statistics – Fisher information of joint distribution of transformed Normal distribution

Suppose $X_1=theta+epsilon_1$ and $$X_i=sqrt{gamma}X_{i-1}+sqrt{1-gamma}epsilon_i+theta(1-sqrt{gamma})$$
Where $gamma in (0,1)$ and $theta$ is the parameter of the model. Also $epsilon_1,epsilon_2,…epsilon_n$ are iid $N(0,1)$.

What is the Fisher information of this model and for what values of $gamma$ does it tensorise. I’ve tried using the Jacobian to find the joint distribution but I’m not sure, especially when determining for which values we have tensorisation. Any help would be much appreaciated.

probability – Given the joint p.d.f., are $X$ and $Y$ independent? What is the conditional p.d.f. of $X$ given $Y=y$

Let $(X,Y)$ be two random variables with joint p.d.f.
$$f_{X,Y}(x,y)=begin{cases}frac{1}{y}e^{-x/y}e^{-y}&text{if }0<x<inftytext{ and }0<y<infty\0&text{otherwise}end{cases}$$

  1. What is the marginal of $Y$?
  2. Are $X$ and $Y$ independent?
  3. What is the conditional p.d.f. of $X$ given $Y=y$?
  4. What is $mathbb{E}(X|Y)$?

My Attempt

  1. The marginal of $Y$ can be found by integrating the joint p.d.f. w.r.t. $x$. So
    $$f_Y(y)=int^infty_0f_{X,Y}(x,y)dx=int^infty_0frac{1}{y}e^{-x/y}e^{-y}dx=left(-e^{-x/y-y}right)^infty_0=e^{-y}$$
  2. To know if they are independent, we must check that $f_{X,Y}(x,y)=f_X(x)f_Y(y)$. This means we need to find the marginal p.d.f. for $X$. My problem is that the integral is quite hard, and even after using an online integral calculator, I was given a gamma function that could not be computed for the bounds $(0,infty)$. Since this is not a calculus class, I was wondering if there is some easier way to check independence that I am missing.
  3. I know the formula for the conditional p.d.f. of $X$ given $Y$:
    $$f_{X|Y}=frac{f_{X,Y}(x,y)}{f_Y(y)}=frac{frac{1}{y}e^{-x/y}e^{-y}}{e^{-y}}=frac{1}{y}e^{-x/y}$$
    My concern is that I was asked for the conditional p.d.f. of $X$ given $Y=y$. Is there a difference between the two or is it just different ways of writing it?
  4. $$mathbb{E}(X|Y)=int^infty_0xf_{X|Y}(x|y)dx$$
    Using the last part, we have
    $$=int^infty_0frac{x}{y}e^{-x/y}dx=left(-(x+y)e^{-x/y}right)^infty_0=y$$

Joint boundedness of solutions of a family of Sturm-Liouville ODE

Let us fix $0 neq lambda in mathbb{R}$. Let us consider the following ODE, on $(0,infty)$: $$ y^{prime prime} (x) + frac{r e^{-x}}{(1+e^{-x})^2} y(x) = -lambda^2 y(x).$$ Here $r ge 1$ is a parameter. Let us consider the solution $e_r (x)$ which satisfies $e_r (x) sim e^{i lambda x}$ as $x to +infty$. How would one approach showing (if this is indeed true) that $$sup_{substack{r ge 1\ x in (0,infty)}} |e_r (x)| < infty.$$ Ideally, I am interested in some wider class of examples, so I am less interested in a "trick" that happens to work in this very particular case, and more in some conceptual approach, but still would like to hear all approaches.

Thank you

list manipulation – How to convert decimals to integers and joint them to a string for exporting

Try

StringJoin[StringTake[StringReplace[ToString[Flatten[Transpose[
  {{a,b,c,d},{0.1,2,0.15,5}}]]],{","->"","."->""," "->""}],{2,-2}],
  "=",ToString[{0.1,2,0.15,5}]]

Start from the innermost nested function, strip that out, run it, see what it does, then look it up in the documentation and see if you can understand why that does what it does and how it gets the input slightly closer to the output that you want, then repeat with each next outer layer.

Then you can export that resulting string to your file.

If you carefully do that then you should greatly improve your skills to be better able to do your next problem.

c++ – How to create a joint, where the rotation of one object is relative to the other?

I would like to create a top down view scene, with one rectangle (top) is on top of an other (base). There is a joint, where these two objects are connected.

The one on the top should be able to freely rotate, but if the base rotates, its rotation should be changed as well, because it should behave as if it is on top of it. If the joint is at the center of both objects, then the top one should rotate like this:

Δ top rotation = Δ time * ( top angular velocity + base angular velocity )

If the joint is not at the center of the objects, then it is more complicated than that. How is it possible to define a joint like that in Box2D?

This is a testbed example with the joint. Notice, that the top one is not rotating:

#include "test.h"

class RelJoint : public Test
{

    b2Body* base;
    b2Body* top;
    b2RevoluteJoint* m_joint;

public:

    RelJoint() {

        m_world->SetGravity(b2Vec2(0.0f, 0.0f));

        {

            b2PolygonShape shape;
            shape.SetAsBox(1.f, 0.5f, b2Vec2(0.0f, 0.0f), 0.0f);

            b2BodyDef bd;
            bd.type = b2_dynamicBody;
            b2Body* body = m_world->CreateBody(&bd);
            body->CreateFixture(&shape, 5.0f);

            body->SetAngularDamping(1.f);

            base = body;
        }

        {

            b2PolygonShape shape;
            shape.SetAsBox(0.5f, 0.25f, b2Vec2(0.0f, 0.0f), 0.0f);

            b2BodyDef bd;
            bd.type = b2_dynamicBody;
            b2Body* body = m_world->CreateBody(&bd);
            body->CreateFixture(&shape, 5.0f);

            top = body;
        }

        b2RevoluteJointDef revoluteJointDef;
        revoluteJointDef.bodyA = base;
        revoluteJointDef.bodyB = top;
        revoluteJointDef.collideConnected = false;

        revoluteJointDef.localAnchorA.Set(0, 0);
        revoluteJointDef.localAnchorB.Set(0, 0);

        m_joint = (b2RevoluteJoint*)m_world->CreateJoint(&revoluteJointDef);

    }

    void Step(Settings& settings) override {
        base->ApplyTorque(3.f, true);

        Test::Step(settings);
    }

    static Test* Create()
    {
        return new RelJoint;
    }

};

static int testIndex = RegisterTest("Joints", "Relative", RelJoint::Create);

pr.probability – Joint distribution of dependent Gaussians and their product

Consider a pair of dependent zero mean unit variance Gaussians, $$X,Y sim mathcal{MVN}left(vec{0},begin{pmatrix}1 & rho \ rho & 1end{pmatrix}right).$$
Their product $Z:=Xcdot Y$ is known to follow the variance gamma distribuiton with density
$$p(z)=frac{1}{pisqrt{1-rho^2}} expleft(frac{rho z}{1-rho^2}right) K_{0}left(frac{vert zvert}{1-rho^2}right),$$ where $K_0$ is the order zero modified Bessel function of the second kind (see Nadaraja and Pogány, 2016).

Is the joint distribution of $X,Y,Z$ known?

The distribution of $W:=aX +bY+cZ$ is also of interest.

Note: This is cross-posted from math.stackexchange.com. Given the recency of the results of which I’m looking for an extension, I think this might be more appropriate here. Let me know in the comments if I’m mistaken!

optimization – Expression for Pareto Frontier with Selection from Joint Distribution

Suppose you’re admitting a fixed fraction of some population using a linear function of their characteristics, can you characterize the Pareto frontier from the correlation of characteristics?

Each person has a vector of characteristics $(x_1,…,x_n)sim N$, drawn from a multivariate Normal density. You are choosing elements using weights, and we want to know the average value of scores for people admitted:

$$E(x_i|sum w_ix_i > k),$$

where $P(sum w_ix_i>k)$ is held constant, i.e. you admit a constant fraction of applicants.

From these conditional expectations we can draw a Pareto frontier: i.e., as you vary the weights $(w_1,ldots,w_n)$ you will map out a space of conditional expectations on each dimension.

The question: can we get a closed-form solution for the Pareto frontier?

When it’s just two-characteristic I think I’ve solved it (the Pareto frontier is an ellipse, the isovalue of the joint density), but I want to find a generalization to n>2, and I guess this problem must be a textbook model in some discipline, but I can’t find the right jargon to get a reference.

Calculating joint probability of three dependent random variables

$$A_N^{(j)}sim Bin(bN, p/N)$$
$$S_N^{(j)}=A_N^{(j)}1_{{A_N^{(j)}lekappa}}$$
$$R_N^{(j)}=1_{{A_N^{(j)}lekappa}}$$
with
$$A_N=sum_{j=1}^{N}A_N^{(j)} ;;and;; S_N=sum_{j=1}^{N}S_N^{(j)} $$

I know that $PBig(frac{1}{N}A_N=aBig)$ is approximately Poisson with parameter $pb$ as $Nrightarrowinfty$.

Any idea how on how to proceed in calculating the equation below?
begin{equation}
P_N(A_N=aN,S_N=sN,R_N=rN)=P_NBig(sum_{j=1}^{N}(A_N^{(j)},S_N^{(j)},R_N^{(j)})=(a,s,r)Big)
end{equation}

Joint Statement by Ripple and SEC

The Securities and Exchange Commission (SEC) and Ripple sent a joint letter to federal judge Analisa Torres from the US District Court for the Southern District of New York, saying there is no possibility of a solution "right now"
The original text is as follows:
— “The parties were met, discussed and the possibility of compromise was discussed, I do not believe that there is a possibility of a solution at the moment. However, if any solution is reached in principle for any defendant, the…

Read more

EZ Releaf CBD: Reviews, Joint Pain Relief, Benefits and Where to Buy?

EZ Releaf CBD will permit the purchaser to ensure that they can have a decent stomach related framework. It won’t allow them to experience the ill effects of the issue of acid reflux, stomach spasms or constipation.It will ensure that the purchaser is getting sound and is having more perseverance. EZ Releaf CBD will permit the shopper to have more energy and stamina.This CBD oil will permit the customer to guarantee that they are having a decent flow of blood in the body. It won’t allow one to experience the ill effects of the issue of high or low circulatory strain rate.Green EZ Releaf CBD will empower the individual to dispose of the issue of uneasiness or endurance.
https://sites.google.com/view/ez-releaf-cbd/
https://sites.google.com/view/ez-releaf-cbd-order/
https://sites.google.com/view/ez-releaf-cbd-buy/