Samuel Reid
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 Apr 19 awarded Popular Question Dec 11 comment Difference Between Axiom of choice and axiom of countable choice. @EricWofsey What if $A \rightarrow \mathbb{C}$ is a continuous linear operator? Dec 11 comment Difference Between Axiom of choice and axiom of countable choice. I believe he is looking for examples of where the difference between the countable and uncountable versions of axiom of choice matter. Relevant answer on mathoverflow: mathoverflow.net/questions/7350/… Dec 11 comment Difference Between Axiom of choice and axiom of countable choice. Do you have a citation for your claim that "the question of whether "every surjection has a pre-inverse" is equivalent to full AC is wildly open."? Is this conjectured anywhere, do you have any evidence for proof or disproof? Dec 8 awarded Notable Question Nov 18 awarded Yearling Nov 17 answered Is the zero matrix in reduced row echelon form? Nov 17 comment Combinatorial proof of an identity involving integer partitions and their conjugates I can probably solve your problem but I'm concerned that there is no $j$ in the formula being indexed by $j$, what is meant here? Littlewood-Richardson coefficients may be relevant. Nov 17 accepted Continuity of Complex Function Nov 17 accepted Cardinality of the Union of an Indexed Collection of Sets Nov 17 accepted Surface Area of a Hypercube Nov 17 accepted Probability of Relatively Prime Integers Sep 28 comment Is $Var(X) = \sum_{y\in D(Y)} Var(X|Y=y) P(Y=y)$, where $D(Y)$ is the domain of $Y$? Hold up. So it IS true that $E(X^2) = \sum E(X^2 | Y=y) P(Y=y)$? Sep 28 accepted Is $Var(X) = \sum_{y\in D(Y)} Var(X|Y=y) P(Y=y)$, where $D(Y)$ is the domain of $Y$? Sep 24 comment Is $Var(X) = \sum_{y\in D(Y)} Var(X|Y=y) P(Y=y)$, where $D(Y)$ is the domain of $Y$? The formula that I was thinking of is right before the "contents" box on that article. I knew about the law of total variance but not this case. Thank you! Sep 24 asked Is $Var(X) = \sum_{y\in D(Y)} Var(X|Y=y) P(Y=y)$, where $D(Y)$ is the domain of $Y$? Sep 23 comment $\text{Cov}[X,Y]$ if $\mathbb{E}[X^2]<\infty$ but $\mathbb{E}[Y^2]=\infty$ Can't you still calculate covariance? You never need $E(Y^2)$ in the covariance calculation. Sep 22 awarded Notable Question Sep 18 awarded Popular Question Aug 15 awarded Notable Question