Draw, in standard position, the angle whose measure is given:

$\frac{4\pi}{3}$

$\frac{4\pi}{3}$

cooloicons62
2022-07-16
Answered

Draw, in standard position, the angle whose measure is given:

$\frac{4\pi}{3}$

$\frac{4\pi}{3}$

You can still ask an expert for help

Alexia Hart

Answered 2022-07-17
Author has **19** answers

$\frac{4\pi}{3}\text{}rad$

Answer:

$\frac{4}{3}=1\frac{1}{3}$

Answer:

$\frac{4}{3}=1\frac{1}{3}$

Jeffrey Jordon

Answered 2022-07-26
Author has **2581** answers

asked 2022-03-04

A lab chemist would like to know if 2 methods produce the exact measurements. To do this, the chemist sampled solutions from a normal population.

A. Dependent

B. Indipendent

A. Dependent

B. Indipendent

asked 2022-07-04

We can consider the sequence space ${\mathbb{R}}^{\mathbb{Z}}$, which becomes a Polish space (complete, separable metric space) when equipped with the product topology. In the theory of stochastic processes, one is frequently interested in probability measures on this space.

A classical result on measure theory in Polish spaces tells us that any probability measure $\mu $ on a Polish space $X$ is tight: given $\u03f5>0$, then there is a compact set ${K}_{\u03f5}\subseteq X$ such that $\mu ({K}_{\u03f5})>1-\u03f5$.

Hence any product probability measure on ${\mathbb{R}}^{\mathbb{Z}}$ is tight.

Say we fix an i.i.d. probability measure $\mu $ on ${\mathbb{R}}^{\mathbb{Z}}$. Precisely, $\mu (\prod {A}_{i})=\prod \nu ({A}_{i})$ for some probability measure $\nu $ on $\mathbb{R}$.

Is there a nice, concrete proof that $\mu $ is tight (independent of $\nu $)?

The reason I ask is tightness is somewhat counter-intuitive here. For instance, if $\nu $ has unbounded support, then, for each $M>0$, the set $\{k\in \mathbb{Z}\phantom{\rule{thinmathspace}{0ex}}\mid \phantom{\rule{thinmathspace}{0ex}}|{x}_{k}|\ge M\}$ is infinite almost surely under $\mu $ (by the 0-1 law). Thinking along those lines, it's not easy to see ${K}_{\u03f5}$ intuitively.

A classical result on measure theory in Polish spaces tells us that any probability measure $\mu $ on a Polish space $X$ is tight: given $\u03f5>0$, then there is a compact set ${K}_{\u03f5}\subseteq X$ such that $\mu ({K}_{\u03f5})>1-\u03f5$.

Hence any product probability measure on ${\mathbb{R}}^{\mathbb{Z}}$ is tight.

Say we fix an i.i.d. probability measure $\mu $ on ${\mathbb{R}}^{\mathbb{Z}}$. Precisely, $\mu (\prod {A}_{i})=\prod \nu ({A}_{i})$ for some probability measure $\nu $ on $\mathbb{R}$.

Is there a nice, concrete proof that $\mu $ is tight (independent of $\nu $)?

The reason I ask is tightness is somewhat counter-intuitive here. For instance, if $\nu $ has unbounded support, then, for each $M>0$, the set $\{k\in \mathbb{Z}\phantom{\rule{thinmathspace}{0ex}}\mid \phantom{\rule{thinmathspace}{0ex}}|{x}_{k}|\ge M\}$ is infinite almost surely under $\mu $ (by the 0-1 law). Thinking along those lines, it's not easy to see ${K}_{\u03f5}$ intuitively.

asked 2022-08-22

A tree is 57 inches tall. How tall is it in feet and inches?

asked 2022-05-21

I am strugling with this exercise:

Let ${f}_{n}(t)=\mathrm{sin}(nt)$. Prove that ${f}_{n}\rightharpoonup \ast 0$ in ${L}^{\mathrm{\infty}}[0,2\pi ]$

The solution given to me reads:

Since the given space is the dual of ${L}^{1}[0,2\pi ]:$ ${L}^{\mathrm{\infty}}[0,2\pi ]=({L}^{1}[0,2\pi ]{)}^{\prime}$

Thinking of the ${f}_{n}(t)$ as linear functionals with domain in ${L}^{1}[0,2\pi ]$

Prove that ${f}_{n}(g)={\int}_{0}^{2\pi}g(t){f}_{n}(t)dt\to 0$ for all $g\in {L}^{1}[0,2\pi ]$

It is enough to prove it holds for $g$ in a set whose span is dense in ${L}^{1}[0,2\pi ]$, like the $\mathrm{\Delta}=\{{1}_{[a,b]}:[a,b]\subset [0,2\pi ]\}$.

That is straightforward since I can easily integrate sin(nt) over [a,b]and notice it converges to 0, but I don't understand why/ I am not convinced that it is enough to do so. I have read elsewhere that " a sequence of continuous functions on a metric space that converges pointwise on a dense subset need not converge pointwise on the full space.",

So why does it work in this particular case? Can you prove it?

I have the following proven propositions, but none of them are exactly what I need, since this problem deals with weak * convergence instead

Proposition 1: X: Banach space, $E\subset X$, such that $\overline{(SpanE)}=X$. Let $\{{g}_{n}\}\subset {X}^{\prime}$, bounded such that $\mathrm{\exists}\underset{n\to \mathrm{\infty}}{lim}{g}_{n}(x)$ $\mathrm{\forall}x\in E$. Then $\mathrm{\exists}g\in {X}^{\prime}$ such that $\underset{n\to \mathrm{\infty}}{lim}{g}_{n}(x)=g(x)$ $\mathrm{\forall}x\in X$

which was actually used to prove:

Proposition 2: X: Banach space, ${x}_{n}\rightharpoonup x$ iff $\{{x}_{n}\}$ is bounded and $\underset{n\to \mathrm{\infty}}{lim}f({x}_{n})=f(x)$ $\mathrm{\forall}f\in \mathrm{\Delta}\subset {X}^{\prime}$ 'such that $\overline{(Span\mathrm{\Delta})}={X}^{\prime}$

Let ${f}_{n}(t)=\mathrm{sin}(nt)$. Prove that ${f}_{n}\rightharpoonup \ast 0$ in ${L}^{\mathrm{\infty}}[0,2\pi ]$

The solution given to me reads:

Since the given space is the dual of ${L}^{1}[0,2\pi ]:$ ${L}^{\mathrm{\infty}}[0,2\pi ]=({L}^{1}[0,2\pi ]{)}^{\prime}$

Thinking of the ${f}_{n}(t)$ as linear functionals with domain in ${L}^{1}[0,2\pi ]$

Prove that ${f}_{n}(g)={\int}_{0}^{2\pi}g(t){f}_{n}(t)dt\to 0$ for all $g\in {L}^{1}[0,2\pi ]$

It is enough to prove it holds for $g$ in a set whose span is dense in ${L}^{1}[0,2\pi ]$, like the $\mathrm{\Delta}=\{{1}_{[a,b]}:[a,b]\subset [0,2\pi ]\}$.

That is straightforward since I can easily integrate sin(nt) over [a,b]and notice it converges to 0, but I don't understand why/ I am not convinced that it is enough to do so. I have read elsewhere that " a sequence of continuous functions on a metric space that converges pointwise on a dense subset need not converge pointwise on the full space.",

So why does it work in this particular case? Can you prove it?

I have the following proven propositions, but none of them are exactly what I need, since this problem deals with weak * convergence instead

Proposition 1: X: Banach space, $E\subset X$, such that $\overline{(SpanE)}=X$. Let $\{{g}_{n}\}\subset {X}^{\prime}$, bounded such that $\mathrm{\exists}\underset{n\to \mathrm{\infty}}{lim}{g}_{n}(x)$ $\mathrm{\forall}x\in E$. Then $\mathrm{\exists}g\in {X}^{\prime}$ such that $\underset{n\to \mathrm{\infty}}{lim}{g}_{n}(x)=g(x)$ $\mathrm{\forall}x\in X$

which was actually used to prove:

Proposition 2: X: Banach space, ${x}_{n}\rightharpoonup x$ iff $\{{x}_{n}\}$ is bounded and $\underset{n\to \mathrm{\infty}}{lim}f({x}_{n})=f(x)$ $\mathrm{\forall}f\in \mathrm{\Delta}\subset {X}^{\prime}$ 'such that $\overline{(Span\mathrm{\Delta})}={X}^{\prime}$

asked 2022-06-20

I am wondering how the Radon-Nikodym derivative is affected by push-forwarding with a random variable. Formally, let $(\mathrm{\Omega},\mathcal{F})$ be a measurable space, and $\mathbb{P}$ and $\mathbb{Q}$ be two probability measures on this space such that $\mathbb{Q}\ll \mathbb{P}$. Radon-Nikodym theorem tells us that there exists a $\mathcal{F}$-measurable function $f:\mathrm{\Omega}\to [0,\mathrm{\infty})$ denoted by $\frac{d\mathbb{Q}}{d\mathbb{P}}$, such that

$\mathbb{Q}(E)={\int}_{E}f(\omega )d\mathbb{P}(\omega )$

for all $E\in \mathcal{F}$.

If somehow we know the expression for $f(\omega )$ already, and $X:\mathrm{\Omega}\to {\mathbb{R}}^{d}$ is a random vector, is there a way to derive the Radon-Nikodym derivative $\frac{d{\mathbb{Q}}^{X}}{d{\mathbb{P}}^{X}}$ for the induced distribution measures ${\mathbb{Q}}^{X}$ and ${\mathbb{P}}^{X}$ on $({\mathbb{R}}^{d},\mathcal{B}({\mathbb{R}}^{d}))$? My rough guess was $\frac{d{\mathbb{Q}}^{X}}{d{\mathbb{P}}^{X}}(X(\omega ))=f(\omega )$ but am not sure how to prove/disprove it.

$\mathbb{Q}(E)={\int}_{E}f(\omega )d\mathbb{P}(\omega )$

for all $E\in \mathcal{F}$.

If somehow we know the expression for $f(\omega )$ already, and $X:\mathrm{\Omega}\to {\mathbb{R}}^{d}$ is a random vector, is there a way to derive the Radon-Nikodym derivative $\frac{d{\mathbb{Q}}^{X}}{d{\mathbb{P}}^{X}}$ for the induced distribution measures ${\mathbb{Q}}^{X}$ and ${\mathbb{P}}^{X}$ on $({\mathbb{R}}^{d},\mathcal{B}({\mathbb{R}}^{d}))$? My rough guess was $\frac{d{\mathbb{Q}}^{X}}{d{\mathbb{P}}^{X}}(X(\omega ))=f(\omega )$ but am not sure how to prove/disprove it.

asked 2022-05-29

Let $\mu $ be a complex measure on ${\mathbb{R}}^{n}$ and $f,g\in {L}^{1}(\mu )$ such that $0<f(x)\le g(x)$ for a.e. $x\in {\mathbb{R}}^{n}$. Then is it true that

$|}{\int}_{{\mathbb{R}}^{n}}fd\mu {\textstyle |}\le {\textstyle |}{\int}_{{\mathbb{R}}^{n}}gd\mu {\textstyle |}\text{?$

I was trying to break each integral first in its real and imaginary part and then their corresponding positive and negative parts, i.e.

${\int}_{{\mathbb{R}}^{n}}fd\mu =Re{\textstyle (}{\int}_{{\mathbb{R}}^{n}}fd\mu {\textstyle )}+iIm{\textstyle (}{\int}_{{\mathbb{R}}^{n}}fd\mu {\textstyle )}$

$=R{e}^{+}{\textstyle (}{\int}_{{\mathbb{R}}^{n}}fd\mu {\textstyle )}-R{e}^{-}{\textstyle (}{\int}_{{\mathbb{R}}^{n}}fd\mu {\textstyle )}+iI{m}^{+}{\textstyle (}{\int}_{{\mathbb{R}}^{n}}fd\mu {\textstyle )}-iI{m}^{-}{\textstyle (}{\int}_{{\mathbb{R}}^{n}}fd\mu {\textstyle )}$

and then wanted to look at corresponding decompositions of the measure and the integral ${\int}_{{\mathbb{R}}^{n}}gd\mu $ and obtain inequalities. But I can't figure them out and proceed from here.

Can someone please help?

$|}{\int}_{{\mathbb{R}}^{n}}fd\mu {\textstyle |}\le {\textstyle |}{\int}_{{\mathbb{R}}^{n}}gd\mu {\textstyle |}\text{?$

I was trying to break each integral first in its real and imaginary part and then their corresponding positive and negative parts, i.e.

${\int}_{{\mathbb{R}}^{n}}fd\mu =Re{\textstyle (}{\int}_{{\mathbb{R}}^{n}}fd\mu {\textstyle )}+iIm{\textstyle (}{\int}_{{\mathbb{R}}^{n}}fd\mu {\textstyle )}$

$=R{e}^{+}{\textstyle (}{\int}_{{\mathbb{R}}^{n}}fd\mu {\textstyle )}-R{e}^{-}{\textstyle (}{\int}_{{\mathbb{R}}^{n}}fd\mu {\textstyle )}+iI{m}^{+}{\textstyle (}{\int}_{{\mathbb{R}}^{n}}fd\mu {\textstyle )}-iI{m}^{-}{\textstyle (}{\int}_{{\mathbb{R}}^{n}}fd\mu {\textstyle )}$

and then wanted to look at corresponding decompositions of the measure and the integral ${\int}_{{\mathbb{R}}^{n}}gd\mu $ and obtain inequalities. But I can't figure them out and proceed from here.

Can someone please help?

asked 2022-05-21

Definition:

Let $(X,\mathcal{A},\mu )$ be a measurable space, an atom of the measure $\mu $ is a set $A\in \mathcal{A}$ with the property that $\mu (A)>0$ and for any $B\in \sigma (A)$ either $\mu (B)=0$, or $\mu (A\setminus B)=0$. If a measure has atoms it is called atomic; in the opposite case, the measure is called non-atomic (or atomeless). A measure is called purely atomic if $X$ can be written as the union of a finite or countable number of atoms.

From the definition of atoms, we get the following corollary:

Corollary:

Every purely atomic measure is an atomic measure.

I am trying to find an example of an atomic measure that is not purely atomic, can anyone help me?

Let $(X,\mathcal{A},\mu )$ be a measurable space, an atom of the measure $\mu $ is a set $A\in \mathcal{A}$ with the property that $\mu (A)>0$ and for any $B\in \sigma (A)$ either $\mu (B)=0$, or $\mu (A\setminus B)=0$. If a measure has atoms it is called atomic; in the opposite case, the measure is called non-atomic (or atomeless). A measure is called purely atomic if $X$ can be written as the union of a finite or countable number of atoms.

From the definition of atoms, we get the following corollary:

Corollary:

Every purely atomic measure is an atomic measure.

I am trying to find an example of an atomic measure that is not purely atomic, can anyone help me?