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1d
comment If $\bar X$ is open, then $X=\bar X$.
This seems trivially false: consider $E=(0,1)$ in the space $X=[0,1]$ with the topology induced by the usual metric, then $\bar E=X$ is open in $X$ but $E\ne\bar E$.
1d
comment Ternary representation of Cantor set
Aaah... and now your last comment changes completely the question. Not sure I want to play this game, sorry.
1d
comment Ternary representation of Cantor set
This is the definition of the Cantor set: given that you are a number, either you have one such binary expansion then you are in the set, or you have none then you are out.
1d
comment Proving that limit of a sequence is 0 from definitions.
Actually, the formulation "there exists a number N∈Z+ such that given ϵ>0 we have ... for all n≥N" is incorrect (and as such it might justly provoke the wrath of your tutor). The correct phrasing is "for every ϵ>0, there exists a number N∈Z+ such that we have ... for all n≥N".
1d
comment Ternary representation of Cantor set
Proof of what? That "one of these two ternary expansions uses only 0s and 2s"? This is even in your question!
1d
comment Square root algorithm. Rudin PMA ch.3 problem 17
Then consider the logarithms and conclude.
1d
comment Ternary representation of Cantor set
Yeah, and one of these two ternary expansions uses only 0s and 2s hence 1/3 is in the Cantor set. Unambiguously. (Repeating a previous comment above.)
1d
comment Square root algorithm. Rudin PMA ch.3 problem 17
Are you asking why $r^{2^n}$ goes to zero more quickly than $s^n$, for every $r$ and $s$ in $(0,1)$?
1d
comment Ternary representation of Cantor set
What is the question exactly?
1d
comment Physical meaning of autonomous system of ODE.
Continuity is irrelevant. The key property here is time invariance, that is: if two equivalent systems with states $y$ and $\bar y$ are such that $y(t)=\bar y(\bar t)$ for some times $t$ and $\bar t$ then $\bar y(s)=y(s+t-\bar t)$ for every time $s$.
1d
comment Is probability mass function (PMF) the “law of X”?
"event U / event V" No, U and V are measurable subsets of the target set while events are measurable subsets of the source set Omega.
1d
revised What does the conditional expectation look like when the $\sigma$-algebra is infinite
added 25 characters in body
1d
comment Determining bounds for change sum of continuous r.v.'s
The most important remark in this context might be that the condition $0<z<2$ need not be imposed from the start, it comes naturally into play. For example $f_{X,Y}(x,z-x)=4x(z-x)$ if $0<x,z-x<1$ and $0$ elsewhere, with no qualification. Later on, the integral with respect to $x$ is over the interval $\max(0,z-1)<x<\min(1,z)$... which is empty unless $0<z<2$. This makes the approach to solving this kind of question, entirely automatized.
1d
comment Properties of independence and conditional independence
Hmmm... Let me advise you to look for simple examples.
1d
comment What's the sample space for a conditional expectation?
@nomen To begin with, $X_0$ is not defined on $F_0$.
2d
comment Properties of independence and conditional independence
"Sorry, I think what you prove is about Q2." ?? Yes, which is why the comment starts by "Re Q2". "You prove that ... not X⊥A,B" ?? One gets P(X=x∣A=a,B=b)=P(X=x∣A=a)=P(X=x∣B=b)=P(X=x). If this is not "X⊥A,B", then what is?
2d
comment Every closed subset of $\mathbb R^n$ has a point that minimizes the distance to a given point $p\in\mathbb R^n$
@BolzWeir "I wonder how OP solved it." Using that every sequence in a compact set has a converging subsequence (as the OP explained), I guess.
2d
revised Law of Large Numbers - utility/difficulty of various versions.
added 57 characters in body
2d
comment Every closed subset of $\mathbb R^n$ has a point that minimizes the distance to a given point $p\in\mathbb R^n$
@idm Yup. $ $ $ $
2d
comment Properties of independence and conditional independence
Re Q2, you should be able to reach the identity $$P(X=x\mid A=a,B=b)=P(X=x\mid A=a)=P(X=x\mid B=b)$$ for every $(x,a,b)$. Hence, $$P(X=x)=\sum_aP(A=a)P(X=x\mid A=a)=\sum_aP(A=a)P(X=x\mid B=b)=P(X=x\mid B=b).$$ By symmetry, $P(X=x)=P(X=x\mid A=a)$, QED.