Stack Exchange Network

Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Visit Stack Exchange
26

By the definition of the correlation coefficient, if two variables are independent their correlation is zero. So, it couldn't happen to have any correlation by accident! $$\rho_{X,Y}=\frac{\operatorname{E}[XY]-\operatorname{E}[X]\operatorname{E}[Y]}{\sqrt{\operatorname{E}[X^2]-[\operatorname{E}[X]]^2}~\sqrt{\operatorname{E}[Y^2]- [\operatorname{E}[Y]]^2}}$$ ...


11

Comment on sample correlation. In comparing two small independent samples of the same size, the sample correlation is often noticeably different from $r = 0.$ [Nothing here contradicts @OmG's Answer (+1) on the population correlation $\rho.]$ Consider correlations between a million pairs of independent samples of size $n = 5$ from the exponential ...


6

A single random variable has a distribution; a sample mean from a random sample is a single random variable. Of course you can only observe its distribution by looking at multiple random samples (such as multiple sample means); then as the number of such samples increases the sample (empirical) cdf will approach the population distribution function. The ...


3

Simple answer: if 2 variables are independent, then the population correlation is zero, whereas the sample correlation will typically be small, but non-zero. That is because the sample is not a perfect representation of the population. The larger the sample, the better it represents the population, so the smaller the correlation you'll have. For an ...


2

With some simple algebra, you obtain the inverse relationship: $$\phi_k = -i \ln Z_k + i \ln A - 2 \pi B.$$ Hence, taking $A$ and $B$ to be constants (which should really be denoted as lower-case), you have: $$\begin{equation} \begin{aligned} \mathbb{E}(\phi_k) &= -i \mathbb{E}(\ln Z_k) + i \ln A - 2 \pi B, \\[10pt] \mathbb{V}(\phi_k) &= \mathbb{V}...


1

Given that $\sigma_t^2$ is part of $I_{t-1}$ because of $$ \sigma_t^2 = \alpha_0 + \sum_{i=1}^p \alpha_i a_{t-i}^2 + \sum_{j=1}^q \beta_j \sigma^2_{t-j} $$ and given $$ \epsilon_t \stackrel{iid}{\sim} \text{WN}(0, 1), $$ couldn't you do $$ \begin{equation} \begin{aligned} E(a_t^2 | I_{t-1}) &= \text{var}(a_t | I_{t-1}) \\ &= \text{var}(\sigma_t \...


1

Suppose $A\in\mathbb R^{m\times n}$ and $B\in\mathbb R^{n\times m}.$ Then $\operatorname{tr}(AB) = \operatorname{tr}(BA).$ The proof of that is routine. So we have $Z\sim N_n(0, I_n)$ and $T\in\mathbb R^{n\times n}.$ Then \begin{align} & \operatorname E(Z'TZ) = \operatorname E(\operatorname{tr}(Z'TZ)) = \operatorname E(\operatorname{tr}(TZZ')) \\[8pt] = ...


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