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Why getting complex number solution to expectation?

Paul Studtmann
Posted 1 year ago

Can someone explain why I am getting a complex number for this computation of an expected value?
POSTED BY: Paul Studtmann
16 Replies
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POSTED BY: Michael Rogers
Jim Baldwin
Jim Baldwin
Posted 1 year ago

My speculation about the appearance of complex numbers is that Mathematica uses the pdf of 1/(1/s+1/t) which contains logs:

dist = TransformedDistribution[1/(1/s + 1/t), {t \[Distributed] UniformDistribution[{-1, 1}], 
    s \[Distributed] UniformDistribution[{-1, 1}]}];
pdf = PDF[dist, st] // FullSimplify

pdf of 1/(1/s+1/t)

Because that random variable can take negative values and there are logs in the pdf, the internal code finds complex numbers even though those complex numbers will eventually "cancel out" outside of that internal code.

So a brute force approach to get the correct result is to do the numerical integeration:

numerator = NIntegrate[r (1/2) (1/2) pdf, {r, -1, 1}, {st, -1, r}, {p, -1, st}]
(* 0.0901967 *)
denominator = NIntegrate[(1/2) (1/2) pdf, {r, -1, 1}, {st, -1, r}, {p, -1, st}]
(* 0.180393 *)
conditionalMean = numerator/denominator
(* 0.5 *)

If one has a few minutes of patience, then Integrate also works:

numerator = Integrate[r (1/2) (1/2) pdf, {r, -1, 1}, {st, -1, r}, {p, -1, st}] // FullSimplify
(* 1/64 (3 + Log[16]) *)
denominator = Integrate[(1/2) (1/2) pdf, {r, -1, 1}, {st, -1, r}, {p, -1, st}] // FullSimplify
(* 1/32 (3 + Log[16]) *)
conditionalMean = numerator/denominator
(* 1/2 *)

Note: If Log[(-1 + st)/st] isn't split into two terms Log[-1 + st] - Log[st], then the imaginary numbers are not an issue. This condition happens when st < -1/2.
POSTED BY: Jim Baldwin

I am using 14.0 and am still getting weird results. I tried with Expectation rather than NExpectation. Using Expectation eliminates the 'invalid input' but still gives a complex number. When I compute: 

Expectation[
 R \[Conditioned] (P < 1/(1 + 1/T) && 
    1/(1 + 1/T) < R), {T \[Distributed] UniformDistribution[{-1, 1}], 
  R \[Distributed] UniformDistribution[{-1, 1}], 
  P \[Distributed] UniformDistribution[{-1, 1}], 
  S \[Distributed] UniformDistribution[{-1, 1}]}]

I get a real number. But when I compute:

Expectation[
 R \[Conditioned] 
  R > 1/(1/S + 1/T) > P, {T \[Distributed] 
   UniformDistribution[{-1, 1}], 
  R \[Distributed] UniformDistribution[{-1, 1}], 
  P \[Distributed] UniformDistribution[{-1, 1}], 
  S \[Distributed] UniformDistribution[{-1, 1}]}]

I get a complex number. The only difference is that the first condition has 1/(1+1/T) where the second condition has 1/(1/S+1/T). So something seems amiss.
POSTED BY: Paul Studtmann
Denis Ivanov
Denis Ivanov
Posted 1 year ago

All your examples works very well with the right order on Mathematica 13.2:

enter image description here
POSTED BY: Denis Ivanov

I don't know if the following is any help:

This syntax seems valid (removing /T from first argument):

NExpectation[
 R \[Conditioned] R > 1/(1 + 1/S) > P,
 {T \[Distributed] UniformDistribution[{-1, 1}],
   R \[Distributed] UniformDistribution[{-1, 1}], 
  P \[Distributed] UniformDistribution[{-1, 1}], 
  S \[Distributed] UniformDistribution[{-1, 1}]}]

But the original throws an error. Maybe it's a bug?

Similarly this is accepted:

NExpectation[
 R \[Conditioned] R > T > S > P,
 {T \[Distributed] UniformDistribution[{-1, 1}], 
  R \[Distributed] UniformDistribution[{-1, 1}], 
  P \[Distributed] UniformDistribution[{-1, 1}], 
  S \[Distributed] UniformDistribution[{-1, 1}]}]

But this is not:

NExpectation[
 R \[Conditioned] R > S > P,
 {T \[Distributed] UniformDistribution[{-1, 1}], 
  R \[Distributed] UniformDistribution[{-1, 1}], 
  P \[Distributed] UniformDistribution[{-1, 1}], 
  S \[Distributed] UniformDistribution[{-1, 1}]}]

Perhaps there is a limitation on the number of variables that can be used in the first argument. Maybe the limitation is due to a bug?

I suggest reporting it to Wolfram. They may be able to help. If it is a bug, it needs to be fixed.
POSTED BY: Michael Rogers
Denis Ivanov
Denis Ivanov
Posted 1 year ago

Screenshot: enter image description here
POSTED BY: Denis Ivanov
Denis Ivanov
Denis Ivanov
Posted 1 year ago

What about 13.2?
Screenshot:

enter image description here

IMHO version 13.2 is the most stable, tested and balanced of the most recent.
POSTED BY: Denis Ivanov
Denis Ivanov
Denis Ivanov
Posted 1 year ago

Not only parentheses, but more correct order of inequality
POSTED BY: Denis Ivanov

I tried your modification in versions 12.0, 14.0, 14.1 and they all give the same complex number. Any idea why (a<b<c) and c>b>a should give different answers?
POSTED BY: Gianluca Gorni

Modifications are

R \[Conditioned] (P < 1/(1/S + 1/T) < R)

instead of

R\[Conditioned]R>1/(1/S+1/T)>P

I double checked this on version 13.2 and it works
POSTED BY: Gianluca Gorni

@Denis's modification was to put parentheses around the condition. (I think.)

However, I, too, still get the "invalid input" error, which seems more significant than the complex number. Basic GIGO: You can't expect the output to make sense if the input does not.
POSTED BY: Michael Rogers
Denis Ivanov
Denis Ivanov
Posted 1 year ago

Modifications are

R \[Conditioned] (P < 1/(1/S + 1/T) < R)

instead of

R\[Conditioned]R>1/(1/S+1/T)>P

I double checked this on version 13.2 and it works
POSTED BY: Denis Ivanov

Same here. I am not sure what the modifications are supposed to be. When I cut and paste the code, I get the same complex number.
POSTED BY: Paul Studtmann

What are the modifications? I get the same complex number also in the modified version. I tried with version 12, and I get the complex number but without error messages.
POSTED BY: Gianluca Gorni

Thank you!
POSTED BY: Paul Studtmann
Denis Ivanov
Denis Ivanov
Posted 1 year ago

After some modifications:

NExpectation[
 R \[Conditioned] (P < 1/(1/S + 1/T) < R), {T \[Distributed] 
   UniformDistribution[{-1, 1}], 
  R \[Distributed] UniformDistribution[{-1, 1}], 
  P \[Distributed] UniformDistribution[{-1, 1}], 
  S \[Distributed] UniformDistribution[{-1, 1}]}]

Out: 0.5
POSTED BY: Denis Ivanov
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