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 Apr17 awarded Popular Question Feb12 awarded Popular Question Oct30 awarded Popular Question Oct22 asked Change of basis from Chebyshev to monomial basis for polynomials Jul2 awarded Curious May30 answered Riemann Integrable Oct24 awarded Yearling Apr16 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. Apr14 answered If $a_n \rightarrow 1$, does $X_n \overset{p}{\rightarrow} X$ implies $a_n X_n \overset{p}{\rightarrow} X$? Apr3 accepted A special case of integrating over a marginal distribution Apr2 revised A special case of integrating over a marginal distribution added 220 characters in body Apr2 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. Apr2 revised A special case of integrating over a marginal distribution Added the discrete case. Apr2 asked A special case of integrating over a marginal distribution Feb24 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. Feb21 comment Prove the topology $\mathcal{T} = \{\emptyset\} \cup \{(-\infty, c) | c \in \mathbb{R}\}$ is Tychonoff Thanks for clearing that up for me! Feb21 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. Feb21 accepted Prove the topology $\mathcal{T} = \{\emptyset\} \cup \{(-\infty, c) | c \in \mathbb{R}\}$ is Tychonoff Feb21 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. Feb21 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 $u$."