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I am trying to get the CDF of a random variable using TransformedDistribution.

Probability[-t v <= y - x <= 
    t v, {x \[Distributed] UniformDistribution[{0, l}], 
    y \[Distributed] UniformDistribution[{0, l}]}] // 
 Simplify[#, l > 0 && t > 0 && v > 0 && t v < l] &

The result is

$$\left\{ \begin{matrix} \frac{t v (2 l-t v)}{l^2} & l\neq 2 t v \\ \frac{2 t^2 v^2}{l^2}+\frac{1}{4} & l=2 t v \\ 0 & \text{True} \\ \end{matrix}\right. $$

Why does Simplify give an answer as a separate case when $l=2tv$? If I plug $l=2 tv$ into the first and second answers, they will come to the same result.

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  • 1
    $\begingroup$ Use FullSimplify instead of Simplify $\endgroup$
    – Bob Hanlon
    Sep 13, 2015 at 3:34
  • $\begingroup$ @BobHanlon FullSimplify gives an answer Rational[3,4], which is the same result if I put l=2tv into the first expression above. So the problem remains. $\endgroup$
    – zwcikyf
    Sep 13, 2015 at 3:50
  • $\begingroup$ This is normal, and I'd venture an optimization - using FullSimplify indeed gives two cases which are equivalent at l=2tv. One requires calculation to resolve, the other doesn't. I'd not want it done another way, particularly when the calc. involved might be time-consuming... $\endgroup$
    – ciao
    Sep 13, 2015 at 5:13

2 Answers 2

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You can simplify the expression using following rewrite rules:

Probability[-t v <= y - x <= 
   t v, {x \[Distributed] UniformDistribution[{0, l}], 
   y \[Distributed] UniformDistribution[{0, l}]}] // 
 Assuming[l > 0 && t > 0 && v > 0 && t v < l, 
   Simplify@# //. {HoldPattern@
        Piecewise[{beg___, {a_, acond_}, mid___, {b_, bcond_}, 
          rest___}, default___] /; 
       FullSimplify[a == b, And @@ Not /@ {beg}[[All, 2]] && acond] :>
       FullSimplify@
       Piecewise[{beg, {b, 
          acond || (! acond && And @@ Not /@ {mid}[[All, 2]] && 
             bcond)}, mid, rest}, default],
     HoldPattern@
        Piecewise[{beg___, {a_, acond_}, mid___, {b_, bcond_}, 
          rest___}, default___] /; 
       FullSimplify[a == b, 
        And @@ Not /@ {beg}[[All, 2]] && ! acond && 
         And @@ Not /@ {mid}[[All, 2]] && bcond] :> 
      FullSimplify@
       Piecewise[{beg, {a, 
          acond || (! acond && And @@ Not /@ {mid}[[All, 2]] && 
             bcond)}, mid, rest}, default]}] &

This rule rewrites all non-default Piecewise regions that FullSimplify can pairwise reason to have the same value under their domain to combined regions, and simplifies the resulting expression. (Pedantic reconstruction of condition chain is a bit messy, I admit.) I can't guarantee this works gracefully in general, but on above case the result is clean:

$$\frac{t v (2 l-t v)}{l^2}$$

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Using a in place of l for readability

pr[a_, t_, v_] = Assuming[
  {a > 0, t > 0, v > 0, t v < a},
  Probability[-t v <= y - x <= t v,
    {x \[Distributed] UniformDistribution[{0, a}],
     y \[Distributed] UniformDistribution[{0, a}]}] //
   FullSimplify]

(* Piecewise[{{(t*v*(2*a - t*v))/a^2, a != 2*t*v}}, 
   Rational[3, 4]] *)

You can overcome the limitation by either extraction

pr[a_, t_, v_] = Assuming[
   {a > 0, t > 0, v > 0, t v < a},
   Probability[-t v <= y - x <= t v,
     {x \[Distributed] UniformDistribution[{0, a}],
      y \[Distributed] UniformDistribution[{0, a}]}] //
    FullSimplify][[1, 
   1, 1]]

(* (t*v*(2*a - t*v))/a^2 *)

Or by including the additional assumption

pr[a_, t_, v_] = Assuming[
  {a > 0, t > 0, v > 0, t v < a, a != 2 t v},
  Probability[-t v <= y - x <= t v,
    {x \[Distributed] UniformDistribution[{0, a}],
     y \[Distributed] UniformDistribution[{0, a}]}] //
   FullSimplify]

(* (t*v*(2*a - t*v))/a^2 *)
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