# 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 Arithmeticedited title 7h revised $C(M)=\{A\in M_n(\mathbb{C}) \mid AM=MA\}$ is a subspace of dimension at least $n$.added 4 characters in body; edited title 7h revised Showing that $1/x$ is NOT Lebesgue Integrable on $(0,1]$added 24 characters in body 12h comment Probability question from GRE subject testOne 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 $\psi_{n+1}(t) = \psi_n(t) + \hat{v}_nA_ne^{\hat{w}_ntf_n}$added 2 characters in body 12h revised Can someone present a visualization of the partitioning of a $L^p$ space into equivalent classes?added 5 characters in body 14h answered Distribution of a product of Multinomials 15h comment Distribution of a product of MultinomialsThis 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 sheetadded 12 characters in body 16h revised Distribution of a product of Multinomialsadded 7 characters in body 16h comment calculate-binomio-newtonThree 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 variablesYou 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 variablesadded 4 characters in body 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 $\displaystyle\vphantom{\frac\int{\displaystyle\int}} v^T v = a^2|1\rangle^T|1\rangle+b^2|0\rangle^T|0\rangle+c^2|{-1} \rangle^T|{- 1}\rangle = a^2|{-1}\rangle|1\rangle+b^2 |0\rangle + c^2 |1\rangle|{-1}\rangle = a^2|0\rangle + b^2|0\rangle +c^2|0\rangle$. ${}\qquad{}$