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Let $X \sim U[0, 1]$ and $Y \sim U[0, 2]$ and assume they are independent.

What is the CDF of $J = \min\{X, Y\}$ and $S = \max\{X, Y\}$?

My attempted solution for min was

$$CDF(J) = 1 - P[J ≥ j] = 1 - P[X \ge j]P[Y \ge j] = 1 - (1-j)(2-j).$$ Obviously this doesn't work since CDF is not non-decreasing from $0 \to 2$.

My attempted solution for max was

$$CDF(S) = P[S \le s] = P[X \le s] P[Y \le s] = F_x(s)F_y(s) = (s)(s/2)$$ (As $F_X(x) = x$ and $F_Y(y) = \frac{y}2$).

However, this solution does not work since the $CDF(S)$ evaluated as $S=2$, is $2$, which is greater than one.

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  • $\begingroup$ $F_x(s)=s$ when $0\le s \le 1$, but $F_x(s)=1$ when $s \ge 1$ and in particular when $s=2$ $\endgroup$ – Henry Nov 14 '17 at 9:09
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Here is the working to find J:\begin{align} P(J \le j) &= 1 -P(J > j)\\ &= 1-P(X > j)P(Y>j)\\ &= \begin{cases} 1-P(X>j)P(Y>j)&,0 \leq j \leq 1 \\1 &,1<j \leq2\end{cases} \end{align}

Since surely, if $j>1$, $P(X>j)=0$.

Try to use similar trick on $S$.

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