# Tag Info

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{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{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