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consider this simple Input:

Probability[g <= m && m <= -g + 1,
  {m, g} \[Distributed] UniformDistribution[{{0, 1}, {0, gg}}]]

which generates a piecewise continuous function:

$$\begin{cases} 1-gg & gg\le\frac{1}{2}\\ \frac{1}{4\,gg} & \text{True} \end{cases}$$

My question is: why does mathematica outputs True rather that gg>1/2?

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closed as off-topic by Daniel Lichtblau, MarcoB, march, RunnyKine, Louis Apr 6 at 17:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Daniel Lichtblau, MarcoB, march, RunnyKine, Louis
If this question can be reworded to fit the rules in the help center, please edit the question.

Because it is the default case, and while g <= 1/2 partitions the space into only two sections, consider the piecewise function found in this answer. Overall, it is just simpler to have a default that is always run if no others match. – rcollyer Jul 19 '13 at 15:05
Note further that the actual output has the form Piecewise[{{1 - gg, gg <= 1/2}}, 1/(4*gg)] (execute Probability[g <= m && m <= -g + 1, {m, g} \[Distributed] UniformDistribution[{{0, 1}, {0, gg}}]] // InputForm to see). – Michael E2 Jul 19 '13 at 15:28
I understand now, thanks. – Emilio Calvano Jul 19 '13 at 18:26
up vote 5 down vote accepted

If you don't like the output format for Piecewise because it contains the default choice True, you may find it clearer to bring the output into a form that uses ConditionalExpression as follows:

p = Probability[
   g <= m && m <= -g + 1, {m, g} \[Distributed] 
    UniformDistribution[{{0, 1}, {0, gg}}]];

probability /. Solve[probability == p, probability, Reals]

$$\left\{ \text{ConditionalExpression}\left[1-\text{gg},\text{gg}<\frac{1}{2}\right],\text{Conditional Expression}\left[\frac{1}{4\text{gg}},\text{gg}>\frac{1}{2}\right]\right\} $$

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