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 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 $u$."