So, I'll explain the problem first. I have two populations that have values for two different channels (c1,c2). I've modeled these using MultinormalDistribution[]. I would like to add some noise to these distributions, which also exists in channels (c1,c2). I've tried doing this using TransformedDistribution[]:

hiDist = MultinormalDistribution[\[Mu]HiErr, \[CapitalSigma]];
loDist = MultinormalDistribution[\[Mu]LoErr, \[CapitalSigma]];
noiseDist = MultinormalDistribution[\[Mu]Noise, \[CapitalSigma]];

hiNDist = 
   h + n, {h \[Distributed] hiDist, n \[Distributed] noiseDist}];
loNDist = 
   l + n, {l \[Distributed] loDist, n \[Distributed] noiseDist}];

This works with regular Normal distributions, but fails when I use it with the Multinormal primitives. Any elegant way to do this?

EDIT 1 Changed a typo.

"Failure" results when I try to call the thing. In place, the TransformedDistribution[] call will just output something like:

    TransformedDistribution[\[FormalX]1 + \[FormalX]2, {\[FormalX]1 \
     20000}, {{1000000, 0}, {0, 
      1000000}}], \[FormalX]2 \[Distributed] 
     2000}, {{1000000, 0}, {0, 1000000}}]}]

However, when I try to use the distribution, e.g.


Mathematica will throw the error:

RandomReal::udist: "The specification TransformedDistribution[[FormalX]> 1+[FormalX]2, {[FormalX]1[Distributed]MultinormalDistribution[{5000,20000},{{1000000,0},{0,1000000}}],[FormalX]2[Distributed]MultinormalDistribution[{2000,2000},{{1000000,0},{0,1000000}}]}] is not a random distribution recognized by the system"

While the code

n1 = NormalDistribution[5000, \[Sigma]];
n2  = NormalDistribution[10000, \[Sigma]];
n3 = TransformedDistribution[
  x + y, {x \[Distributed] n1, y \[Distributed] n2}]

Gives the expected

NormalDistribution[15000, 1000 Sqrt[2]]
  • $\begingroup$ Could you detail "fails" as used here? Also providing the code of what you actually tried would be helpful. BTW I assume hiDist should be loDist in the second line, otherwise both distributions are identical. $\endgroup$ Mar 14, 2014 at 16:34
  • $\begingroup$ Updated my question. And yep, that was a typo. $\endgroup$
    – Ted C
    Mar 14, 2014 at 21:30

1 Answer 1


Try, e.g. (I just plugged in values from mu and sigma from your error text:

dist = MultinormalDistribution[{5000, 20000}, {{1000000, 0}, {0, 1000000}}]
noise = MultinormalDistribution[{2000, 2000}, {{1000000, 0}, {0, 1000000}}]

transformed = 
 TransformedDistribution[{a + c, b + d}, {{a, b} \[Distributed] dist,
                                          {c, d} \[Distributed] noise}]


  • $\begingroup$ I see! So you have to explicitly describe each component of the output? $\endgroup$
    – Ted C
    Apr 7, 2014 at 18:05
  • $\begingroup$ @ted: It appears that way. Some of the more esoteric corners of the Mathematica probability functionality sometimes need manual "tweaking". When I have to do such, I always spot-check it against expected values... $\endgroup$
    – ciao
    Apr 7, 2014 at 21:40

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