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Yury
Yury
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Let $\varepsilon \to 0$ and $M= \sqrt{(1-\varepsilon)/\varepsilon}$. Define $X$ and $Y$ as follows:

  • w.p. $\frac{1-\varepsilon}2$: $X=Y=1$
  • w.p. $\frac{1-\varepsilon}2$: $X=Y=-1$
  • w.p. $\frac{\varepsilon}2$: $X=M$ and $Y=-M$
  • w.p. $\frac{\varepsilon}2$: $X=-M$ and $Y=M$

Then $\mathbf{E}[XY] = (1-\varepsilon) - \varepsilon M^2 = 0$. Also, $\mathbf{E}|X-Y| = \varepsilon \cdot (2M) = 2 \sqrt{\varepsilon(1-\varepsilon)}$. Finally, $\mathbf{E}[X] = \mathbf{E}[Y] = 0$.

However, $\mathbf{E}|\tilde X- \tilde Y| \geq \mathbf{Pr}(\tilde X = - \tilde Y) \cdot 2\geq \frac{(1-\varepsilon)^2}{2} \cdot 2 = (1-\varepsilon)^2$.

On a separate note, observe that $\mathbf{E}|\tilde X - \tilde Y| = \Theta(\mathbf{E}|\tilde X| + \mathbf{E}|\tilde Y|)$.

Let $\varepsilon \to 0$ and $M= \sqrt{(1-\varepsilon)/\varepsilon}$. Define $X$ and $Y$ as follows:

  • w.p. $\frac{1-\varepsilon}2$: $X=Y=1$
  • w.p. $\frac{1-\varepsilon}2$: $X=Y=-1$
  • w.p. $\frac{\varepsilon}2$: $X=M$ and $Y=-M$
  • w.p. $\frac{\varepsilon}2$: $X=-M$ and $Y=M$

Then $\mathbf{E}[XY] = (1-\varepsilon) - \varepsilon M^2 = 0$. Also, $\mathbf{E}|X-Y| = \varepsilon \cdot (2M) = 2 \sqrt{\varepsilon(1-\varepsilon)}$. Finally, $\mathbf{E}[X] = \mathbf{E}[Y] = 0$.

However, $\mathbf{E}|\tilde X- \tilde Y| \geq \mathbf{Pr}(\tilde X = - \tilde Y) \cdot 2\geq \frac{(1-\varepsilon)^2}{2} \cdot 2 = (1-\varepsilon)^2$.

Let $\varepsilon \to 0$ and $M= \sqrt{(1-\varepsilon)/\varepsilon}$. Define $X$ and $Y$ as follows:

  • w.p. $\frac{1-\varepsilon}2$: $X=Y=1$
  • w.p. $\frac{1-\varepsilon}2$: $X=Y=-1$
  • w.p. $\frac{\varepsilon}2$: $X=M$ and $Y=-M$
  • w.p. $\frac{\varepsilon}2$: $X=-M$ and $Y=M$

Then $\mathbf{E}[XY] = (1-\varepsilon) - \varepsilon M^2 = 0$. Also, $\mathbf{E}|X-Y| = \varepsilon \cdot (2M) = 2 \sqrt{\varepsilon(1-\varepsilon)}$. Finally, $\mathbf{E}[X] = \mathbf{E}[Y] = 0$.

However, $\mathbf{E}|\tilde X- \tilde Y| \geq \mathbf{Pr}(\tilde X = - \tilde Y) \cdot 2\geq \frac{(1-\varepsilon)^2}{2} \cdot 2 = (1-\varepsilon)^2$.

On a separate note, observe that $\mathbf{E}|\tilde X - \tilde Y| = \Theta(\mathbf{E}|\tilde X| + \mathbf{E}|\tilde Y|)$.

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Yury
Yury
  • 726
  • 3
  • 9

Let $\varepsilon \to 0$ and $M= \sqrt{(1-\varepsilon)/\varepsilon}$. Define $X$ and $Y$ as follows:

  • w.p. $\frac{1-\varepsilon}2$: $X=Y=1$
  • w.p. $\frac{1-\varepsilon}2$: $X=Y=-1$
  • w.p. $\frac{\varepsilon}2$: $X=M$ and $Y=-M$
  • w.p. $\frac{\varepsilon}2$: $X=-M$ and $Y=M$

Then $\mathbf{E}[XY] = (1-\varepsilon) - \varepsilon M^2 = 0$. Also, $\mathbf{E}|X-Y| = \varepsilon \cdot (2M) = 2 \sqrt{\varepsilon(1-\varepsilon)}$. Finally, $\mathbf{E}[X] = \mathbf{E}[Y] = 0$.

However, $\mathbf{E}|\tilde X- \tilde Y| \geq \mathbf{Pr}(\tilde X = - \tilde Y) \cdot 2\geq \frac{(1-\varepsilon)^2}{2} \cdot 2 = (1-\varepsilon)^2$.