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0

The energy distance[1] and the kernel two-sample test[2] provide two-sample tests for arbitrary distributions, defined using samples. In your case, you might be able to arbitrarily create sample sets that are uniformly distributed, then measure the distributional distance (by energy distance or Maximum Mean Discrepancy [MMD]) between candidate sets and ...


2

This is known as the probability integral transform. For $0<t<1$ we have \begin{align} F_Y(t) &= \mathbb P(Y\leqslant t)\\ &= \mathbb P(F_X(X)\leqslant t)\\ &= \mathbb P(X\leqslant F_X^{-1}(t))\\ &= F_X(F_X^{-1}(t))\\ &= t, \end{align} so that $Y$ is uniformly distributed over $(0,1)$. Note that when $X$ is not continuous the map $...


1

You want to find the probability density function; not the cumulative distribution function. $$\begin{align}f_{X^3}(u) &=\dfrac{\mathrm d~~~}{\mathrm d~u}F_{X^3}(u)\\[1ex] &=\dfrac{\mathrm d~~~}{\mathrm d~u}\mathsf P(X\leq u^{1/3})\\[1ex] &=\dfrac{\mathrm d~\tfrac 13u^{1/3}}{\mathrm d~u~~~}\mathbf 1_{u^{1/3}\in[0..3]}\\[1ex]&=\tfrac 19 u^{-2/...


0

By the LOTUS, we have $$E(\vert X - Y \vert) = \int_{0}^{1}\int_{0}^{1} \vert x - y \vert \, \mathrm{d}x \,\mathrm{d}y.$$ The $\vert x - y\vert$ term inside the integral takes one of the two forms $|x-y| = x-y$ for $x - y \geq 0$ and $|x -y| = y - x$ for $x - y < 0.$ So we split the region of interest into two regions (see attached image). 1 In the ...


0

$$\begin{align} & E(|X - Y|) = \\ &= \int_0^1 \int_0^1 |x - y| \ dx dy \\ &= \int_0^1 \int_y^1 (x - y) \ dxdy + \int_0^1 \int_0^y (y - x) \ dxdy \text{$\qquad$ Seperate for $y<x$ and $x<y$} \\ &= 2 \int_0^1 \int_y^1 (x - y) \ dxdy = 1/3 \text{$\qquad$ Since they are equal, compute one of them and times 2} \\ \end{align}$$


-2

So much time has passed, but at least now we know it can be done: https://en.wikibooks.org/wiki/Statistics/Distributions/Discrete_Unif


1

I don't understand why one would use MGF for this . You get it from definition: $P(Y<c)=P(X<\frac {c-a} {b-a}) =\frac {c-a} {b-a}$ for $c$ between $a$ and $b$ and this is the definition of uniform distribution.


0

For part b) you've incorrectly taken $\ p=\frac{1}{2}\ $. If $\ A\ $ is the event $\ \left\{ 0.3\le p\le0.7\right\}\ $, and $\ B\ $ the event that $6$ out $10$ tosses come up heads, then \begin{align} P\left(A\cap B\right) &= \int_{0.3}^{0.7} P\left(B\left\vert\, p=x\right.\right)dx\ ,\ \text{and}\\ P\left(B\right) &= \int_0^1 P\left(B\left\vert\, p=...


0

The correct value of the triple integral is $\int_0^{1}\int_0^{1-x_3} \int_{x_2+x_3} ^{1} dx_1dx_2dx$ which evaluates to $\frac 1 6$.


0

We may directly compute the density of $Y:=X_2+X_3$ by convolution: $$ (f_{X_2}\star f_{X_3})(t) = \begin{cases} \int_0^t \ \mathsf dt = t,& 0\leqslant t<1\\ \int_{t-1}^1 \ \mathsf dt = 2-t,& 1\leqslant t\leqslant 2. \end{cases} $$ We can then compute \begin{align} \mathbb P(X_1>Y) &= \mathbb P(X_1>Y,0\leqslant Y <1) + \mathbb P(X_1&...


2

Euler's formula states that, for any planar graph, $f - e + v = 2$, where $f$ is the number of faces (what you call regions, with an important caveat to be discussed at the end of this post), $e$ is the number of edges, and $v$ is the number of vertices. Each line necessarily intersects the circle at exactly $2$ distinct points (the case of a tangent line ...


0

Following the hint from antkam, With $L$ lines intersecting $P$ points, we have number of regions at $R=1+L+P$. $E(P(L))$ can be calculated by linearity of expectation, which means for any two lines, there is $\frac{1}{3}$ probability that they intersect and we have $C^2_L$ combinations. So, $E(R(L))=1+L+\frac{C^2_L}{3}$.


6

Hint: Are they each uniform on their support? What are the highest and lowest possible values of $3-4U$ ? What are the highest and lowest possible values of $4(1-U)$ ? Do they have the same distribution?


1

If $X \sim Y$ and $X' \sim Y'$ we cannot say $X+X' \sim Y+Y'$ without some independence assumptions. In this case $1-U$ and $cU$ are not independent (unless $c=0$) so the argumnet fails.


2

The idea is to show that $f$ is the rectangle function convolved with itself, and then to show that the convolution can be expressed as a polynomial on each $(k,k+1)$. Let $\operatorname{sinc}x = { \sin x \over x}$, and use $\hat{g}(\omega) = \int_{-\infty}^\infty g(t) e^{-i\omega t} dt$ to denote the Fourier transform of an $L^1$ function $g$. If $\hat{g}$...


1

Let $Z=(X+Y)^2$. The variance of $Z$ is $E(Z^2)-E(Z)^2$. Now \begin{align} E(Z^2)=E((X+Y)^4)&=E(X^4+4X^3Y+6X^2Y^2+4XY^3+Y^4)\\ &=E(X^4)+4E(X^3)E(Y)+6E(X^2)E(Y^2)+4E(X)E(Y^3)+E(Y^4) \end{align} using linearity of expectation and independence. To proceed further use $E(X^k)=E(Y^k)=1/(k+1)$ (by a simple integration). The calculation of $E(Z)$ is similar,...


0

This is essentially a change of variable. Since $X$ takes values between $0$ and $L$, the random variable $Y=X(1-X)$ takes values between $0$ and $1/4$ (if $L \geq 1/2$) or between $0$ and $L(1-L)$ (if $L \leq 1/2$). I'll deal with the second cas. For any bounded measurable function $f$, $$\mathbb{E} (f(Y)) = \mathbb{E} (f(X(1-X))) = \frac{1}{L} \int_0^L ...


0

Comment: The first step is to determine the support of $Y.$ Then the so-called 'PDF method' or 'transformation method' should work nicely. Maybe you can start on this, edit results into your Question, and let us know of any specific difficulties you encounter. It seems worthwhile to look at two kinds of examples: The following brief simulation in R (for $L ...


0

In hyperbolic Geometry the area of a circle is given by the equation $$a=4{\pi}\left(Rsinh\left(\frac{r}{2R}\right)\right)^2$$ with $a$ being the area and $r$ being the radius of the circle, and $R$ being a constant for any hyperbolic plane. In order to generate random points inside the circle that follow a uniform distribution, I can first generate ...


0

We can use the following: for each permutation $\sigma$ from $\{1,\dots,n\}$ to itself, $$\tag{*} \mathbb E\left[Y_1+\dots+Y_k\mid Y_1+\dots+Y_n\right]=\mathbb E\left[Y_{\sigma(1)}+\dots+Y_{\sigma(k)}\mid Y_1+\dots+Y_n\right], $$ which is due to the fact that the vectors $\left(Y_1+\dots+Y_k,Y_1+\dots+Y_n\right)$ and $\left(Y_{\sigma(1)}+\dots+Y_{\sigma(k)},...


1

Given $U\sim Unif(0,1)$, $X=F^{-1}(U)$, to show that $X$ has CDF $F$, we check \begin{align} P(X\le c)&=P(F^{-1}(U)\le c)\\&=P(U\le F(c))\\&=F(c) \end{align}


0

Consider the analogous situation in spherical geometry: We have a circle (a spherical cap) with radius $r$ and centre $\vec c\in\mathbb R^3,\;\vec c\cdot\vec c=1$. The cap is the set of all $\vec x\in\mathbb R^3$ such that $$\vec x\cdot\vec x=1,\quad\cos r\leq\vec x\cdot\vec c\leq1.$$ Following the idea from Wikipedia (Marsaglia), generate some random ...


1

Yes, the authors are saying 'linear' for 'affine'. If $X$ has uniform distribution on $(0,1)$ then $Y=X^{2}$ does not have uniform distribution on $(0,1)$: $P(Y \leq y)= P(X \leq \sqrt y)=\sqrt y$ for $0<y<1$. Note that $X^{2} <X$. $X^{2}$ assigns higher probabilites than $X$ for values near $0$.


1

Rather than computing densities, you can work with moments: because the $X_{(k)}$ and the $V_j$ all take values in $(0,1)$, their joint moments fully determine their distributions. For example, it's easy to check that for a positive integer $p$, $$ \Bbb E[V_1^p]=2\int_{0<u<v<1}u^pv^{-p}\,du\,dv=2\int_0^1\int_0^v u^pv^{-p}\,du\,dv={1\over p+1}. $$ As ...


0

Since $X$ and $Y$ are iid Uniform Random variables, the joint density of $(X, Y)$ is $$f(x,y)=1, \qquad 0<x<1, \qquad 0<y<1 $$ Given that $Z=|X-Y|.$ Assume another variable $U=Y.$ The above set of transformation from $$S_{x,y}=\{(x,y): 0<x<1, 0<y<1 \}$$ to $$S_{Z,U}=\{ (z,u): 0<z<1,0<u<1, z+u\leq 1 \} $$ is not one-to-...


2

What does it mean that $N>n$? It means that for $k=2,3,...,n$ we can't have $U_k>U_{k-1}$, otherwise $N$ would be at most $k$. So $N>n$ if and only if $U_1\geq U_2\geq...\geq U_n$. Now, since the random variables are continuous we have $P(U_1\geq U_2\geq...\geq U_n)=P(U_1>U_2>...>U_n)$.


2

I don't know what you mean by shifted, but assuming that $Y_i$s all have the same parameter for their distribution, the random you are looking for will be binomial with parameters (p, c) with $p = P(X_i + Y_i \leq 1)$. This is because you can think of each event $\{X_i + Y_i \leq 1 \}$ as an independent trial (say a coin toss), and you will do it $c$ times. ...


2

If you look at the graph of $\tan(x)$, you see that you have two cases: If $y<0$, then: $$P(Y\le y) = P(\tan(X)\le y)=\\P(-\pi/2\lt X\le \arctan(y))+P(\pi/2\lt x\le\arctan(y)+\pi)$$ If $y\ge0$, then: $$P(Y\le y) = P(\tan(X)\le y)=\\P(-\pi\lt X\le \arctan(y)-\pi)+P(-\pi/2\lt x\le\arctan(y))+P(\pi/2\lt x\lt\pi)$$ In both cases: $$F_Y(y)=P(Y\le y) = \...


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