# Tag Info

For a counterexample: let $X,Y$ be independent random variables with $X\sim \operatorname{Unif}\{-1,1\}$ and $Y\sim \operatorname{Unif}\{-1,0,1\}$. Then we have $\mathbb{E}[X] = \mathbb{E}[Y] = 0$, $$\mathbb{E}[\lvert X\rvert] = \mathbb{E}[X^2] = \operatorname{Var} X = 1,$$ and $$\mathbb{E}[\lvert Y\rvert] = \mathbb{E}[Y^2] = \operatorname{Var} Y = ... 2 Let it be that X,Y both have mean 0 and both have variance 1. Then your claim takes the form:$$\mathbb E|XY|=1$$If X,Y are moreover independent then it takes the form:$$\mathbb E|X|\mathbb E|Y|=1$$If moreover X,Y have equal distribution then it takes the form:$$\mathbb E|X|=1$$Can you find a random variable X having mean 0, variance 1 ... 2 Spearman's rank correlation is calculated by ranking bivariate normal variables {X_i},\;{Y_i} as variables {x_i},\;{y_i}. Pearson's correlation between the ranked variables is then given by:$$\rho = \frac{\sum_i(x_i-\bar{x})(y_i-\bar{y})}{\sqrt{\sum_i (x_i-\bar{x})^2 \sum_i(y_i-\bar{y})^2}}Since there are no ties, the x's and y's both consist of ... 1 I think this is just Cauchy's inequality: \begin{align}\big|E[(X-E[Y])(Y-E[Y])]\big| &= \left| \int_\Omega (X(\omega) - e_x) \, (Y(\omega) - e_y) \, \mathrm d \omega \right| \\&\le \left| \int_\Omega (X(\omega) - e_x)^2 \, \mathrm d \omega \right|^{1/2} \, \left| \int_\Omega (Y(\omega) - e_y)^2 \, \mathrm d \omega \right|^{1/2} = \sigma_X \, ... 1 Cor(X,Y)=\dfrac{E(XY)-E(X)E(Y)}{\sqrt{(E(X^2)-(E(X))^2)(E(Y^2)-(E(Y))^2)}} Now plug in the values. 1 The bound doesn't hold for the altered formula. If X is a standard normal variable, then \operatorname{Cov}(X,X)=\operatorname{Var}(X)=1 while d_X= E|X| =\sqrt{\frac2\pi} (derivation here). Therefore\frac{\operatorname{Cov}(X,X)}{ d_X d_X}=\frac\pi2,$$which is greater than 1. 1 The last equality in the first line of your equation is wrong. Note that you have$$u(t)u(t+\tau)=\begin{cases}u(t),&\tau>0\\u(t+\tau),&\tau<0\end{cases}$$Now compute the integral for both cases (\tau>0 and \tau<0), and you'll get the desired result. 1 Firstly. If Y=0.5+0.6X and \mathsf {Var}(X)=\sigma^2 , then \mathsf{Var}(Y)=0.36\sigma^2 Because \mathsf {Var}(a+bX) ~=~ b^2~\mathsf{Var}(X) when a,b are constants. Similarly: \mathsf {Cov}(a+bX, c+dX) ~=~ bd~\mathsf{Var}(X) Revisit all your calculations. Secondly The correlation coefficient is defined as:$$\mathsf {Corr}(U,V) ~=~ ...