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3

a) $Y^{2}=X^{2}$ so that $\mathbb{E}Y^{2}=\mathbb{E}X^{2}=1$. $X$ is symmetric and consequently $Y$ is symmetric. That implies $\mathbb{E}Y=0$ so that $\text{Var}Y=\mathbb EY^2=1$. Then: $$\rho\left(X,Y\right):=\mathbb{E}\left(\frac{X-\mu_{X}}{\sigma_{X}}\right)\left(\frac{Y-\mu_{Y}}{\sigma_{Y}}\right)=\mathbb{E}XY$$ Now work this out. b) If $a$ ...

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Let $X$ be $-1$, $0$, or $1$, each with probability $1/3$, and let $Y=X^2$. Then $X$ and $Y$ are uncorrelated, but they are no independed. Independent random variables for which a correlation exists are always uncorrelated. Their correlation exists only if both of their variances are finite. I don't know what, if anything, it would mean to say that the ...

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When you are dealing with two discrete random variables you can calculate the conditional probabilities and simulate directly from there. The following doesn't require that the two variables $X$, and $Y$ have the same distribution, just that they both have $n$ outcomes (for simplicity of exposition). Whatever your correlation is, use that to determine the ...

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Simplest example I can think of: Let $X_1, X_2, X_3$ be iid Bernoulli with success probability $\theta = 1/2.$ Then $Y_1 = X_1 + X_2$ and $Y_2 = X_2 + X_3$ are both $Binom(2, 1/2),$ but they are correlated. Because you mention 'generating' here is a simple simulation in R: x1 = rbinom(10^6, 1, 1/2) x2 = rbinom(10^6, 1, 1/2) x3 = rbinom(10^6, 1, 1/2) ...

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The correlation between two random variables $X, Y$ is given by: $$\rho = \frac{Cov(X, Y)}{\sqrt{Var(X)}\sqrt{Var(Y)}}$$ In your case, \begin{eqnarray*} \rho &=& \frac{Cov(f_i - f_j, y)}{\sqrt{Var(f_i - f_j)}\sqrt{Var(y)}} \\ &=& \frac{Cov(f_i,y) - Cov(f_j, y)}{\sqrt{Var(f_i - f_j)}\sqrt{Var(y)}} \\ &=& \frac{Cov(f_i,y) ...

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