1
$\begingroup$

I'm trying to find Probability involving 3 random variates from the same distribution, but Mathematica does not seem to solve it.

Probability[ z1 > 1.325*((z2)^2 + (z3)^2)^(1/2), {z1 \[Distributed] 
NormalDistribution[0, 1],   z2 \[Distributed] NormalDistribution[0, 1], 
z3 \[Distributed] NormalDistribution[0, 1]}]

What am I missing?

I also tried:

Probability[ z1 > 1.325*((z2)^2 + (z3)^2)^(1/2), {z1, z2, z3}
\[Distributed] NormalDistribution[0, 1]]
$\endgroup$
  • 3
    $\begingroup$ NProbability[ z1 > 1325/100*Sqrt@((z2)^2 + (z3)^2), {z1 \[Distributed] NormalDistribution[0, 1], z2 \[Distributed] NormalDistribution[0, 1], z3 \[Distributed] NormalDistribution[0, 1]}] $\endgroup$ – Dr. belisarius Nov 30 '14 at 20:08
3
$\begingroup$

As this is simple enough, you could use definitions :

dist = MultinormalDistribution[{0, 0, 0}, IdentityMatrix[3]] ;
prob[a_] = Integrate[PDF[dist, {z1, z2, z3}], 
                     {z2, -Infinity, Infinity}, 
                     {z3, -Infinity, Infinity}, 
                     {z1, a Sqrt[z2^2 + z3^2], Infinity}]

(* ConditionalExpression[1/4 (2 - 2/Sqrt[1 + 1/a^2]), Re[a] > 0] *)   

Then

prob[1.325]
(* 0.100905 *)
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.