Tag Info

Hot answers tagged

7

Let $X$ be a standard normal random variable and let $Y = X^2$. Then, since $E(X) = E(X^3) = 0$, we have $E(XY) = E(X^3) = 0 =E(X)E(Y).$ However, they are not independent: $$P(0<X<1,Y>1) = 0 \neq P(0<X<1)P(Y>1)$$ For a simple discrete example, let $X_1$ and $X_2$ be independent random variables each taking values in $\{0,1\}$ with ...


3

Given iid random variables $X_i, i = 1, \ldots, n$. Define $Z_j = \frac{X_j}{\sum_{i=1}^n X_i}.$ Clearly, $$\sum_{j=1}^n Z_j = 1.$$ Take variance on both sides, we get $$ \begin{split} Cov(\sum_{j=1}^n Z_j, \sum_{j=1}^n Z_j) &= 0. \end{split} $$ We know $$ \begin{split} Cov(\sum_{j=1}^n Z_j, \sum_{j=1}^n Z_j) &= \sum_{j=1}^n Var(Z_j) + 2\sum_{j ...


3

You are putting the cart before the horse. Two objects are not considered to be dissimilar because they are orthogonal under some inner product; rather, the inner product is chosen so that two dissimilar objects would have small inner product. Going back to the vector space example: inner products can be taken relative to any symmetric positive definite ...


2

Well a related formula for random variables that sum to a constant $c$ is the following: Suppose $Y_1, \ldots, Y_N$ are random variables that satisfy $\sum_{i=1}^N Y_i = c$. Then: \begin{align} c &= \sum_{i=1}^NE[Y_i] \\ c^2 &= \sum_{i, j} E[Y_i]E[Y_j] \: \: (*) \end{align} Also: \begin{align} c^2 &= \sum_{i, j} Y_i Y_j\\ c^2 &= ...


1

I don't think so. $\begin{align} \rho & = \mathsf{Corr}(X,Y) \\[1ex] & = \dfrac{\mathsf E\Big(\big(X-\mathsf E(X)\big)\big(Y-\mathsf E(Y)\big)\Big)}{\mathsf E\Big(\big(X-\mathsf E(X)\big)^2\Big)^{1/2} \;\mathsf E\Big(\big(Y-\mathsf E(Y)\big)^2\Big)^{1/2}} \\[1ex] & =\frac{\mathsf ...


1

Actaually, perpendicularity is defined in terms of the inner product. You can define different inner products that give a very different notion of perpendicularity; in fact another inner product may turn some "almost parallel" vectors into orthogonal ones. But on the other hand tis means: Whatever otherwise "externally justified" idea of ...


1

There are two parts to your question. For the second part, assuming X(t) and Y(t') are independent, requires the cross spectral density to be zero. The cross spectral density is the Fourier transform of the cross correlation function. The cross correlation is the ensemble average of the time-shifted product of X(t) and Y(t'), and if these are independent ...


1

Assuming that $$y = \frac{A_{i}}{ln[x]^{3}} + B_{i}$$ is the right model to use, the general approach you used is very correct for me (at least in its principle). For each value of $z$, you adjusted the $A_z$ and $B_z$ and plotting the values of these coefficients as a function of $z$ you observed regular trends and you decided to fit them using a cubic ...



Only top voted, non community-wiki answers of a minimum length are eligible