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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
Jun
20
awarded  Popular Question