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Let $E$ be uniformly distributed between $0$ and $10$. The cumulative distribution function of $\max(E-6,0)$ is: $$F(x) = \left\{ \begin{array}{ll} 0 & \mbox{if } x < 0 \\ \frac{6+x}{10} & \mbox{if } 0 \leq x \leq 4 \\ 1 & \mbox{if } x > 4 \end{array} \right. $$

How does one arrive at this result?

Max Muller
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1 Answers1

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Define $X = \max(E - 6, 0)$. Because $X \geqslant 0$, for any $x < 0$, there is $P(X \leqslant x) = 0$.

For $0 \leqslant x \leqslant 4$,\begin{align*} P(X \leqslant x) &= P(X \leqslant x \mid E \leqslant 6) P(E \leqslant 6) + P(X \leqslant x \mid E > 6) P(E > 6)\\ &= P(0 \leqslant x \mid E \leqslant 6) P(E \leqslant 6) + P(E - 6 \leqslant x \mid E > 6) P(E > 6)\\ &= P(E \leqslant 6) + P(E - 6 \leqslant x, E > 6)\\ &= \frac{6}{10} + \frac{x}{10} = \frac{x + 6}{10}. \end{align*}

For $x > 4$, $P(X \leqslant x) \geqslant P(X \leqslant 4) = 1 \Rightarrow P(X \leqslant x) = 1$.

Ѕᴀᴀᴅ
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