$(\sum_{i=1}^{n}|x_i|^p)^{1/p}\leq \sum_{i=1}^n |x_i|$ I know can be deduced from Jensen inequality and convexity of $x \mapsto x^p$ for positive integer $p$. is there a simpler way to deduce this?
Hi everyone. On the following question I think I proved the impossibility of having two perfectly negatively correlated $\rho=-1$ Geometric Brownian Motion series. I hope you could take a look and tell me if you find any mistakes on the proof. Thanks beforehand.
Suppose in the above theorem b) does not hold. What's the conclusion then of the theorem? Only that $f_x$ and $f^y$ are measurable for a.e. $x$ and for a.e. $y$ respectively?
In the last para "and two sections with this property coincide on an open nbd of x" it's not clear to me that how they arrived at this consequence from axiom 1
I have a question but when I have to find the absolute extremes on a circumference and I parametrize, if $\theta \in [0,2\pi]$ , then in the end I will also have to check the value of the function at $0, π/2 , π , 3π/2, 2π$?
That is, manually
Because while reviewing the exercises I noticed that here: Calculate the function definition set: $f(x, y) = \arcsin(x^2-y-1)$ and calculate the absolute extrema in the circle with center (0, −1) and and unit radius.
Setting the first derivative = 0, I find i $\theta = \pi/2, 3\pi/2, 7\pi/6, 11\pi/6$
So doing the calculations I find the absolute minimum at $\pi/2$ and the absolute maximum at $3\pi/2$ , if I were to also consider $0, \pi$ then I also find the missing ones
So I was thinking that maybe in domains like squares, triangles, rectangles etc., i consider the vertices, while in the circle I have to consider $0, \pi/2, \pi , 3\pi/2, 2\pi$?
theorem about taking quotients $a_{n+1}/a_n$ for power series doesn't work here since your series $\sum a_nx^n$ is such that $a_n = 0$ infinitely many times
the alternative here is to use the formula $\lim_n \sqrt[n]{a_n} = \lim_n (1/n!)^{1/n^2}$ and say, Stirling approximation $n!\sim (n/e)^n \sqrt{2\pi n}$ to show $\lim_n (1/n!)^{1/n^2} = \lim_n \sqrt[n]{n/e} \cdot (2\pi n)^{1/2n^2} = 1$
don't give me random links, give me words and your justification for why you think something is or isn't true and WHAT is that you think is or isn't true
In order to treat the series $\sum \frac{x^{n^2}}{n!}$ as a power series $\sum a_nx^n$ we need to define $a_n = \begin{cases} \frac{1}{m!} & \text{if }n = m^2 \\ 0 & \text{otherwise} \end{cases}$
@SoumikMukherjee let $U, V$ be as in axiom $1$. If $s$ is a section with the properties listed, then there exists neighbourhood $U_0$ of $x$ such that $s:U_0\to V$ from continuity of $s$. Since $\pi\circ s$ is an identity map, we can take inverse of $\pi$ restricted to $V$ and compose it to show $s$ must be obtained as inverse of $\pi$ restricted to $V$ in a neighbourhood of $x$.
Regarding a recent question that I posed here, it just occurred to me that folks who find the statement $0\not\in\left\{x\in\Bbb R:\frac1x>1\right\}$ unacceptable perhaps shouldn't accept $\forall x{\in}\mathbb R{\setminus}\{0\}\;\frac1x\ne5$ either, since it merely abbreviates $\forall x{\in}\mathbb R\;(x\ne0\implies\frac1x\ne5)$ and asserts, in particular, the illegal statement $0\ne0\implies\frac10\ne5.$
@ryang that is the formal way, sure. But writing $\forall_{x\in S} P(x)$ when $P$ involves a function $f$ say, should be interpreted as abbreviating $\forall_{x\in S} f$ is well-defined and $P(x)$
sorry not $f$ is well-defined... you know, say $f$ explcitly depends on variable $x$, then it should be interpreted as $x\in \text{dom}(f)$ and $P(x)$
@Jakobian The whole context of that definition is that that abbreviation is informal; anyhow, i'm not following why you think $\forall x{\in}\mathbb R{\setminus}\{0\}\;\frac1x\ne5$ doesn't abbreviate $\forall x{\in}\mathbb R\;(x\ne0\implies\frac1x\ne5).$ In any case—and I'm possibly misunderstanding you—but ultimately aren't you selectively making concessions (for $\forall x{\in}\mathbb R{\setminus}\{0\}\;\frac1x\ne5$ but not for $0\not\in\left\{x\in\Bbb R:\frac1x>1\right\}\;).$
I have a confusion about defining probability measures. Suppose our sample space consists of infinite sequences with values in $\{0,1\}$ and our set of events will be generated by $A_i=\{i\text{th coordinate is }1\}$ where we define $P(A_i)=1/2$ for any $i\in\mathbb{N}$. How does one actually show countable additivity for such a function? I cannot see how this follows from just our constant function definition.
@Simd No, not exactly, but it gives you good places to look if you can apply some other theory (e.g. monotonicity).
Though it is also worth noting that for $c \in (0,1)$, the expression $c^{r^2}$ is decreasing in $r$, and can be made arbitrarily close to zero by choosing $r$ big enough. So you are not going to have an easy time finding a minimum.
@Jakobian Oh, I misread the question as $(c^r)^r$.
Nevermind.
In the case of $(cr)^r$, go back to my original statement: you can't just minimize over the reals and hope that it will work by rounding, but if you minimize over the reals and then apply a monotonicity argument, you can get there.
Typing (c*r)^r is somewhat confusing, as no one bothers with multiplication symbols most of the time. Your expression is (cr)^r---the extra symbol threw me.
In the case of $(cr)^r$, go back to my original statement: you can't just minimize over the reals and hope that it will work by rounding, but if you minimize over the reals and then apply a monotonicity argument, you can get there.
Some very quick experimentation with GeoGebra seems to give a counterexample with $c=0.1474$. In that case, minimizing over the reals gives $r_{\text{min}} \approx 2.496$, which rounds down to $2$. But $(0.1474\cdot 2)^2 \approx 0.0869$, while $(0.1474\cdot 3)^3 \approx 0.0865$.
@Jakobian I have assumed that @Simd knows how to minimize the expression over the reals---they are asking if minimizing over the reals and then rounding will always give the minimum over $\mathbb{N}$.
So the minimum over $\mathbb{R}$ will be at $1/(c\mathrm{e})$. The minimum over $\mathbb{N}$ is then what you get be either rounding that up or down, depending on which gives the smaller result.
(which is the monotonicity argument I was suggesting above).
But you can't just round $1/(c\mathrm{e})$ to nearest integer and expect it to always work out.
Some very quick experimentation with GeoGebra seems to give a counterexample with $c=0.1474$. In that case, minimizing over the reals gives $r_{\text{min}} \approx 2.496$, which rounds down to $2$. But $(0.1474\cdot 2)^2 \approx 0.0869$, while $(0.1474\cdot 3)^3 \approx 0.0865$.
@ephe If you just say $P(A_i) = 1/2$ you did not define the probability for all events What you probably want to say is that $P(A_i) = 1/2$ and the events $A_i$ must be independent. Even here we don't give explicitly the values of $P$ for all events, so the question should be not "Prove that $P$ is $\sigma$-additive", but "Prove that there exists a unique $\sigma$-additive $P$ satisfying these conditions"
And I guess in this case you need to give a complete description of events and their probabilities. If I'm not mistaken, any event here will be a countable union of sets of the following …
Hi everyone, sorry to bother you, could you kindly explain why this edit was rejected? The post does not meet Math SE guidelines at all, since it is not written in English, nor in MathJax syntax. Where is the error?
Here's an exercise. Compute $\int \chi_D \, d(\mu\times\nu)$ where $\mu$ is Lebesgue measure on $[0,1]$ and $\nu$ counting measure on $[0,1]$ and $D$ the diagonal in $[0,1]\times[0,1]$. I have already computed iterated integrals, and they disagree. Is it possible to use Fubini's theorem to conclude that $\chi_D\notin L^1$ and hence the integral is infinite? What confuses me is that in Fubini's theorem we require the measure spaces to be $\sigma$-finite, and one of them isn't here.
All of you have probably seen various images of Mandelbrot set about a billion times already, right? But I still want to share this result with you, because I think it is interesting since it is displayed in console window (which means no graphics, only text). And the source code of the program responsible for generating that result is only 151 bytes long, so it nicely shows how mathematics can enable a simple algoritm to generate a complex result!
BTW here is the code in C which generated this result (I am not the author of the code, and I found it on the Internet with a note that the author is unknown):
@Jakobian $\mu\times\nu$ is only defined on the algebra of finite disjoint union of rectangles, namely by $\mu\times\nu(E)=\sum_1^n\mu(A_i)\nu(B_i)$. We can extend this by Caratheodory's extension theorem to the product sigma algebra $\mathcal M\otimes\mathcal N$ and get a unique measure provided $\mu,\nu$ are $\sigma$-finite.
But I guess what you are after is some kind of definition in terms of infimum of a set.
E.g. $$\mu\times\nu(E)=\inf\left\{\sum_1^\infty\mu(A_i)\nu(B_i):\text{ something something}\right\}.$$
@Jakobian and? This very exercise is meant to demonstrate that Fubini-Tonelli's theorem fails when the spaces are not $\sigma$-finite, but I don't see how to compute that integral of $\chi_D$ with respect to the product measure.
In mathematics, given two measurable spaces and measures on them, one can obtain a product measurable space and a product measure on that space. Conceptually, this is similar to defining the Cartesian product of sets and the product topology of two topological spaces, except that there can be many natural choices for the product measure.
Let
(
X
1
,
Σ
1
)
{\displaystyle (X_{1},\Sigma _{1})}
and...
According to wikipedia there is a minimal product measure, I'd try with that
@Jakobian yeah, I believe that's the case. I think a descriptive argument would work kind of. Something like...the "smallest" cover of $D$ of rectangles would be $\{\{x\}\times\{x\}:x\in[0,1]\}$. Since $[0,1]$ is uncountable, we get infinity immediately, or?
@psie I'm replacing "something something" here by "$\{E_n\}_1^\infty$ a sequence of finite disjoint union of rectangles which cover $E$". Or equivalently, simply "$\{A_i\times B_i\}_1^\infty$ a sequence of rectangles which cover $E$".
@Jakobian I think I misunderstood Folland. When he writes $\mu\times\nu$ he has a specific object in mind, namely the restriction of the outer measure induced by the premeasure we start with on the algebra of finite disjoint union of rectangles: $\pi(E)=\sum_1^n\mu(A_i)\nu(B_i)$. This is well defined as one can show. Absent $\sigma$-finiteness, it is true that there might be other measures that also extend the premeasure. But when he asks about $\mu\times\nu(D)$, it's a specific measure I think.
Since the measure spaces in the problem are not $\sigma$-finite, there might be other measures such that the measure of every measurable rectangle is the product of the measures of its sides. But when he writes $\mu\times\nu$, its the restriction of the outer measure induced by the premeasure.
@Jakobian probably my last question on this, but is it true that \begin{align} \mu\times \nu(D) &=\inf\left\{\sum_1^\infty \mu\times \nu(E_n) : D\subset \bigcup_1^\infty E_n,\{E_n\}_1^\infty\subset\mathcal{A} \right\}\\ &=\inf\left\{ \sum_1^\infty \mu(A_n)\nu(B_n) : D\subset \bigcup_1^\infty A_n\times B_n\right\},\end{align}where $\mathcal A$ is the finite disjoint union of rectangles?
The first set seems more natural to me as the definition of $\mu\times\nu$ and I don't see how it equals the second.
@Jakobian the first set seems more natural to me due to the above, i.e. every premeasure induces an outer measure according to equation (1.12), but you see, the sets that cover our set are from the algebra, and these are finite disjoint unions of rectangles.