Question

5.16 Correlation coefficient Recall that Cov(X, Y) can be arbitrarily high or low depending on the magnitude of X and Y. In practice, it is common to use a normalized version on covariance called the correlation coefficient, $\rho(X, Y)$, where $\rho(X,Y) = \frac{Cov(X, Y)}{\sigma_X \sigma_Y}$, where $\sigma_X$ and $\sigma_Y$ represent the standard deviations of X and Y respectively. Prove that the magnitude of Cov(X, Y) is bounded, specifically $-1 \leq \rho(X,Y) \leq 1$. [Hint 1: It helps to use the Cauchy-Schwarz inequality from Exercise 5.15.] [Hint 2: Start by working with V = (X-E [X]) and W = (Y-E [Y]).]

          5.16 Correlation coefficient
Recall that Cov(X, Y) can be arbitrarily high or low depending on the
magnitude of X and Y. In practice, it is common to use a normalized
version on covariance called the correlation coefficient, $\rho(X, Y)$, where
$\rho(X,Y) = \frac{Cov(X, Y)}{\sigma_X \sigma_Y}$,
where $\sigma_X$ and $\sigma_Y$ represent the standard deviations of X and Y respectively.
Prove that the magnitude of Cov(X, Y) is bounded, specifically
$-1 \leq \rho(X,Y) \leq 1$.
[Hint 1: It helps to use the Cauchy-Schwarz inequality from Exercise 5.15.]
[Hint 2: Start by working with V = (X-E [X]) and W = (Y-E [Y]).]

5.16 Correlation coefficient
Recall that Cov(X, Y) can be arbitrarily high or low depending on the
magnitude of X and Y. In practice, it is common to use a normalized
version on covariance called the correlation coefficient, ρ(X, Y), where
ρ(X,Y) = (Cov(X, Y))/(),
where and represent the standard deviations of X and Y respectively.
Prove that the magnitude of Cov(X, Y) is bounded, specifically
-1 ≤ρ(X,Y) ≤ 1.
[Hint 1: It helps to use the Cauchy-Schwarz inequality from Exercise 5.15.]
[Hint 2: Start by working with V = (X-E [X]) and W = (Y-E [Y]).]

Added by Eduardo B.

Question

Please give Ace some feedback