How correlation coefficient is Independent of Units of Measurement?

Coradossi7xod
2022-03-27
Answered

How correlation coefficient is Independent of Units of Measurement?

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asked 2022-07-10

Suppose that ${\lambda}_{1}(A)\ge {\lambda}_{2}(A)\ge \cdots \ge {\lambda}_{n}(A)$ are (real) eigenvalues of a Hermitian matrix $A$ and denote the empirical measure by ${L}_{A}:=\frac{1}{n}\sum _{i=1}^{n}{\delta}_{{\lambda}_{i}(A)}$. Let ${F}_{A}$ to be distribution function related to the (counting) measure ${L}_{A}$. We have the following integration by parts formula

$\int g\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}{L}_{A}-\int g\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}{L}_{B}=\int ({F}_{A}-{F}_{B})\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}g,$

in which $g\text{}:\mathbb{R}\to \mathbb{R}$ is a bounded function with bounded total variation. I am curious whether the integration by parts in this case is connected to Abel's summation by parts, and if so, can anyone explicitly present the connection?

$\int g\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}{L}_{A}-\int g\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}{L}_{B}=\int ({F}_{A}-{F}_{B})\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}g,$

in which $g\text{}:\mathbb{R}\to \mathbb{R}$ is a bounded function with bounded total variation. I am curious whether the integration by parts in this case is connected to Abel's summation by parts, and if so, can anyone explicitly present the connection?

asked 2022-05-24

Let $({X}_{n}{)}_{n\ge 0}$ be a sequence of random variables and $\tau ,t$ stopping times with respect to the sequence $({X}_{n}{)}_{n\ge 0}$

$\begin{array}{rl}\{\tau +t=n\}=\{\tau +t=n\}\cap \{t\le n\}& =\bigcup _{k=0}^{n}\{\tau +t=n\}\cap \{t=k\}\\ & =\bigcup _{k=0}^{n}\{\tau =n-k\}\cap \{t=k\}.\end{array}$

As $\{\tau =n-k\}\in {\mathcal{F}}_{n-k}\subseteq {\mathcal{F}}_{n}$ and $\{t=k\}\in {\mathcal{F}}_{k}\subseteq {\mathcal{F}}_{n}$ for any $k\le n$, this implies that $\{\tau +t=n\}\in {\mathcal{F}}_{n}$, and so $\tau +t$ is a stopping time.

Now, my question is, let assume that I am considering $\tau -t$ I know that in general, $\tau -t$ is not a stopping time. However, if I were to consider my birthday this year (a stopping time), which is a deterministic stopping time. At any time, I know exactly when my birthday occurs. Also, I know two days before my birthday i.e, $\tau -2$. What kind of a formulated counterexample will show that $\tau -2$ is indeed a stopping time in this setting.

$\begin{array}{rl}\{\tau +t=n\}=\{\tau +t=n\}\cap \{t\le n\}& =\bigcup _{k=0}^{n}\{\tau +t=n\}\cap \{t=k\}\\ & =\bigcup _{k=0}^{n}\{\tau =n-k\}\cap \{t=k\}.\end{array}$

As $\{\tau =n-k\}\in {\mathcal{F}}_{n-k}\subseteq {\mathcal{F}}_{n}$ and $\{t=k\}\in {\mathcal{F}}_{k}\subseteq {\mathcal{F}}_{n}$ for any $k\le n$, this implies that $\{\tau +t=n\}\in {\mathcal{F}}_{n}$, and so $\tau +t$ is a stopping time.

Now, my question is, let assume that I am considering $\tau -t$ I know that in general, $\tau -t$ is not a stopping time. However, if I were to consider my birthday this year (a stopping time), which is a deterministic stopping time. At any time, I know exactly when my birthday occurs. Also, I know two days before my birthday i.e, $\tau -2$. What kind of a formulated counterexample will show that $\tau -2$ is indeed a stopping time in this setting.

asked 2022-03-23

Four student measured the same volume of a glass of water. They all made three trials and summarized their results in the given table. Suppose the accepted volume of the glass of water is 4.20 mL. Analyze the table below and answer the following questions.

5. Who has a precise but not accurate measurement? Why?

6. Who has an accurate but not precise measurement? Why?

7. Who has an accurate and precise measurement? Why?

$\begin{array}{cc}& \text{Volume of Water in the Glass (mL)}\\ \text{}& \begin{array}{|ccccc|}\hline \text{Trial}& \text{Andrew}& \text{William}& \text{Diane}& \text{Lianne}\\ 1& 6.90& 7.60& 4.32& 4.24\\ 2& 5.01& 7.62& 4.30& 4.26\\ 3& 8.80& 7.65& 4.29& 4.25\\ \hline\end{array}\text{}\end{array}$

5. Who has a precise but not accurate measurement? Why?

6. Who has an accurate but not precise measurement? Why?

7. Who has an accurate and precise measurement? Why?

asked 2022-05-20

Guiding question:Should measure theory be learned before functional analysis or should it be the other way around?

Perhaps there is no largely agreed upon answer to this so I'll ask:

More specific question: What connections are there between the two subjects that might make a person choose to study one before the next?

All feedback is appreciated.

Perhaps there is no largely agreed upon answer to this so I'll ask:

More specific question: What connections are there between the two subjects that might make a person choose to study one before the next?

All feedback is appreciated.

asked 2022-05-14

We want to prove the following implication :

$A\in {\mathcal{F}}_{S}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}A\cap \{S\le T\}\in {\mathcal{F}}_{T}$

Since $A\in {\mathcal{F}}_{S}$, we have $A\in {\mathcal{F}}_{\mathrm{\infty}}$ and $A\cap \{S\le t\}\in {\mathcal{F}}_{t}$.

We will then show that

$A\cap \{S\le T\}\in {\mathcal{F}}_{T}$

Meaning :

$A\cap \{S\le T\}\cap \{T\le t\}\in {\mathcal{F}}_{t}$

We have :

$A\cap \{S\le T\}\cap \{T\le t\}=A\cap \{S\le t\}\cap \{T\le t\}\cap \{S\wedge t\le T\wedge t\}$

The ''chosen'' set $\{S\wedge t\le T\wedge t\}$, according to the book, guarantees that the stopping time $S$ will be less or equal to $T$, and at the same time, the chosen set is in fact ${\mathcal{F}}_{t}$-measurable, which helps a lot in the proof

It's not very clear to me where it did come from.

I understand that

$\{S\le T\}\ne \{S\le t\}\cap \{T\le t\},\phantom{\rule{1cm}{0ex}}\mathrm{\forall}t\in {\mathbb{R}}_{+}$

but the exact choice of $\{S\wedge t\le T\wedge t\}$ is still not obvious.

$A\in {\mathcal{F}}_{S}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}A\cap \{S\le T\}\in {\mathcal{F}}_{T}$

Since $A\in {\mathcal{F}}_{S}$, we have $A\in {\mathcal{F}}_{\mathrm{\infty}}$ and $A\cap \{S\le t\}\in {\mathcal{F}}_{t}$.

We will then show that

$A\cap \{S\le T\}\in {\mathcal{F}}_{T}$

Meaning :

$A\cap \{S\le T\}\cap \{T\le t\}\in {\mathcal{F}}_{t}$

We have :

$A\cap \{S\le T\}\cap \{T\le t\}=A\cap \{S\le t\}\cap \{T\le t\}\cap \{S\wedge t\le T\wedge t\}$

The ''chosen'' set $\{S\wedge t\le T\wedge t\}$, according to the book, guarantees that the stopping time $S$ will be less or equal to $T$, and at the same time, the chosen set is in fact ${\mathcal{F}}_{t}$-measurable, which helps a lot in the proof

It's not very clear to me where it did come from.

I understand that

$\{S\le T\}\ne \{S\le t\}\cap \{T\le t\},\phantom{\rule{1cm}{0ex}}\mathrm{\forall}t\in {\mathbb{R}}_{+}$

but the exact choice of $\{S\wedge t\le T\wedge t\}$ is still not obvious.

asked 2022-04-21

In a study, the data you collect is wage per hour. What is the level of measurement?

a. ratio

b. nominal

c. ordinal

4. interval

a. ratio

b. nominal

c. ordinal

4. interval

asked 2022-06-17

State the number of possible triangles that can be formed using the given measurements, then sketch and solve the triangles, if possible.

$m\mathrm{\angle}A={48}^{\circ},c=29,a=4$

$m\mathrm{\angle}A={48}^{\circ},c=29,a=4$