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Mar
3
revised Sigma algebra generated by a set
added 79 characters in body
Mar
3
answered Sigma algebra generated by a set
Mar
3
comment Closed under equality
this is known as the identity of indiscernibles
Mar
2
comment Are all integers fractions?
@Carl Mummert . In my comment to an answer I ended up with "It is equal because there is a mathematical convention that makes it equal.", which I think it is incorrect in light of your answer. So they are not equal as mathematical entities. Are we left to say they are equal as a language convention? That is when we say Z is a subset of Q or z=z/1 for z integer, we make a language convention (shortcut, compromise)? Thanks
Mar
2
comment Are all integers fractions?
I think it's more than that. One can ask why c is the same as c/1? One might answer because we see them as embedded in the quotient ring Q, that is c is the same as equivalence class of (c,1) . Or one can maybe say because "ordinary" division is preserved, that is c divided by 1 is c. In other words, the answer because "it is a subset" or because "c=c/1" is not sufficient. It is equal because there is a mathematical convention that makes it equal.
Mar
2
comment Limit of a function
Yes, what Berci says
Mar
2
comment Limit of a function
well why not choose epsilon such that l>epsilon>0 . Such epsilon exists
Mar
2
comment The number of positive integers
@Brian M. Scott not quite ok at the end, more justification is needed
Feb
11
comment Questions on atoms of a measure
Does this work? Suppose the intersection has probability 1, then its complement has probability 0. If we choose only the countable set, the probability of complement of intersection , which is union of compelments has probability 1. Then there exist one element with probability 1. Then its complement has probability 0. Contradiction with the choise of elements in intersection
Feb
11
comment Questions on atoms of a measure
in lemma's proof, why cant the intersection of all elements in sigma with probability 1 have probability 0?
Feb
8
comment How should I understand the $\sigma$-algebra in Kolmogorov's zero-one law?
I refer to "atomic" as used in Lemma.
Feb
8
comment How should I understand the $\sigma$-algebra in Kolmogorov's zero-one law?
May I know how you define "atomic" here? Thanks
Jun
8
awarded  Caucus
Feb
19
awarded  Yearling
Feb
14
awarded  Talkative
Feb
5
comment the expectation of a random variable of a random variable
@Tim Unless you meant the collection of p.d.f's Z|x is a random variable, which is an interesting question and I think it is true. But perhaps a sidetrack.
Feb
5
comment the expectation of a random variable of a random variable
@Tim $E(Z|X)$ is a random variable and a mapping of X, not the p.m.f. of Z|X=x . Note that h(z) above is the p.m.f. of Z|X=x . Did you write h(z) explicitly? I don't want to spoil the hint for Shuai
Feb
5
comment the expectation of a random variable of a random variable
@Tim I think it's like this $\delta(a)(z)=1$ if $z=a$ and $\delta(a)(z)=0$ if $z\neq a$ and then define $h(z)=x\delta(1)(z)+(1-x)\delta(0)(z)$ So $\delta(a)$ is a function that depends on $a$-you can call it alternatively and more clearly $\delta_a$
Feb
5
comment the expectation of a random variable of a random variable
you can start to write h(z)=... explicitly as a function. Also, $E[X] = \int xf_xdx$.
Aug
5
awarded  Citizen Patrol