513 reputation
318
bio website
location Canada
age 37
visits member for 3 years, 6 months
seen May 4 at 23:59

To see Latex in chat: go to http://www.math.ucla.edu/~robjohn/math/mathjax.html Click on bookmark every time I enter the Chat

To hide text for hints: >!


Mar
31
comment Square of absolute value of a function different than square of function
I will think on it later. I was not familiar with complex measurable functions.
Feb
14
comment Intuition/Real-life Examples: Pairwise Independence vs (Mutual) Independence
But this does not explain the intuition. I think any two pair are independent since in addition to what they have in common, there is something extraneous (in our case a different (temporally) toss event) that makes them different. When you put two together, then you know what happened in the two temporal events, and that influences anything here.
Feb
4
comment Prove $A+ \emptyset = A, A+A = \emptyset$, and $A +A' = U$ using the definition of $A+B$
So you need to define at least what is empty set and what is A minus empty set rigorously.
Feb
4
comment Prove $A+ \emptyset = A, A+A = \emptyset$, and $A +A' = U$ using the definition of $A+B$
You did not base your arguments on axioms or some kind of definitions. When you say "This because, if you "throw away" the empty set form A, the result will be again A, whichever A is (we need it again under c))." it is a tautology, i.e. "something is like that because it is like that"
Dec
15
comment What is $(-1)^{\frac{2}{3}}$?
thank you, it makes sense
Dec
15
comment What is $(-1)^{\frac{2}{3}}$?
about 8) we can see it as a function with the codomain C^3, no?
Sep
2
comment Find $f_{X|Y}(x|y)$ given $f_{Y|X}(y|x) = I_{x,x+1}(y)$ and $f_{X}(x) = I_{0,1}(x)$.
To be honest, I calculated $f(y_0)$ and $f(x|y_0)$ through integrals, and then I noticed it is uniform. I think we might have the following result: if $X$ is uniform, then for any other random variable $Y$, $X|y_0$ is uniform.
Sep
2
comment $x_1=0,\,x_{2n}=\frac{x_{2n-1}}{2},\,x_{2n+1}=x_{2n}+\frac{1}{2}$ Find $\lim \sup {x_n}$ and $\lim \inf {x_n}$
I don't see how this helps-it's basically the same thing as the original question
Sep
2
comment trigonometry and integral properties
You actually lack the question. What you show is an equality. I reckon the exercise is "Show/prove that ..."
May
28
comment Sigma algebra generated by a set
@UnadulteratedImagination No, you can have any number of atoms.
Mar
17
comment Show: grad f (x + y) = grad f (x) + grad f (y)
saying x,y are in $R^2$ already implies x,y (hence x+y) are in the domain of $f$ and $g$.
Mar
14
comment Determining the dimensions of the null space and column space of a matrix
"use this to find rank A" can it get more simpler than that?
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?