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  • 0 posts edited
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  • 39 votes cast
Apr
17
awarded  Popular Question
Feb
12
awarded  Popular Question
Oct
30
awarded  Popular Question
Oct
22
asked Change of basis from Chebyshev to monomial basis for polynomials
Jul
2
awarded  Curious
May
30
answered Riemann Integrable
Oct
24
awarded  Yearling
Apr
16
comment If $a_n \rightarrow 1$, does $X_n \overset{p}{\rightarrow} X$ implies $a_n X_n \overset{p}{\rightarrow} X$?
A deterministic sequence is a special case of a random sequence. Each element in the sequence is a random variable with a degenerate (dirac) law.
Apr
14
answered If $a_n \rightarrow 1$, does $X_n \overset{p}{\rightarrow} X$ implies $a_n X_n \overset{p}{\rightarrow} X$?
Apr
3
accepted A special case of integrating over a marginal distribution
Apr
2
revised A special case of integrating over a marginal distribution
added 220 characters in body
Apr
2
comment A special case of integrating over a marginal distribution
Thank you for pointing out that I missed off the $f(z)$. However, I am not sure that the actual fact is wrong. Don't we have $F(x|y) = \int_z F(x|y, z) f(z) dz = \int_z F(x|y', z) f(z) dz = F(x|y')$, where $f$ is the marginal distribution of $z$. The intuition is as follows. By the assumption in the first display $F(x|y, z) = F(x|y', z)$ for all $y, y', z$. Thus if we integrate against the marginal distribution of $z$ on both sides, then equality is preserved.
Apr
2
revised A special case of integrating over a marginal distribution
Added the discrete case.
Apr
2
asked A special case of integrating over a marginal distribution
Feb
24
comment If $A$ and $B$ are $n\times n$ matrices, prove that $|(A^TB)|^2\leq|A^TA||B^TB|$; when is this an equality?
This looks like the Cauchy Schwarz inequality to me.
Feb
21
comment Prove the topology $\mathcal{T} = \{\emptyset\} \cup \{(-\infty, c) | c \in \mathbb{R}\}$ is Tychonoff
Thanks for clearing that up for me!
Feb
21
comment Prove the topology $\mathcal{T} = \{\emptyset\} \cup \{(-\infty, c) | c \in \mathbb{R}\}$ is Tychonoff
Hm, I agree with your intuition @ThomasE.; however, Royden says that "For a continuous mapping $f$ of a topological space $X$ to a topological space $Y$, by the definition of the subspace topology, the restriction of $f$ to a subspace of $X$ is also continuous." The proof is not given. Have I misunderstood this statement? It comes straight after the open set characterization of continuity. I tried to write a proof but the way I tried didn't work.
Feb
21
accepted Prove the topology $\mathcal{T} = \{\emptyset\} \cup \{(-\infty, c) | c \in \mathbb{R}\}$ is Tychonoff
Feb
21
comment Prove the topology $\mathcal{T} = \{\emptyset\} \cup \{(-\infty, c) | c \in \mathbb{R}\}$ is Tychonoff
? = (y-x)/(c-x)? So as $y \to x$, $f(y) \to 0$ and as $y \to c$, $f(y) \to 1$. Then $f$ is continuous on $\mathbb{R}$ with the Euclidean metric and so a restriction of $f$ to this topology is also continuous. But now I wonder about the definition of the Tychonoff separation property on page 227 of Royden/Fitzpatrick.
Feb
21
comment Prove the topology $\mathcal{T} = \{\emptyset\} \cup \{(-\infty, c) | c \in \mathbb{R}\}$ is Tychonoff
Royden's defines a topological space $(X, \mathcal{T})$ to have the Tychonoff propert if "For each two points $u$ and $v$ in $X$, there is a neighborhood of $u$ that does not contain $v$ and a neighborhood of $v$ that does not contain $u$."