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2

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 ...


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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 \, ...


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$Cor(X,Y)=\dfrac{E(XY)-E(X)E(Y)}{\sqrt{(E(X^2)-(E(X))^2)(E(Y^2)-(E(Y))^2)}}$ Now plug in the values.


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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 $1$.


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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.


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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) ~=~ ...



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