I am trying to find the characteristic function of a sum of "named" distributions, though I am interested in how to use Mathematica to find the characteristic function of $$f(X_1,\ldots, X_n),$$ where $X_i$ are some given random variables.

A toy problem is finding the characteristic function of $X+Y$, where $X$ and $Y$ are iid $\mathcal{N}(0,1)$. The following code:

\[ScriptCapitalD] =  TransformedDistribution[x + y, {x, y} \[Distributed] NormalDistribution[]]
CharacteristicFunction[\[ScriptCapitalD], \[Lambda]]

Produces the output

CharacteristicFunction[TransformedDistribution[\[FormalX]1 + \[FormalX]2, {\[FormalX]1, \[FormalX]2} \[Distributed] NormalDistribution[0, 1]], \[Lambda]]

which is obviously not what's desired.

It seems that CharacteristicFunction only works with named distributions that ship with Mathematica. Is this correct? If not, how can I make it work with a transformed distribution?

(I know that there are other ways to find the characteristic function of $X+Y$ above; it was only used for illustration purposes.)

  • $\begingroup$ What answer do you expect in the example you gave? $\endgroup$
    – RunnyKine
    Oct 27 '13 at 1:00
  • $\begingroup$ The characteristic function should be $e^{-\lambda^2}$ if I am not mistaken. $\endgroup$
    – Kevin
    Oct 27 '13 at 2:43

Here's a way to get the characteristic function. First, get the transformed distribution (in this case the sum) and then take the Characteristic function.

maxD = TransformedDistribution[x + y, {x \[Distributed] NormalDistribution[0, 5], 
                                       y \[Distributed] NormalDistribution[0, 3]}];
CharacteristicFunction[maxD, t]

E^(-17 t^2)

You can supply symbolic values instead of the fixed values, and it also works fine. For example:

CharacteristicFunction[TransformedDistribution[x + y, 
      {x \[Distributed] NormalDistribution[m1, s1], 
       y \[Distributed] NormalDistribution[m2, s2]}], t]

E^(I (m1 + m2) t - 1/2 (s1^2 + s2^2) t^2)

It also works for some (simple) functions. Both (x+y)^2 and x y work fine, but I gave up waiting for (x+y)^n.


The problem relates to the specfiication of the transformed distribution. {x,y}[Distributed]NormalDisribution[] is incorrect. NormalDistribution[] is a univariate distribution. Assuming X and Y are iid you could either use the syntax of bill s:

maxD = TransformedDistribution[x + y, {x \[Distributed] NormalDistribution[0, 5], 
                                       y \[Distributed] NormalDistribution[0, 3]}];

Or for this specific example binormal distribution with zero correlation:

d = BinormalDistribution[{0, 0}, {1, 1}, 0];
t = TransformedDistribution[x + y, {x, y} \[Distributed] d];
CharacteristicFunction[t, \[Lambda]]

This yields:



  • $\begingroup$ Ah I see; should have read the documentation more closely. $\endgroup$
    – Kevin
    Oct 27 '13 at 19:55
  • $\begingroup$ It's ok. I have made same or similar mistakes in past... $\endgroup$
    – ubpdqn
    Oct 28 '13 at 1:48

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