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Two continuous, independent random variables X and Y each take a random value uniformly distributed between 0 and 8. What is the probability that the difference between them is greater than 4?

Intuitively, I think it's 1/2, because it seems like for every case where the difference is greater than 4, there is a mirrored case where the difference is lesser than 4.

Any ideas on how to prove or am I just wrong?

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  • $\begingroup$ There are various different probability distributions on the interval from $0$ to $8$. $\endgroup$ – Michael Hardy Apr 23 '14 at 18:03
  • $\begingroup$ There are various different dependence relationships beteween probability distributions on the interval from 0 to 8. $\endgroup$ – Did Apr 23 '14 at 18:04
  • $\begingroup$ You probably want to write are independent. $\endgroup$ – Did Apr 23 '14 at 18:13
  • $\begingroup$ To solve this, draw an $8\times8$ square: $0 \leq x \leq 8$ and $0 \leq y \leq 8$. Draw lines for $|x-y| = 4$. Measure areas. $\endgroup$ – user3294068 Apr 23 '14 at 20:46
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For this, we know that $x\in\left[0,8\right]$ and $y\in\left[0,8\right]$, so we just need to have $\left|x-y\right|>4$. This only occurs if $x\in\left(4,8\right]$ and $y\in\left[0,4\right)$ or visa versa with $y\in\left(4,8\right]$ and $x\in\left[0,4\right)$.

Therefore. \begin{align*} \mathbb{P}\left(\left|X-Y\right|>4\right) &= \mathbb{P}\left(\left(X>4\cap Y<4\right)\cup\left(Y>4\cup X<4\right)\right) \\ &=\mathbb{P}\left(X>4\cap Y<4\right)+\mathbb{P}\left(Y>4\cup X<4\right)\quad\text{by additivity} \\ &=\mathbb{P}\left(X>4\right)\mathbb{P}\left(Y<4\right)+\mathbb{P}\left(Y>4\right)\mathbb{P}\left(X<4\right)\quad\text{by independence} \\ &=\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{2} \\ &=\frac{1}{2} \end{align*}

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