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6

If you need to generate $n$ correlated Gaussian distributed random variables $$\bf Y \sim \mathcal N(\bf \mu, \Sigma)$$ where $\textbf{Y} = (Y_1,\dots,Y_n)$ is the vector you want to simulate, $\mu =(\mu_1,\dots, \mu_n)$ the vector of means and $\Sigma$ the given covariance matrix, you first need to simulate a vector of uncorrelated Gaussian random ...

5

Yes, although the restriction that all entries are between $-1$ and $1$ is implied by the other properties (and so is not needed). Let $\Sigma$ be a $n \times n$, symmetric, positive semidefinite matrix with $1$'s along the main diagonal. First, $\Sigma$ is a covariance matrix. Since $\Sigma$ is symmetric and positive semidefinite, $\Sigma$ has a ...

5

For any two random variables: $$\text{Var}(X+Y) =\text{Var}(X)+\text{Var}(Y)+2\text{Cov}(X,Y).$$ If the variables are uncorrelated (that is, $\text{Cov}(X,Y)=0$), then $$\tag{1}\text{Var}(X+Y) =\text{Var}(X)+\text{Var}(Y).$$ In particular, if $X$ and $Y$ are independent, then equation $(1)$ holds. In general $$\text{Var}\Bigl(\,\sum_{i=1}^n X_i\,\Bigr)= ... 4 To do maximum likelihood you would need to make an assumption about the distribution of x. But in any case your system is underidentified, and you cannot obtain a unique in some optimal sense set of values for the four coefficients. To see this clearly, take the natural logarithms (you can do this since you assume \beta >\alpha):$$\ln y = ...

4

The covariance is bilinear, hence $$\mathrm{Corr}(V_1,V_2)=\frac{\mathrm{Cov}(V_1,V_2)}{\sqrt{\mathrm{Var}(V_1)\mathrm{Var}(V_2)}}=\frac{\sum\limits_{j=1}^n\sum\limits_{k=1}^na_{1j}a_{2k}\mathrm{Cov}(X_j,X_k)}{\sqrt{\mathrm{Var}(V_1)\mathrm{Var}(V_2)}},$$ where, for $i=1$ and $i=2$, $$... 4 Let A be an n\times n matrix of your desired form. Let us form the matrix of all ones, call it J. Then we can represent A as$$A=\rho J - (\rho - 1)I$$The determinant of the above matrix is$$\det(A) = \rho^n\det\left(J - \frac{\rho - 1}{\rho}I\right)$$Letting \lambda = \frac{\rho - 1}{\rho}, the latter determinant is precisely the characteristic ... 4 By definition, two random variables X and Y satisfy your equation if and only if their covariance is zero. Thus, any pair of dependent random variables with zero covariance give the counterexample you're looking for. For instance, let X be uniformly distributed between -1 and 1, and let Y = X^2. Then$$\mathrm E[XY] = \mathrm E[X^3] = 0 = ...

3

I have one idea about this procedure you may have found helpful. In the linked post, the answer advises you to draw samples randomly until you reach the desired correlation. However, I guess it might take some time if you draw these samples independently of the original sequence. Let us use another trick - namely we construct a random sequence from the ...

3

The formula is $$r = b_1 \frac{s_x}{s_y},$$ where $r$ is the correlation, $b_1$ is the slope, and $s_x$ and $s_y$ are the standard deviations of the independent $(x)$ and dependent $(y)$ variables, respectively. A reference is Wikipedia's page on simple linear regression. See the formula for $\hat{\beta}$.

3

Empirical characteristics of a typical cell $\widehat C$ are usually defined by the ergodic limits $$E(\varphi(\widehat C))=\lim_{R\to\infty}\frac1{|\mathcal C_R|}\sum\varphi(C)\cdot[C\in \mathcal C_R],$$ defined for every suitable function $\varphi$, where $\mathcal C_R$ is the almost surely finite collection of cells $C$ such that $C\subseteq B_R$ or ...

3

The standard example for this is considering a continuous, uniformly distributed variable $X$ on $[-1,1]$ and then looking at $X$ and $X^{2}$. They are uncorrelated, but obviously dependent. The Wikipedia article on "Uncorrelated" cites this example and gives another, more explicit, discrete example with calculations. As is mentioned in the Wikipedia ...

3

A picture is (to me) essential. Draw the line $x+y=1$, the line $y=x+1$. Our pairs $(X,Y)$ live in the region that lies below each of these two lines, and above the $x$-axis. The region is a triangle of area $1$. It has corners $(1,0)$, $(0,1)$, and $(-1,0)$. The hard part is now finished. (a) We want $E(XY)-E(X)E(Y)$. By symmetry we have $E(XY)=0$ and ...

3

You can use the fact, that correlations can be understood as cosines between vectors from the common origin. Then apply the arccos-function, and check, whether all possible pairwise sums are greater than the third angle, such that they make a tetraeder. I get [acos(0.9),acos(0.8),acos(0.1)] %1695 = [0.451026811796, 0.643501108793, 1.47062890563] The sum ...

3

Assume without loss of generality that the random variables $A$, $B$, $C$ are standard, that is, with mean zero and unit variance. Then, for any $(A,B,C)$ with the prescribed covariances, $$\mathrm{var}(A-B+C)=\mathrm{var}(A)+\mathrm{var}(B)+\mathrm{var}(C)-2\mathrm{cov}(A,B)-2\mathrm{cov}(B,C)+2\mathrm{cov}(A,C),$$ that is, $$... 3 I am afraid this is a non-answer: probably. But imagine tossing a fair coin 100 times. Let A be the number of heads in the first 50 tosses, B the number of heads in the full 100 tosses, and C the number of heads in the last 50 tosses. Then A and B are (weakly) positively correlated, as are B and C, but A and C have correlation ... 3 Correlations are cosines of angles in L^2 hence, if \varrho_{AB}\geqslant 0 and \varrho_{BC}\geqslant 0, then$$ \varrho_{AC}\geqslant\varrho_{AB}\varrho_{BC}-\sqrt{1-\varrho_{AB}^2}\cdot\sqrt{1-\varrho_{BC}^2}. $$Thus, if \varrho_{AB}\geqslant c and \varrho_{BC}\geqslant c with c\geqslant0, then$$ \varrho_{AC}\geqslant2c^2-1. $$For example, ... 3 You said it yourself: if \mathcal F contains every complex exponential function and if X and Y are such that \rho_\mathcal F(X,Y)=0, then \mathrm E(\mathrm e^{\mathrm i(xX+yY)})=\mathrm E(\mathrm e^{\mathrm ixX})\mathrm E(\mathrm e^{\mathrm iyY}) for every (x,y) in \mathbb R^2. This means the Fourier transform of the distribution \mathrm ... 3 The ratio of two random variables does not in general have a well-defined variance, even when the numerator and denominator do. A simple example is the Cauchy distribution which is the ratio of two independent normal random variables. As Sivaram has pointed out in the comments, the formula you have given (with the correction noted by Henry) is for the ... 3 As others have noted, the formula you provide is incorrect. For general distributions, there is no closed formula for the variance or a ratio. However, you can approximate it by using some Taylor series expansions. The results are presented (in strange notation) in the this pdf file. You can work it out exactly yourself by constructing the Taylor series ... 3 We always have |\rho| \le 1. If the joint distribution of X and Y is concentrated on a straight line with positive slope, i.e. Y = a X + b for some constants with a > 0, then \rho = 1; if it is concentrated on a straight line with negative slope, \rho = -1; if X and Y are independent \rho = 0. In other cases \rho tells you, in a ... 3 I'm going to assume that S is already known; otherwise, you can just choose S=I and you're done--except for an m/n issue here that I do not think can be avoided; you'll see below. Let \tilde{Z} be an n\times m random matrix with elements independently drawn from \mathcal{N}(0,1).$$\textbf{E}(\tilde{Z}_i\tilde{Z}_j^T) = \begin{cases} I & i=j ...

3

Hint: In order for the correlation to be well-defined we must assume that $X$ and $Y$ are not degenerate ($X$ being degenerate meaning that $X=a$ almost surely). Now, show that $$\exists a,b\in\mathbb{R}:Y=aX+b\quad\text{a.s.}\iff \exists a\in\mathbb{R}\setminus\{0\}:\,\mathrm{Var}(Y-aX)=0.$$ Then you just need to show that $$... 3 Obtain this paper, written by T.S. Ferguson, a professor at UCLA (his page is here). It derives the joint asymptotic distribution for the sample mean and sample median. To be specific, let \hat X_n be the sample mean and \mu the population mean, Y_n be the sample median and \mathbb v the population median. Let f() be the probability density of ... 3 You cannot conclude that \mathrm{Cov}(X,Y)\geq 0 in general. To see this, let X\sim\mathrm{bin}(1,p) and Y=1-X, then obviously P(X\geq 0)=P(Y\geq 0)=1 but$$ \mathrm{Cov}(X,Y)=\mathrm{Cov}(X,1-X)=-\mathrm{Cov}(X,X)=-\mathrm{Var}(X)<0 $$if p\in (0,1). In general you have the bound (see e.g. this answer):$$ |\mathrm{Cov}(X,Y)|\leq ...

2

What you need is the double expectation formula: $$\DeclareMathOperator{\E}{E} \E(X) = \E \E (X|Y)$$ In the double expectation, the inner expectation is a function of $Y$, that is, a random variable, and the outer expectation then is then taken over the distribution of $Y$. Assume $\E X = \E Y = 0$ (we can do that without loss of geneality, since ...

2

For the last (and important) edit/answer see the text after the last horizontal line I'm pretty sure, the formula for the correlation as given by Wikipedia helps you a lot: https://secure.wikimedia.org/wikipedia/en/wiki/Correlation Especially the sample correlation coefficient. As well it might be helpful to use the linearity of the expected value: ...

2

If two random variables are linearly dependent then you will find $r=\pm 1$. To be linearly dependent means that one is a linear function of the other: $$Y = a + bX$$ However, you can have a linear correlation without linear dependence, for example $$Y = a + bX + \epsilon$$ where $\epsilon$ is some other random variable which is independent from $X$. In ...

2

This is a direct consequence of Rearrangement Inequality. All you need to assume WLOG is that the elements of your set are ordered. This assumption does not affect your problem statement as it is. Then, application of rearrangement inequality gives the result you desire. [1] http://en.wikipedia.org/wiki/Rearrangement_inequality

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