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Mathematics Wolfram Language Statistics and Probability

Probability that two correlated Normal random variables have the same sign?

Greg Sternberg

Posted 1 year ago

When computing the probability for two correlated (zero mean) Normal random variables, I'm getting conflicting results. One of them seems nonsensical and potentially indicative of a bug in Mathematica. In the attached notebook, I define a MultiNormalDistribution and ask Mathematica to compute the probability that xy > 0. I believe this should lead to the same result as the probability that (x > 0 && y > 0) \|\| (x < 0 && y < 0). The latter result is quite sensible whereas the probability that the product is positive is very odd. Does this make sense? Attachments:* SignAgreementBug.nb

POSTED BY: Greg Sternberg

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Jim Baldwin

Posted 1 year ago

It's more than odd. The two approaches you give should give the same result but only the second one below does give the right answer. (You should inform Wolfram, Inc.) FullSimplify[Probability[x y >= 0, {x, y} \[Distributed] MultinormalDistribution[{{1, \[Rho]}, {\[Rho], 1}}]], Assumptions -> -1 <= \[Rho] <= 1] (* 1/2 ) 2 Probability[x >= 0 && y >= 0, {x, y} \[Distributed] MultinormalDistribution[{{1, \[Rho]}, {\[Rho], 1}}]] // FullSimplify ( 1/2 + ArcSin[\[Rho]]/\[Pi] ) An alternative way that does work with `xy` is the following: distxy = TransformedDistribution[x y, {x, y} \[Distributed] MultinormalDistribution[{{1, \[Rho]}, {\[Rho], 1}}]]] 1 - CDF[distxy, 0] // FullSimplify (* 1/2 + ArcSin[\[Rho]]/\[Pi] *)

It's more than odd. The two approaches you give should give the same result but only the second one below does give the right answer. (You should inform Wolfram, Inc.)

FullSimplify[Probability[x y >= 0, {x, y} \[Distributed] MultinormalDistribution[{{1, \[Rho]}, {\[Rho], 1}}]],
   Assumptions -> -1 <= \[Rho] <= 1]
(* 1/2 *)

2 Probability[x >= 0 && y >= 0, {x, y} \[Distributed] MultinormalDistribution[{{1, \[Rho]}, {\[Rho], 1}}]] // FullSimplify
(* 1/2 + ArcSin[\[Rho]]/\[Pi]  *)

An alternative way that does work with x*y is the following:

distxy =  TransformedDistribution[x y, {x, y} \[Distributed]  MultinormalDistribution[{{1, \[Rho]}, {\[Rho], 1}}]]]
1 - CDF[distxy, 0] // FullSimplify
(* 1/2 + ArcSin[\[Rho]]/\[Pi] *)

POSTED BY: Jim Baldwin

Greg Sternberg

Posted 1 year ago

Thanks Jim! Will do.

POSTED BY: Greg Sternberg

Gosia Konwerska

Gosia Konwerska, Wolfram Research, Inc.

Posted 1 year ago

Thank you for catching the discrepancy - it looks like it is an issue with `Probability`, I will investigate.

POSTED BY: Gosia Konwerska

Greg Sternberg

Posted 6 months ago

Hi Gosia Konwerska, I just purchased a new license for Mathematica 14.0.0 and observe the same issue is still present. Has there been any progress on this? Thanks, Greg Sternberg

POSTED BY: Greg Sternberg

Gosia Konwerska

Gosia Konwerska, Wolfram Research, Inc.

Posted 6 months ago

Unfortunately has not been resolved yet...

POSTED BY: Gosia Konwerska

Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 6 months ago

It seems to me that the problem lies with `Integrate` with parameters, rather than with `Probability`. In the following examples the result depends on how we specify the integration region and on the assumptions on the parameter: dist = MultinormalDistribution[{{1, \[Rho]}, {\[Rho], 1}}]; pdf = PDF[dist, {x, y}]; reg = ImplicitRegion[a >= 0 && b >= 0, {a, b}]; int1 = Integrate[pdf, {x, y} \[Element] reg, Assumptions -> -1 < \[Rho] < 1] int2 = Integrate[pdf, {x, y} \[Element] reg, Assumptions -> -1 < \[Rho] < 0] int3 = Integrate[pdf, {x, 0, Infinity}, {y, 0, Infinity}, Assumptions -> -1 < \[Rho] < 1] int4 = Integrate[pdf, {x, 0, Infinity}, {y, 0, Infinity}, Assumptions -> -1 < \[Rho] < 0] {int1, int2, int3, int4} /. \[Rho] -> -1/2

It seems to me that the problem lies with Integrate with parameters, rather than with Probability. In the following examples the result depends on how we specify the integration region and on the assumptions on the parameter:

dist = MultinormalDistribution[{{1, \[Rho]}, {\[Rho], 1}}];
pdf = PDF[dist, {x, y}];
reg = ImplicitRegion[a >= 0 && b >= 0, {a, b}];
int1 = Integrate[pdf,
  {x, y} \[Element] reg,
  Assumptions -> -1 < \[Rho] < 1]
int2 = Integrate[pdf,
  {x, y} \[Element] reg,
  Assumptions -> -1 < \[Rho] < 0]
int3 = Integrate[pdf,
  {x, 0, Infinity}, {y, 0, Infinity},
  Assumptions -> -1 < \[Rho] < 1]
int4 = Integrate[pdf,
  {x, 0, Infinity}, {y, 0, Infinity},
  Assumptions -> -1 < \[Rho] < 0]
{int1, int2, int3, int4} /. \[Rho] -> -1/2

POSTED BY: Gianluca Gorni

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