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I would like to numerically calculate the probability of an event involving a random vector. Specifically:

NProbability[(X > (Y/Z)), {X, Y, Z} \[Distributed] MultinormalDistribution[{0, 0, 0}, {{1, .5, .5}, {.5, 1, .5}, {.5, .5, 1}}]]

When entering the above code, the output gives (after a convergence warning)

0.408295

If instead I enter the code:

NProbability[(X*Z) - Y > 0, {X, Y, Z} \[Distributed] MultinormalDistribution[{0, 0, 0}, {{1, .5, .5}, {.5, 1, .5}, {.5, .5, 1}}]]

I get (also after a convergence warning)

0.585629

Why the difference?

Note 1. The convergence warning message is discussed in another post here. As the accepted answer in the linked post shows, simulations suggest that speed of convergence should not be responsible for this.

Note 2. I am running Mathematica 10.2, which seems to struggle to produce symbolic shorthands for TransformedDistribution (see comments here).

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    $\begingroup$ That's because the statements X > (Y/Z) and (X*Z) - Y > 0 are not equivalent (negative numbers are involved). Try it with simulations and you'll get pretty much the same differences. $\endgroup$ – JimB Feb 8 '17 at 16:55
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X > (Y/Z) is not the same as X*Z - Y > 0 since Z is not always positive.

NProbability[(X > (Y/Z)), {X, Y, Z} \[Distributed] 
   MultinormalDistribution[{0, 0, 
     0}, {{1, .5, .5}, {.5, 1, .5}, {.5, .5, 1}}]] // Quiet

(*  0.408295  *)

NProbability[(Z > 0 && X*Z - Y > 0) || (Z < 0 && X*Z - Y < 0), {X, Y, 
    Z} \[Distributed] 
   MultinormalDistribution[{0, 0, 
     0}, {{1, .5, .5}, {.5, 1, .5}, {.5, .5, 1}}]] // Quiet

(*  0.408295  *)
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  • $\begingroup$ @EOO - it doesn't assume that Z > 0, that assumption would be equivalent to NProbability[(Z > 0 && X > (Y/Z)), {X, Y, Z} \[Distributed] MultinormalDistribution[{0, 0, 0}, {{1, .5, .5}, {.5, 1, .5}, {.5, .5, 1}}]] // Quiet which evaluates to 0.254846 $\endgroup$ – Bob Hanlon Feb 8 '17 at 17:54

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