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3

Your effort was correct, though it could be simpler. You just used the wrong formula for correlation. Here's my shot. Let $H_1,H_2,H_3$ be the indicators of a Head on the relevant toss.   There are independent events, with \begin{align}\mathsf E(H_n)=&~\tfrac 1 2\\[1ex]\mathsf {Var}(H_n)=&~\tfrac 1 4 \\[1ex] \mathsf ... 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$$Can you find a random variable X having mean 0, variance 1 ... 2 For a counterexample: let X,Y be independent random variables with X\sim \operatorname{Unif}\{-1,1\} and Y\sim \operatorname{Unif}\{-1,0,1\}. Then we have \mathbb{E}[X] = \mathbb{E}[Y] = 0,$$\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

The pdf of $X=(X_1,\,X_2)$ is $$f(x_1,x_2)=\begin{cases} \frac{1}{4} & \text{for } (x_1,x_2)\in Q=\{(-1,0), (1,0), (0,-1), (0,1)\}\\ 0 & \text{otherwise} \end{cases}$$ and the variable $X=(X_1,X_2)$ can be represented in tabular form $$\begin{pmatrix} (X_1,X_2)\\ f(x_1,x_2) \end{pmatrix}= \begin{pmatrix} (-1,0) & (1,0) & (0,-1) & ... 1 X_1 X_2=0 so E[X_1 X_2]=0 and thus, since E[X_1]=E[X_2]=0, the covariance and correlation must be zero. But, for example, X_1=1 \implies X_2=0 so they are not independent 1 The bound doesn't hold for the altered formula. If X is a standard normal variable, then \operatorname{Cov}(X,X)=\operatorname{Var}(X)=1 while d_X= E|X| =\sqrt{\frac2\pi} (derivation here). Therefore$$\frac{\operatorname{Cov}(X,X)}{ d_X d_X}=\frac\pi2, which is greater than $1$.

1

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.

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