Improve Your Algebra Skills with Practice Problems

Suppose $X$ is a real random variable with some density function $f_X(x)$. Let $\mathbf{Y}=\mathbf{g}(X)$, where $\mathbf{g}(.):\mathbb{R}\to <!--\; \to \; -->{\mathbb{R}}^n$.
Is there a general way to represent the density $f_{\mathbf{Y}}(\mathbf{y})$?
For example, consider a simple case, $X\sim <!--\; \sim \; -->Uniform(0,1)$. $(Y_1,Y_2)=(X+1,X+2)$. Is there a closed form expression for the density of $(Y_1,Y_2)$ in ${\mathbb{R}}^2$?
Edit:
Dirac measures are permitted. In the above simple case, I presume the density could be
$f_{(Y_1,Y_2)}(y_1,y_2)=\{\begin{array}{lll}\mathrm{\delta <!--\; \delta \; -->}(y_1-<!--\; -\; -->y_2+1)& & 1\le <!--\; \le \; -->y_1\le <!--\; \le \; -->2\text{ and }2\le <!--\; \le \; -->y_2\le <!--\; \le \; -->3\\ 0& & \text{else}\end{array}$

Referring to the Doob's Maximal inequality we have that letting $M_t$ be a continous martingale, $\mathrm{\lambda <!--\; \lambda \; -->}>0$, $p>1$ and $T>0$ we have $P(\underset{0\le <!--\; \le \; -->t\le <!--\; \le \; -->T}{sup}|M_t|>\mathrm{\lambda <!--\; \lambda \; -->})\le <!--\; \le \; -->\frac{1}{{\mathrm{\lambda <!--\; \lambda \; -->}}^p}\mathbb{E}[|M_T{|}^p]$. I did not understand if the sup is the sup of the events ($|M_t|>\mathrm{\lambda <!--\; \lambda \; -->}$ : $t\in <!--\; \in \; -->[0,T]$) or if it is the sup of the random variable $|M_t|$ itself.

How do you write the expression for A number y added to 2 is greater than 7?

An equivalence relation is a binary relation that fulfills reflexivity, symmetry and transitivity.
A partially ordered set is a set equipped with a binary relation that fulfills reflexivity, antisymmetry and transitivity.
For a partially ordered set, some elements of the set might be incomparable, i.e. $x\le <!--\; \le \; -->y$ or $y\le <!--\; \le \; -->x$ may both not hold. That's why we use the term"partially". An example would be $\subset <!--\; \subset \; -->$ as a binary relation and the set M:={{1},{2},{1,2}}. Obviously, {1} and {2} are incomparable.
Question: Can also elements of a set, equipped with an equivalence relation, be incomparable?

Reading this document, when measuring a thing we have
$\text{relative error}=\frac{\text{absolute error}}{\text{value of thing measured}}.$
Let's say we have a measurement of the height of an object as 45.2m with error of $\pm <!--\; \pm \; -->0.05m$.
At that point our relative error would be $\frac{0.05m}{45.2m}$.
However, if a person who measures heights always rounds his measurement to three significant digits would it change the relative error?
I am not sure if this is correct, but in other words would the answer 45.2m+0.05m=45.25m rounded to 45.3m as the approximate value. After that would th e absolute error be 0.1m, so that the relative error would be $\frac{0.1}{45.2}$?

I am wondering if there is a basic unit in math.
$3cm\times <!--\; \times \; -->2cm=6c{m}^2$
The cm is a unit of measurement, but what about:
$3cm\times <!--\; \times \; -->2=6cm$
Should the 2 have a unit of counting? Like I have 2 (amount of stuff) 3 cm strings.
Should it be something like:
$3cm\times <!--\; \times \; -->2m{u}^0=6cm$
(where mu is a "mathematical" unit)
The unit is to the power of 0 because if it was not it would do this:
$3mu\times <!--\; \times \; -->2mu=6m{u}^2$
And I don't think this works right, but $m{u}^0$ seems to:
$3m{u}^0\times <!--\; \times \; -->2m{u}^0=6m{u}^0$
Is any of this correct, or is it all wrong?

What happens when you add $x$ to $\frac{1}{3}x$?
I am dealing with an equation that requires me to add $x$ to $\frac{1}{3}x$:
$x+\frac{1}{3}x$ = ??
I know this might be simple to any of you on this site, because you are all asking questions with symbols I have never seen, but this is confusing to me.
I guess one way of thinking about this is - You are adding $x$ to $yx$, right? Or just adding another $\frac{1}{3}$?
The complete equation that I am working on is [- don't laugh at its simplicity ;)]:
$\frac{2}{3}b+5=20-<!--\; -\; -->b$
So, when worked out... I got:
$\frac{2}{3}b+b=15$
And this is where I get stuck.

The accounting process involves all of the following except
a. Analyzing and interpreting financial reports.
b. Recording non?quantifiable economic events.
c. Communicating financial information to users by preparing financial reports.
d. Identifying economic events that are relevant to the business.
Which of the following would not be considered an external user of accounting data for the GHI Company?
a. Creditors.
b. Customers.
c. Management.
d. Taxing authority representative.

What is the divisibility rule of 6?

Hard problem with fractions
I can't solve the following problem.
A person is x years old. Find his age if the following is true. In a group of x people each one started taking pictures of each of the others. At some point we know that more than $\frac{1}{2}$ of the people has taken exactly $\frac{1}{2}$ of all of their pictures (which are $x-<!--\; -\; -->1$ for every one of them), at the same time more than $\frac{1}{3}$ of them has taken exactly $\frac{1}{3}$ of all of their pictures, and at the same time more than $\frac{1}{7}$ of them has taken exactly $\frac{1}{7}$ of their pictures.

Euler's Continued Fraction Theorem for fractions
How can I use Euler's Continued Fraction Theorem to find the continued fraction expansion for a (ordinary, finite) fraction via its (terminating or recurring) decimal expansion, rather than via the more obvious Euclid's Algorithm?
The examples I'm thinking of are $5/7$ and $3/8$ each of which can be thought of as a series in a power of $10$
$5/7=0.\stackrel{¯<!--\; ¯\; -->}{571428}=\underset{k=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}571428({10}^{-<!--\; -\; -->6}{)}^k$
and
$3/8=0.375=3\ast <!--\; \ast \; -->{10}^{-<!--\; -\; -->1}+7\ast <!--\; \ast \; -->{10}^{-<!--\; -\; -->2}+5\ast <!--\; \ast \; -->{10}^{-<!--\; -\; -->3}.$
I'd like to use these coefficients and expansion points together with Euler's Theorem to show the continued fraction expansions
$5/7=[0;1,2,2]$
and
$3/8=[0;2,1,2].$

Let $a,b,c>0$ then show taht $\frac{9}{a+b+c}\le <!--\; \le \; -->2(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})\le <!--\; \le \; -->\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ .

I am trying to understand how to average certain data, interpret the result of that, and potentially convert the results to percentages. I have four categories for the data:
1: (0-50%): Not Well
2: (50-75%): Okay
3: (75-90%): Well Done
4: (90-100%): Excellent
This is being used in a rating system. Say, for instance, we have 3 ratings each of category 4, and 7 ratings of category 3: {3, 3, 3, 3, 3, 3, 3, 4, 4, 4}. I don't believe doing a straight arithmetic mean will be useful in this case since category 1 is larger than category 2, category 2 is larger than category 3, and category 3 is larger than category 4. For instance, what would a result of 3.3 mean if we took the arithmetic mean? I'm trying to understand how to interpret the results of the data, but am at a loss for how to do this. Is there a way to convert the results back to one of the 4 categories or a percentage?
Edit: I think I've made some progress, but I could be off the mark. Taking a few different examples...
{1,1,1,1,1,1,1,1,4,4}.
They got a rating of 1 80% of the time, and a rating of 4 20% of the time .8*1+.2*4 = 1.6, possibly a Not Well result?
{1,1,1,3,3,4,4,4,4,4}: .3*1+.5*4+.2*3 = 2.9, maybe an Okay result?
{1,1,1,1,2,2,2,4,4,4}: .4*1+.3*2+.3*4 = 2.2, maybe an Okay result?
The problem with using that method is that Excellent will only come up if every rating is a 4. I am not sure how to fix that or if what I've tried is the correct method.

How do you write an inequality that expresses the reason the lengths 5 feet, 10 feet, and 20 feet could not be used to make a triangle?

Geometrical interpretations of SVD
I'm a bit confused by the various geometrical/visual interpretations of SVD or better I'm wondering how to reconcile them.
Transformations : As explained here, the 3 matrices produced by the SVD can be interpreted as rotation, scaling, rotation.
Projection on an axis (dimensionality reduction): As explained here, SVD enables to capture the most of the variance in the data by selecting the right orthogonal axes.
My (very naive) question is: are these 2 interpretations related visually or not at all, and if yes, how? Or are they 2 unrelated applications of SVD?
As far as I understand, the transformations of the matrices (1) can be applied to anything/any object, so it's not about the data mentioned in (2) so I would say it's not related but I may be wrong (because in (2) we also talk about rotations...).
Please note that I'm looking for the intuition with as few math as possible (I don't know much linear algebra as you may have guessed). Many thanks.

I have the following task from my textbook:
Let ${\left\{f_n\right\}}_{n\in <!--\; \in \; -->\mathbb{N}}$ be a sequence of measurable function on M with
$f_n\to <!--\; \to \; -->f\text{ a.s., }$
where f is also a measurable function. Here, a.s. means almost surely (= almost everywhere).
Show: if there exists a nonnegative measurable function g satisfying the following conditions:
$\left|f_n\right|\le <!--\; \le \; -->g\text{ a.s. for all }n\in <!--\; \in \; -->\mathbb{N}$
and
${\int <!--\; \int \; -->}_M{g}^{p}d\mathrm{\mu <!--\; \mu \; -->}<\mathrm{\infty <!--\; \infty \; -->}\text{ for one }p>0,$
then
${\int <!--\; \int \; -->}_M{|f_n-<!--\; -\; -->f|}^{p}d\mathrm{\mu <!--\; \mu \; -->}\to <!--\; \to \; -->0\text{ for }n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}.$
These notations remind me a little bit of the Hölder inequality.
What I have got:
I thought that we have
$|f_n{|}^p\le <!--\; \le \; -->g^{p}a.s.$
Thus
${\int <!--\; \int \; -->}_M|f_n{|}^{p}d\mathrm{\mu <!--\; \mu \; -->}\le <!--\; \le \; -->{\int <!--\; \int \; -->}_M{g}^{p}d\mathrm{\mu <!--\; \mu \; -->}\le <!--\; \le \; -->infty$
And as
$f_n\to <!--\; \to \; -->f$
then the rest follows?

How does this numerators and denominators relate with the fraction?
Suppose if we have two fractions $\frac{a}{b}$ and $\frac{c}{d}$ then how are their values related with the fraction $\frac{a+c}{b+d}$?
I have observed this inequality: $\frac{a}{b}\le <!--\; \le \; -->\frac{a+c}{b+d}\le <!--\; \le \; -->\frac{c}{d}$
Does this hold true $\mathrm{\forall <!--\; \forall \; -->}a,b,c,d\in <!--\; \in \; -->N$ ? Is there any proof?
I am sorry if this question is too basic, but I want help regarding this.

I wish to determine all functions $f$ of bounded variation on [0,1] such that
$f(x)+(TV(f_{[0,x]}){)}^{1/2}=1\text{for all }x\in <!--\; \in \; -->[0,1]\text{and}{\int <!--\; \int \; -->}_0^{1}f(x)dx=1/3.$
Here $TV(f_{[a,b]})$ denotes the total variation of $f$ on [a,b].
So looking at what needs to be done, it is clear to me that the function f(x) = 1/3 satisfies the integral equation however, I have no intuition on how to proceed with the equation that involves the total variation or how to bring it all together.

Let $O=\{(x_1,x_2)\in <!--\; \in \; -->{\mathbb{R}}^2:0<x_1=x_2<1\}$, $I=\{N\in <!--\; \in \; -->\mathcal{B}:\mathrm{\lambda <!--\; \lambda \; -->}((0,1)\cap <!--\; \cap \; -->N)=0\}$, where $\mathrm{\lambda <!--\; \lambda \; -->}$ is the Lebesgue measure and $\mathcal{B}$ is the Borel sigma algebra.
Furthermore let $\mathrm{\mu <!--\; \mu \; -->}(A)=\{\begin{array}{ll}0& \mathrm{\exists <!--\; \exists \; -->}{A}_1,A_2\in <!--\; \in \; -->I,\text{ and }\mathrm{\exists <!--\; \exists \; -->}{B}_1,B_2\in <!--\; \in \; -->\mathcal{B},\text{ so }A\subseteq <!--\; \subseteq \; -->(A_1\times <!--\; \times \; -->B_1)\cup <!--\; \cup \; -->(A_2\times <!--\; \times \; -->B_2)\\ 1& \text{otherwise}\end{array}$
Show that $\mathrm{\mu <!--\; \mu \; -->}(O)=1$.
My thought is a contradiction argument. If $\mathrm{\mu <!--\; \mu \; -->}(O)=0$, then there must exist $N_1,N_2\in <!--\; \in \; -->I$ and B1,B2∈B such that $O\subseteq <!--\; \subseteq \; -->(N_1\times <!--\; \times \; -->B_1)\cup <!--\; \cup \; -->(N_2\times <!--\; \times \; -->B_2)$. But since $O$ is a line from the point (0,0) to the point (1,1), there cannot be a set with Lebesguemeasure 0 on the interval (0,1) on either axis. There is therefore no way to "describe" the line, because it goes from 0 to 1 on both axis, and therefore there does not exist $N_1,N_2\in <!--\; \in \; -->I$ so $O\subseteq <!--\; \subseteq \; -->(N_1\times <!--\; \times \; -->B_1)\cup <!--\; \cup \; -->(N_2\times <!--\; \times \; -->B_2)$
My problem is that I understand why is the measure $\mathrm{\mu <!--\; \mu \; -->}(O)=1$, I just can't seem to find the right proof technique.

Interpreting the differences of two log normal distributions:
I have read a couple of posts, and did not see the exact interpretation, I apologize in advance if this is not in the right location
Purpose: I am preparing a paper on the distribution of litter densities along the shore line in freshwater environnements. The data is collected by hand and the individual pieces are classified and counted by volunteers. These programs exist in many countries and there are fairly large data sets.
The units are expressed as 'pieces of trash/meter(or foot) of shoreline.
Assumptions:
- The data is collected in the same manner
- The volunteers have the same motivations
- There is no (under counting or over counting)
- Accuracy is basically the same across the spectrum
- The math is correct
- The graph below represents the graph of two sets of Data:
MCBP is regional results for Lake Geneva (Switzerland) n=100 samples
SLR results from the 'The Swiss Litter Report' n=365 samples The following code was used to calculate the distributions and present the graphs from a DataFrame in pandas/python 3.6:
df['Density] = df['Total']/df['Length']
df['Logs'] = df['Density'].apply(np.log)#<- skewed data(get it close to norm)
mu, sigma = stats.norm.fit(df['Logs'])
#repeat for df2 to get the second curve
#Build histograms for the two data sets
#plot the two disributions where x = df['Logs']
#and y = stats.norm.pdf(x, loc=mu, scale=sigma)
The resulting two distributions
mu for the the SLR disribution is 0.1564617, which is equal to the 5th percentile of the MCBP distirbution.
I am interpreting this as meaning:
There is a 5% oprobability that a sample from MCBP will be less than the average from SLR.
There is a 95% probability that a sample taken from the MCBP region will be greater than the national average
In general I can expect litter densities to be greater in the MCBP region than in the SLR region
Is this interpretation correct? (It does correlate with observations)