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10 cze 2018 · If you know that $\rho_{X Y} = \sigma_{X Y} / (\sigma_X \sigma_Y)$ is zero, then $\sigma_{X Y}$ must be zero. Note that the covariance of two independent variables is $\sigma_{X Y} = E[(X-EX)(Y-EY)] = E[X Y] - E[X] E[Y] = 0$, because by independence $E[X Y] = E[X] E[Y]$.
29 sty 2017 · If $X$ and $Y$ are two random variables that can only take two possible states, how can I show that $Cov(X,Y) = 0$ implies independence? This kind of goes against what I learned back in the day tha...
19 gru 2021 · Consider three normally distributed random variables, $X,Y,Z$ where $Cov(X,Y)<0$ and $Cov(X,Z)>0$. Can we say anything about the sign of $Cov(Y,Z)$? Intuitively, $Y$ goes down when $X$ goes up.
24 kwi 2022 · \( \cov(X, Y) = 0\), \(\cor(X, Y) = 0\). \( X \) and \( Y \) are independent. \(\cov(X, Y) = \frac{a^2}{9}\), \(\cor(X, Y) = \frac{1}{2}\). \( X \) and \( Y \) are dependent. \(\cov(X, Y) = 0\), \(\cor(X, Y) = 0\). \( X \) and \( Y \) are dependent.
We know that variance measures the spread of a random variable, so Covariance measures how two random random variables vary together. Unlike Variance, which is non-negative, Covariance can be negative or positive (or zero, of course).
5 lut 2021 · It is possible for two variables to be dependent but have zero covariance. For example, suppose we first sample a real number x from a uniform distribution over the interval [−1,1]. We next sample a random variable s. With probability 12, we choose the value of s to be 1. Otherwise, we choose the value of s to be −1.
\(E[X] = E[Y] = 0\) and \(\text{Var} [X] = \text{Var} [Y] = 1/3\) This means the \(\rho = 1\) line is \(u = t\) and the \(\rho = -1\) line is \(u = -t\). a. By symmetry, \(E[XY] = 0\) (in fact the pair is independent) and \(\rho = 0\). b. For every pair of possible values, the two signs must be the same, so \(E[XY] > 0\) which implies \(\rho > 0\).
