How to measure 1/3 cup without measuring cup?

mangicele4s
2022-09-17
Answered

How to measure 1/3 cup without measuring cup?

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asked 2022-05-14

I'm using a bayesian framework to estimate the pose of a moving vehicle with a single camera. The state vector of the vehicle is

$X=\left[\begin{array}{c}x\\ y\\ \theta \end{array}\right]$

where x and y are the coordinates in the plan and $\theta $ is the heading angle. The state vector X is constatntly updated in the prediction step.

In the correction step, an image retrieval algorithm compares the current image to a database of images by a measure based on pixel by pixel differences. Then, it returns the best match which is an image of index $j\in N$ together with its pose, the measurement

$z=\left[\begin{array}{c}{x}_{m}\\ {y}_{m}\\ {\theta}_{m}\end{array}\right]$

Let's say my current image is actually the image of index j=1000 but the algorithm returns j=1005, how to evaluate the uncertainty? How to calculate the covariance matrix of this measurement?

$X=\left[\begin{array}{c}x\\ y\\ \theta \end{array}\right]$

where x and y are the coordinates in the plan and $\theta $ is the heading angle. The state vector X is constatntly updated in the prediction step.

In the correction step, an image retrieval algorithm compares the current image to a database of images by a measure based on pixel by pixel differences. Then, it returns the best match which is an image of index $j\in N$ together with its pose, the measurement

$z=\left[\begin{array}{c}{x}_{m}\\ {y}_{m}\\ {\theta}_{m}\end{array}\right]$

Let's say my current image is actually the image of index j=1000 but the algorithm returns j=1005, how to evaluate the uncertainty? How to calculate the covariance matrix of this measurement?

asked 2022-06-16

Let $\{{f}_{n}\}$ be a sequence of measurable functions on a domain $E$ and p be a positive finite real number such that

(a) $\{{f}_{n}\}$ converges to a measurable function $f$ almost everywhere, and;

(b) $\underset{n\to \mathrm{\infty}}{lim}(||{f}_{n}|{|}_{p}-||f|{|}_{p})=0.$

Prove that $||{f}_{n}-f|{|}_{p}\to 0.$

I believe what we have to show is that

$\underset{n\to \mathrm{\infty}}{lim}{\int}_{E}|{f}_{n}-f{|}^{p}=0.$

My approach is to hope to invoke the Dominated Convergence theorem, i.e., prove that there exists some $g\in {L}^{1}(E)$ such that $|{f}_{n}|\le g$ for almost everywhere for all $n$, then we can have

${\int}_{E}|{f}_{n}-f|\to 0.$

Then by (a), we can find an integer $K$ such that

$|{f}_{k}-f|<1$

for all $k\ge K$. Then

$0\le {\int}_{E}|{f}_{k}-f{|}^{p}\le {\int}_{E}|{f}_{k}-f|.$

Since the integral on the right approaches 0, we conclude that the same is true for the middle term and we will be done.

However, I have no idea how to obtain the $g$ as mentioned before, as well as use condition (b) given in the question. Some help would be appreciated.

(a) $\{{f}_{n}\}$ converges to a measurable function $f$ almost everywhere, and;

(b) $\underset{n\to \mathrm{\infty}}{lim}(||{f}_{n}|{|}_{p}-||f|{|}_{p})=0.$

Prove that $||{f}_{n}-f|{|}_{p}\to 0.$

I believe what we have to show is that

$\underset{n\to \mathrm{\infty}}{lim}{\int}_{E}|{f}_{n}-f{|}^{p}=0.$

My approach is to hope to invoke the Dominated Convergence theorem, i.e., prove that there exists some $g\in {L}^{1}(E)$ such that $|{f}_{n}|\le g$ for almost everywhere for all $n$, then we can have

${\int}_{E}|{f}_{n}-f|\to 0.$

Then by (a), we can find an integer $K$ such that

$|{f}_{k}-f|<1$

for all $k\ge K$. Then

$0\le {\int}_{E}|{f}_{k}-f{|}^{p}\le {\int}_{E}|{f}_{k}-f|.$

Since the integral on the right approaches 0, we conclude that the same is true for the middle term and we will be done.

However, I have no idea how to obtain the $g$ as mentioned before, as well as use condition (b) given in the question. Some help would be appreciated.

asked 2022-05-28

For a measure space $(\mathrm{\Omega},\mathcal{F},\mu )$ : $\mu $ is positve and finite, show the following: The set function

$\nu (A)={\int}_{A}fd\mu $

defines a new positive measure on $\mathcal{F}$ equivalent to $\mu $, iff $f$ is strictly positive $\mu $-a.e and $\mu $-integrable.

The forward direction obviously holds true from the Radon-Nikodym theorem. I'm having trouble proving the last tip of the reverse direction. Starting with a funcion $f$ as described above, It's not too hard to show that $\nu $ is indeed a measure. It's also trivial that $\nu \ll \mu $. But how can we rigorously prove that $\mu \ll \nu $ ? I'm sure we've all used the argument that when for a strictly positive $f$, it holds:

${\int}_{A}fd\mu =0$, then $\Rightarrow $$\mu (A)=0$,

but what would be a formal and stringent proof of this property/claim ?

$\nu (A)={\int}_{A}fd\mu $

defines a new positive measure on $\mathcal{F}$ equivalent to $\mu $, iff $f$ is strictly positive $\mu $-a.e and $\mu $-integrable.

The forward direction obviously holds true from the Radon-Nikodym theorem. I'm having trouble proving the last tip of the reverse direction. Starting with a funcion $f$ as described above, It's not too hard to show that $\nu $ is indeed a measure. It's also trivial that $\nu \ll \mu $. But how can we rigorously prove that $\mu \ll \nu $ ? I'm sure we've all used the argument that when for a strictly positive $f$, it holds:

${\int}_{A}fd\mu =0$, then $\Rightarrow $$\mu (A)=0$,

but what would be a formal and stringent proof of this property/claim ?

asked 2022-04-28

Level of Measurement What is the level of measurement of these data (nominal, ordinal, interval, ratio)? Are the original unrounded arrival times continuous data or discrete data?

-30 -23 14 -21 -32 11 -23 28 103 -19 -5 -46

-30 -23 14 -21 -32 11 -23 28 103 -19 -5 -46

asked 2022-05-21

If we take a Cartesian system with length over length then the vector magnitude has length as unit of measurement. I get that. Let's say we have cups of coffee as the x axis and cost in dollars as the y axis. If we take a point (1cup, 1dollar) and we create the point vector $u=\{1\text{}cup,1\text{}dollar\}$ then we can calculate the vector length which is: $|u|=\sqrt{{1}^{2}\text{}cup+{1}^{2}\text{}dollar}$.

1. what is the result?

2. what unit of measurement will it have?

3. how can we interpret the vector length with that unit of measurement? If the result is $\sqrt{2}\ast \sqrt{cu{p}^{2}+dolla{r}^{2}}$ how can we interpret this?

1. what is the result?

2. what unit of measurement will it have?

3. how can we interpret the vector length with that unit of measurement? If the result is $\sqrt{2}\ast \sqrt{cu{p}^{2}+dolla{r}^{2}}$ how can we interpret this?

asked 2022-05-21

Let $(X,\mathcal{A},\mu )$ be a measure space and we consider a measurable positive function $f:X\to [0,+\mathrm{\infty}]$. I already proved that if the Lebesgue integral of $f$ on $X$ is finite, that is

${\int}_{X}f\phantom{\rule{thickmathspace}{0ex}}\text{d}\mu <+\mathrm{\infty},$

then $\mu (\{f=+\mathrm{\infty}\})=0$, that is $f$ is finite almost everywhere.

Now, let $f:X\to [-\mathrm{\infty},+\mathrm{\infty}]$ an $\mu $- integrable function, that is

${\int}_{X}{f}_{+}\phantom{\rule{thickmathspace}{0ex}}\text{d}\mu <+\mathrm{\infty}\phantom{\rule{1em}{0ex}}{\int}_{X}{f}_{-}\phantom{\rule{thickmathspace}{0ex}}\text{d}\mu <+\mathrm{\infty},$

then

${\int}_{X}|f|\phantom{\rule{thickmathspace}{0ex}}\text{d}\mu ={\int}_{X}({f}_{+}+{f}_{-})\phantom{\rule{thickmathspace}{0ex}}\text{d}\mu <\mathrm{\infty},$

and therefore $|f|$ is finite ae, so is a measurable positive function.

From this can I conclude that $f$ itself is finite almost everywhere?

Notation.

${f}_{+}:=max\{f(x),0\}$ and ${f}_{-}:=max\{-f(x),0\}$

${\int}_{X}f\phantom{\rule{thickmathspace}{0ex}}\text{d}\mu <+\mathrm{\infty},$

then $\mu (\{f=+\mathrm{\infty}\})=0$, that is $f$ is finite almost everywhere.

Now, let $f:X\to [-\mathrm{\infty},+\mathrm{\infty}]$ an $\mu $- integrable function, that is

${\int}_{X}{f}_{+}\phantom{\rule{thickmathspace}{0ex}}\text{d}\mu <+\mathrm{\infty}\phantom{\rule{1em}{0ex}}{\int}_{X}{f}_{-}\phantom{\rule{thickmathspace}{0ex}}\text{d}\mu <+\mathrm{\infty},$

then

${\int}_{X}|f|\phantom{\rule{thickmathspace}{0ex}}\text{d}\mu ={\int}_{X}({f}_{+}+{f}_{-})\phantom{\rule{thickmathspace}{0ex}}\text{d}\mu <\mathrm{\infty},$

and therefore $|f|$ is finite ae, so is a measurable positive function.

From this can I conclude that $f$ itself is finite almost everywhere?

Notation.

${f}_{+}:=max\{f(x),0\}$ and ${f}_{-}:=max\{-f(x),0\}$

asked 2022-05-27

I have to prove the following:

Let $X,Y$ be countable sets. Show that $\mathcal{P}(X)\otimes \mathcal{P}(Y)=\mathcal{P}(X\times Y)$.

I'm not sure if in my case, $\mathcal{P}(X)\otimes \mathcal{P}(Y)$ is a product $\sigma $-algebra and thus defined as $\mathcal{P}(X)\otimes \mathcal{P}(Y)=\sigma (\{{B}_{1}\times {B}_{2}|{B}_{1}\in \mathcal{P}(X),{B}_{2}\in \mathcal{P}(Y)\})$.

I know that the power set of a set is a $\sigma $-algebra. So, it would make sense that $\mathcal{P}(X)\otimes \mathcal{P}(Y)$ is a product $\sigma $-algebra as defined above.

Can somebody confirm this or explain it if it's not the case?

Let $X,Y$ be countable sets. Show that $\mathcal{P}(X)\otimes \mathcal{P}(Y)=\mathcal{P}(X\times Y)$.

I'm not sure if in my case, $\mathcal{P}(X)\otimes \mathcal{P}(Y)$ is a product $\sigma $-algebra and thus defined as $\mathcal{P}(X)\otimes \mathcal{P}(Y)=\sigma (\{{B}_{1}\times {B}_{2}|{B}_{1}\in \mathcal{P}(X),{B}_{2}\in \mathcal{P}(Y)\})$.

I know that the power set of a set is a $\sigma $-algebra. So, it would make sense that $\mathcal{P}(X)\otimes \mathcal{P}(Y)$ is a product $\sigma $-algebra as defined above.

Can somebody confirm this or explain it if it's not the case?