Відео

Welcome!

Right! Thanks for pointing it out. c.f., Wikipedia: en.wikipedia.org/wiki/Geometric_process

Thanks!

I'm not sure what you mean exactly. Why don't you make a video and let me know when you figure it out?

The intent of my statement is that I want a way to study the Uniform Convergence of the above series of functions

Most welcome!

It is P(X) = P(X1, X2, ..., Xn) (joint probability density), not P(Xi)*P(Xj). Note Xi and Xj may not be independent.

@@BruneiMathClub Thank you so much. I finally get it.

That 1/2 in the covariance constraint is not essential. It's there mostly for an aesthetic reason (it looks nicer after differentiation). You get the same result without the 1/2 factor (try it!), as it can be absorbed in the Lagrange multipliers (γ's).

@@BruneiMathClub Yes indeed. Thank you for your reply and fantastic videos! I’ve been working on the exercise of the Pattern Recognition and Machine Learning book and your videos helped a lot!

@@BruneiMathClub BTW you can also evaluate the stationary point in full matrix form using the trace operator for the quadratic term, which I find is pretty neat.

For example, the regression problem can be cast as finding a projection onto a subspace "generated" by a dataset. Future videos will explain such applications.

It's in quite a few textbooks. For example, in "Elements of Information Theory" by Cover and Thomas, See also the Wikipedia page: en.wikipedia.org/wiki/Jensen%27s_inequality

@@BruneiMathClub thank's a lot.

You're welcome! God bless you, too.

Glad it was helpful!

Glad it was helpful!

Thanks. Shorts can be helpful sometimes.

You are welcome.

Thanks and welcome!

Thanks!

You really love duality!

@@BruneiMathClub Yes, duality means that there is a 4th law of thermodynamics. Anything which is dual to entropy is by definition the 4th law of thermodynamics:- Syntropy (prediction) is dual to increasing entropy -- the 4th law of thermodynamics! Teleological physics (syntropy) is dual to non teleological physics (entropy). Syntax is dual to semantics -- languages, communication. If mathematics is a language then it is dual. All observers make predictions to track targets, goals and objectives and this is a syntropic process -- teleological. The Einstein reality criterion:- "If, without in any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of reality corresponding to that quantity." (Einstein, Podolsky, Rosen 1935, p. 777) Internet Encyclopedia of Philosophy:- www.iep.utm.edu/epr/ According to Einstein reality is predicted into existence -- a syntropic process, teleological. Your brain/mind creates models or predictions of reality hence your mind is syntropic (convergent). Here is a video of some well known physicists talking about duality, watch at 1 hour 4 minutes:- ua-cam.com/video/UjDxk9ZnYJQ/v-deo.html Mathematics is full of dualities (see next comment). Once you accept the 4th law this means that that there is a 5th law of thermodynamics:- Symmetry is dual to conservation -- the duality of Noether's theorem. Duality is a symmetry and it is being conserved according to Noether's theorem. Energy is duality, duality is energy -- the 5th law of thermodynamics! Potential energy is dual to kinetic energy -- gravitational energy is dual.

@@BruneiMathClub Here are some examples of duality in mathematics:- Points are dual to lines -- the principle of duality in geometry. The point duality theorem is dual to the line duality theorem. Homology is dual to co homology -- the word co means mutual and implies duality. Sine is dual to cosine or dual sine -- perpendicularity. Sinh is dual to cosh -- hyperbolic functions. Addition is dual to subtraction (additive inverses) -- abstract algebra. Multiplication is dual to division (multiplicative inverses) -- abstract algebra. Integration (summations, syntropy) is dual to differentiation (differences, entropy). Convergence (syntropy) is dual to divergence (entropy). Injective is dual to surjective synthesizes bijective or isomorphism. The word isomorphism actually means duality. Subgroups are dual to subfields -- the Galois correspondence. Positive is dual to negative -- electric charge or numbers. Positive curvature is dual to negative curvature -- Gauss, Riemann geometry. Curvature or gravitation is dual. "Perpendicularity in hyperbolic geometry is measured in terms of duality" -- Professor Norman J. Wildberger, universal hyperbolic geometry:- ua-cam.com/video/EvP8VtyhzXs/v-deo.html All observers have a syntropic or hyperbolic perspective of reality. The tetrahedron is self dual. The cube is dual to the octahedron. The icosahedron is dual to the dodecahedron. Waves are dual to particles -- quantum duality or pure energy is dual. Symmetric wave functions (Bosons, waves) are dual to anti-symmetric wave functions (Fermions, particles) -- the spin statistics theorem. Bosons are dual to Fermions -- atomic duality. Pure energy is dual and it is being conserved -- the 5th law of thermodynamics!

@@BruneiMathClub Concepts are dual to percepts -- the mind duality of Immanuel Kant. Mathematicians create new ideas or concepts all the time from their perceptions, measurements, observations or intuitions -- they are using duality! The bad news is that Immanuel Kant has been completely ignored for over 200 years and this is why you need new laws of physics! Antinomy (duality) is two truths that contradict each other -- Immanuel Kant. Enantiodromia is the unconscious opposite or opposame (duality) -- Carl Jung.

You're welcome.

Almost, but not exactly. They differ at x = 0. For f(x) = |x|, f'(0) = 1, whereas signum(0) = 0.

You are welcome.

Wow, I'm thrilled to hear that! Thanks, and enjoy your study.

Is it?

Do you mean that although x_i and x_j are not independent, after some transformation, why z_i and z_j seem to be independent? That's because, after the orthogonal transformation, z_i and z_j ARE independent. This is one of the peculiar properties of the multivariate normal distribution: We can always transform it into a joint distribution of independent variables by some orthogonal transformation. By the way, regarding the question on MSE (Math StackExchange), I think that's a different issue, and it has a mistake (the Borel set S should also be transformed along with the variables).

@@BruneiMathClub Thank you for taking time to respond, I do see where the derivation went wrong now. But may I still ask how to derive the marginal distribution of each component X_i without resorting to the moment generating function? (As most sources do to get the distribution of the linear transformed AX+b) The initial attempt was to simplify the joint distribution by a sequence of change of variables and hopefully to be able to evaluate P(X_i in S_i) = P(X in R x ... x S_i x ... x R), but since this works only when the components are independent, it doesn't seem to apply in the general case.

@@user-hr8uj4qw4k I don't have an answer now, but I recommend you try a brute-force calculation for the bivariate case to get a feel for it.

@@BruneiMathClub Thank you for your advice.

Thanks!

No problem.

That's exactly what I'm preparing now. Thanks for the suggestion!

That's an interesting question! There are many ways to answer this, but it's basically the same as the univariate case. I will make a video about this in the near future. Stay tuned.

@@BruneiMathClub Sure! I figured they would be similar, I just wanted to know where covariance comes into play

That's a deep question! I suppose you are referring to the example right after the definition of L2 convergence. Countably many points do not contribute to the integral. In this case, the set of points {x_n = (1/2)^{1/n}} is countable (has the same cardinality as the set of natural numbers). Therefore, they contribute nothing to the integral. You will learn this when you study the theory of Lebesgue integrals (measure theory). Roughly speaking, the "length" of a single point is zero, so the area (contribution to the integral) = height (fn - f)x0 = 0. If you add the lengths of countably many points, it's still zero. For this reason, we say the set {x_n = (1/2)^{1/n}} is a *null set*. That's why they contribute nothing to the integral.

I hope this helps.

Many many thanks!

Well, good luck!

Glad it helped!

Yes, the multivariate version looks scary at first. But it becomes easier if you carefully compare and recognize the similarities with and differences from the univariate version. After all, it's the most natural and straightforward generalization.

@@BruneiMathClub Is there a great visualisation for the 4d gaussian haha. I know it's very hard or almost impossible to do. Probably just assign the 4th dimension to color and it will work. Just wondering, is there a simpler kernel than the gaussian that produces similar results. I know bandwidth matters a lot in this case, but the paper I read about this topic did not elaborate on the alternative kernels that we can use. There are square kernels, triangular kernel, etc. The paper also said that the shape of kernels don't matter, since the pdf will still be approximated with enough data. The exception is the bandwidth. Different bandwidths result in different limiting functions. How do we compute the bandwidth for the triangular kernel, square kernel, and others.

Right. You don't have to worry about it in practice. It just itches.

Who doesn't?

Showing a subset of a field is closed under addition and multiplication is not enough to show the subset is a (sub)field. For example, Z (integers) is a subgroup of R (real numbers). Z is closed under addition and multiplication, yet it is not a field. If you know some group theory, you can see a field (K, +, ×) as two groups (K, +) and (K\{0}, ×), then apply the *subgroup criterion* to each of them. That is, a subset S of a group is a (sub)group if and only if for all a, b ∈ S, ab ∈ S and a^{-1} ∈ S.

Thanks! I suppose you mean p_{n_0}(t) = e^{-lambda * n_0 * t}? That is, the left-hand side is p_{n_0}(t), not p_{n_0+1}(t)? For p_{n_0 + 1}(t), we have p_{n_0+1}(t) = n_0*e^{-λ*n_0*t}(1 - e^{-λ*t}). See around 18:25

@@BruneiMathClub Sorry, I meant that I cant solve p_{n_0+1}(t) = n_0*e^{-λ*n_0*t}(1 - e^{-λ*t}). I tried to do in the same way like in poisson process, but the product rule does not apply right in the same way than in poisson process, and I find it impossible to solve.

@@marrey7 I see. You are right. The method I showed for the Poisson process doesn't work for the birth process. In this case, you can use a more standard method for solving inhomogeneous linear ODEs: the method of variable parameters. For example, see ua-cam.com/video/wJyJaDwt48w/v-deo.html

@@BruneiMathClub Oh thank you so much