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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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