0\rangle + c^2 |1\rangle|{-1}\rangle = a^2|0\rangle + b^2|0\rangle +c^2|0\rangle$. ${}\qquad{}$
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