Improve Your Algebra Skills with Practice Problems

I consider 2 linear Kalman filters (KF) that follow periodic cycles as shown here: KF(1) : Receive 1 measurement at time $k$, No measurements at time $k+1$, Receive 1 measurement at time $k+2$, No measurements at time $k+3$, etc.
KF(2) : Receive measurements at every time step.
In KF(1), whenever a measurement is missing, I simply apply time-update and ignore the measurement update step. Simulations show that in the steady-state, tr($P_{k|k}$) (the trace of the a-posteriori error covariance matrix computed after the reception of a measurement) is slightly greater in KF(1) than in KF(2), which intuitively makes sense. I would like to quantify this difference, or at least find upper and lower bounds for their ratio. Any leads or suggestions on where to look at or how to proceed?

I am not that familiar in proving measurability of a function. The definition I know is that the preimage of measurable sets has to be measurable again. But how to prove that seems difficult to me at the first glance. I do have the following exercise: Let $(B_t)$ be a Brownian motion on $(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{A},\mathbb{P})$ with continuous trajectories/paths. Consider the product space
$([0,\mathrm{\infty <!--\; \infty \; -->}),{\mathcal{B}}_{[0,\mathrm{\infty <!--\; \infty \; -->})},\mathrm{\lambda <!--\; \lambda \; -->})\otimes <!--\; \otimes \; -->(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{A},\mathbb{P})$
where $\mathrm{\lambda <!--\; \lambda \; -->}$ is the Lebesgue measure. Show that the mapping $(t,\mathrm{\omega <!--\; \omega \; -->})\to <!--\; \to \; -->(B_t(\mathrm{\omega <!--\; \omega \; -->}))$($t\ge <!--\; \ge \; -->0$) defined from that space to $({\mathbb{R}}^{[0,\mathrm{\infty <!--\; \infty \; -->})},{\mathcal{B}}_{{\mathbb{R}}^{[0,\mathrm{\infty <!--\; \infty \; -->})}})$ is measurable. How can I prove this? I think the continuity is important, but I dont know how to involve it. Any help is appreciated.

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) $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}(||f_n|{|}_p-<!--\; -\; -->||f|{|}_p)=0.$
Prove that $||f_n-<!--\; -\; -->f|{|}_p\to <!--\; \to \; -->0.$
I believe what we have to show is that
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\int <!--\; \int \; -->}_E|f_n-<!--\; -\; -->f{|}^p=0.$
My approach is to hope to invoke the Dominated Convergence theorem, i.e., prove that there exists some $g\in <!--\; \in \; -->L^1(E)$ such that $|f_n|\le <!--\; \le \; -->g$ for almost everywhere for all $n$, then we can have
${\int <!--\; \int \; -->}_E|f_n-<!--\; -\; -->f|\to <!--\; \to \; -->0.$
Then by (a), we can find an integer $K$ such that
$|f_k-<!--\; -\; -->f|<1$
for all $k\ge <!--\; \ge \; -->K$. Then
$0\le <!--\; \le \; -->{\int <!--\; \int \; -->}_E|f_k-<!--\; -\; -->f{|}^p\le <!--\; \le \; -->{\int <!--\; \int \; -->}_E|f_k-<!--\; -\; -->f|.$
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.

Ok, so this is a bit of an odd question. It is sort of a physics question, but I felt it was more appropriate here. Feel free to suggest another venue.
Let's just look at a simple example. We'll use kilograms as an example unit of measurement. Let's say we have three objects, A, B, and C with masses 11 kg, 12 kg, and 23 kg. It is clear there is an additive relationship here that has a real physical meaning. The third object really does have the equivalent amount of matter as the first two combined. This real relationship is played out with how physical laws play out. We can go down the philosophical rabbit hole here, but I don't think it is productive.
A measurement in a given unit must be equivalent to that number times the unit, i.e. $11kg=11\times <!--\; \times \; -->(1kg)$. Of course, multiplication can be defined in terms of addition. So we actually require that
$11kg=\underset{\text{11 terms}}{\underset{⏟<!--\; ⏟\; -->}{1kg+1kg+\cdots <!--\; \cdots \; -->1kg}}.$
In a real physical sense, the amount of stuff in eleven objects each with 1 kg mass really is equivalent to the amount of stuff in one object with 11 kg of mass.
So here is the main question: Is number of kilograms beyond 10 kilograms a valid unit of measurement? Let us denote it $k{g}_{10}$
Now our objects have "masses" $1k{g}_{10}$, $2k{g}_{10}$, and $13k{g}_{10}$. These aren't actually masses, they are "masses beyond 10kg". The additive relationship from our original unit of kg does not carry over to our new "unit" of $k{g}_{10}$.
Can we even say that $1k{g}_{10}$ + $1k{g}_{10}$ = $2k{g}_{10}$? What are we actually adding here? We are not adding kilograms! If I have an object that has mass 11 kg ($1k{g}_{10}$) and I increase its mass by 1 kg, it now has mass 12 kg ($2k{g}_{10}$) so it would seem that $1k{g}_{10}$ + 1 kg = $2k{g}_{10}$.
So, maybe this is a stupid thing to worry about, but it just seems that we should have rigorous set of rules about what constitutes a valid unit of measurement. Those rules would include something about additivity and the relation to the thing being measured and not being able to just arbitrarily set zero anywhere.
I would argue that $k{g}_{10}$ is not technically a valid unit of measurement, and that this is reasonable since the unit of measurement is actually kilograms, which is a true unit of measurement (arithmetic operations on it make sense in terms of the physical thing it is actually measuring), but for whatever reason, we might be interested only in kilograms beyond 10.

What is the ' factorization of 63?

I am a newby to Kalmar filters, but after some study, I think I understand how it works now. For my application, I need a Kalmar filter that combines the measurement input from two sources. In the standard Kalmar filter, that is no problem at all, but it assumes that the measurement inputs from the two sensors are available at the same times. In my application, there is one new measurement from sensor 'b' for every 13 measurements of sensor 'a'.That is, 12 out of 13 times, the measurement of sensor 'b' is missing.
How would you handle that normally? Do you simply use the predicted measurements values as substitute for the missing ones? Does that not lead to overconfidence in the missing measurements? How else can it be handled?

Let $n\in <!--\; \in \; -->N$, $(X,A,\mathrm{\mu <!--\; \mu \; -->})$ be a measure space and $E_1,...,E_n\in <!--\; \in \; -->A$. For each $j\in <!--\; \in \; -->\{1,...,n\}$ define
$C_j:=\{x\in <!--\; \in \; -->X|x\in <!--\; \in \; -->E_k\text{for exactly }j\text{ indices},k\in \{1...,n\}\}.$
I want to find a general expression for all the $C_j$ since I need to prove that each $C_j$ is a measurable set, I made an example and it is as follows
With $n=3$
$C_1=(E_1\cap <!--\; \cap \; -->E_2^c\cap <!--\; \cap \; -->E_3^c)\cup <!--\; \cup \; -->(E_1^c\cap <!--\; \cap \; -->E_2\cap <!--\; \cap \; -->E_3^c)\cup <!--\; \cup \; -->(E_1^c\cap <!--\; \cap \; -->E_2^c\cap <!--\; \cap \; -->E_3)$
$C_2=(E_1\cap <!--\; \cap \; -->E_2\cap <!--\; \cap \; -->E_3^c)\cup <!--\; \cup \; -->(E_1\cap <!--\; \cap \; -->E_2^c\cap <!--\; \cap \; -->E_3)\cup <!--\; \cup \; -->(E_1^c\cap <!--\; \cap \; -->E_2^c\cap <!--\; \cap \; -->E_3)$
$C_3=\underset{k=1}{\overset{3}{\bigcap <!--\; \bigcap \; -->}}{E}_k$
So for $C_n$ we already have a general expression:
$C_n=\underset{k=1}{\overset{n}{\bigcap <!--\; \bigcap \; -->}}{E}_k$
for $C_1$:
$C_1=\underset{j=1}{\overset{n}{\bigcup <!--\; \bigcup \; -->}}(E_j\cap <!--\; \cap \; -->\underset{k\ne <!--\; \ne \; -->j}{\overset{n}{\bigcap <!--\; \bigcap \; -->}}{E}_k^c)$
I was thinking of introducing permutations but actually I don't know if it works, someone can guide me a bit to express the $C_j$ sets in a manageable way

Revised based on comments.
Measure-theoretic probability is used in finance (among other fields).
However, was wondering where unbounded measures are used, outside of mathematics and physics. For example, is there a use for unbounded measures in finance or biology or business?
Apologies for a broad question -- there really is not 'right answer' but hoping this group will have suggestions. Many thanks!

How to simplify $(a^2+ab+b^2)/(a+\sqrt{ab}+b)$
How can I simplify as much as possible:
$\frac{a^2+ab+b^2}{a+\sqrt{ab}+b}$
Also, first post here, looking forward to sticking around!

What are the measurement units of $\mathrm{\omega <!--\; \omega \; -->}$ in this equation and why
$Y(x,t)=A\cos <!--\; \; -->(kx-<!--\; -\; -->\mathrm{\omega <!--\; \omega \; -->}t)?$

Consider the model
${𝑋}_{𝑖}\ast$
In practice we measure ${𝑋}_{𝑖}\ast$ by ${𝑋}_{𝑖}$ such that
a) ${𝑋}_{𝑖}\ast <!--\; \ast \; -->+3$
b) $5{𝑋}_{𝑖}\ast <!--\; \ast \; -->$
What will be the effect of these measurement errors on estimates of true ${𝛽}_0$ and ${𝛽}_1$? In terms of biasedness, BLUE, consistency, efficiency?

When solving inequality equations, we have to change the sign of the inequality when we divide/multiply by a negative number. And I get why. Because your making one side positive and one side negative.
But let's say you were multiplying by say $(x^3+1)/x>1$ or something. When you multiply both sides by x, why don't you need to flip the sign? Because there's the possibility x is negative! Why don't you need to construct to equations, one where the sign is the same, and one where the sign is flipped.
As an extension, let's say you were multiplying both sides of $(x^3+1)/-<!--\; -\; -->x>1$ by -x. Do you need to change the sign too?

If $F$ is a continuous distribution function on $(\mathbb{R},\mathcal{B},{\mathrm{\mu <!--\; \mu \; -->}}_{\mathcal{L}})$ with distribution ${\mathrm{\mu <!--\; \mu \; -->}}_F$, use Fubini's theorem to show that
1. ${\int <!--\; \int \; -->}_{\mathbb{R}}F(x)d{\mathrm{\mu <!--\; \mu \; -->}}_F(x)=\frac{1}{2}$
2. If $X_1,X_2$ are i.i.d random variables with common distribution $F$, then $P(\{X_1\le <!--\; \le \; -->X_2\})=1/2$ and $\text{E}(F(X_1))=1/2$.

My Attempt:
I don't really understand bs_math's answer so I have been trying to write my own. I just deleted an attempt here that was (I think) completely nonsensical. I am working on another attempt.
For example, I don't understand what's going on in line 4 of bs_math's answer.

Let $f_j\in <!--\; \in \; -->L^P\cap <!--\; \cap \; -->L_{loc}^1$ and $g\in <!--\; \in \; -->L^q$ where $p,q$ are conjugates and $||f_j|{|}_p<<!--\; <\; -->M$ for all $j$ . I have that ${\int <!--\; \int \; -->}_S{f}_j\to <!--\; \to \; -->0$ for all $S\subset <!--\; \subset \; -->A$ (measurable) where $A$ is a compact subset of $\mathbb{R}$ and want to show ${\int <!--\; \int \; -->}_A{f}_{j}g\to <!--\; \to \; -->0$.
Since $A$ is bounded, $g\in <!--\; \in \; -->L^q⟹<!--\; ⟹\; -->g\in <!--\; \in \; -->L^{\mathrm{\infty <!--\; \infty \; -->}}$ so I want to write down something like $|{\int <!--\; \int \; -->}_A{f}_{j}g|\le <!--\; \le \; -->||g|{|}_{L^{\mathrm{\infty <!--\; \infty \; -->}}}|\int <!--\; \int \; -->f_j|$. But this is false. I can only use the tools that I know if I control $\int <!--\; \int \; -->|f|$, but I've already come up with counterexamples to show we have no control over that so considering $\int <!--\; \int \; -->|f_{j}g|$ is useless. Another idea was to use Cauchy schwarz but we need not have $\int <!--\; \int \; -->f_j^2\to <!--\; \to \; -->0$

$e^{itH}$ notation
recently I saw the notation $e^{itH}$, and just wondering how should I interpret it?
In my understanding, $u(t,x)=e^{itH}{u}_0$ is, for example, a solution to Schrodinger-type equation $i{\mathrm{\partial <!--\; \partial \; -->}}_{t}u=-<!--\; -\; -->Hu$ with the initial data $u_0$. In case $H=\mathrm{\Delta <!--\; \Delta \; -->}$, the solution to Schrodinger equation is known to involve the Schrodinger kernel in the integrand. In such case, does $e^{itH}$ is a short-hand notation for the operator involving the Schrodinger kernel?
Or should I interpret $e^{itH}$ as the Taylor series with $H^k$ terms involved? In this case, does the (operator) series converge once applied to the element in the domain of H?
Also, I would be very glad to get a reference to read more on this type of operators. Thank you very much!

A killer whale has eaten 75 pounds of fish today. It needs to eat at least 140 pounds of fish each day A bucket holds 15 pounds of fish. Identify an inequality that represents how many more buckets z of fish the whale needs to eat?

Interpreting Entropy
All you data scientists will probably know the entropy equation:
$H(p)=-<!--\; -\; -->\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{p}_i\cdot <!--\; \cdot \; -->{\log }_2<!--\; \; -->p_i$
And, using this, I was messing around with some compression, and calculated the entropy for a set of probablilties $\{0.3,0.2,0.2,0.1\}$, which came out as about 2.246.
This doesn't make sense to me, because if Entropy $\propto <!--\; \propto \; -->$ 1/Compression, then I've done the impossible by compressing data with these proportions.
I find myself confused as to how to interpret this value any other way. Is it bits per arbitrary unit? Am I simply wrong?

If $f=f(x)$ is absolutely continuous on [0,1] such that $f^{\prime }\in <!--\; \in \; -->L^2([0,1])$ and $f(0)=0$. Prove then that
$\underset{x\to <!--\; \to \; -->0^{+}}{lim}\frac{f(x)}{x^{\frac{1}{2}}}=0$
The hint says to use Fundamental theorem of calculus but I don't see where to use that.
I know if $f^{\prime }\in <!--\; \in \; -->L^2([0,1])$ then
$({\int <!--\; \int \; -->}_{[0,1]}|f^{\prime }(x){|}^{2}dx{)}^{\frac{1}{2}}<+\mathrm{\infty <!--\; \infty \; -->}.$
I don't know what to do from here though. Any hints would be appreciated. Am I merely using L'Hôpital's rule, since I get 0/0?

I will appreciate any help.
Consider the following function:
$g\left(x\right)=\{\begin{array}{ll}\frac{1}{x},& \text{if}x\in <!--\; \in \; -->(1,\mathrm{\infty <!--\; \infty \; -->})\\ 0,& \text{if}x\in <!--\; \in \; -->(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},1]\end{array}$
Determine the set $\{p\in <!--\; \in \; -->[0,\mathrm{\infty <!--\; \infty \; -->}]:g\in <!--\; \in \; -->{\mathcal{L}}^p\left(\mathrm{\lambda <!--\; \lambda \; -->}\right)\}$. I know that the set is equal to $(1,\mathrm{\infty <!--\; \infty \; -->}]$. But I am not sure how to calculate the cases $p=0$ and $p=\mathrm{\infty <!--\; \infty \; -->}$.
In my textbook we have the following definitions for $p=0$ and $p=\mathrm{\infty <!--\; \infty \; -->}$:
${\mathcal{L}}^{\mathrm{\infty <!--\; \infty \; -->}}\left(\mathrm{\mu <!--\; \mu \; -->}\right)=\{f\in <!--\; \in \; -->\mathcal{M}\left(\mathcal{E}\right):\mathrm{\exists <!--\; \exists \; -->}R>0:|f|\le <!--\; \le \; -->R\mathrm{\mu <!--\; \mu \; -->}\text{-almost everywhere}\}$
and
${\mathcal{L}}^0\left(\mathrm{\mu <!--\; \mu \; -->}\right)=\{f\in <!--\; \in \; -->\mathcal{M}\left(\mathcal{E}\right):\underset{t\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\mathrm{\mu <!--\; \mu \; -->}(\{|f|\ge <!--\; \ge \; -->t\})=0\}$
where $\mathcal{M}\left(\mathcal{E}\right)=\{f:X\to <!--\; \to \; -->\mathbb{R}:f\text{is}\mathcal{E}\text{-}\mathcal{B}\left(\mathbb{R}\right)\text{- measurable}\}$.

How do you solve ${\displaystyle x-14\le 23}$ ?