I am trying to find the characteristic function for Johnson's SU distribution by integrating the probability density function with Exp[I*t*x] but Mathematica is returning the input itself.

As the characteristic function always exists, I'm not able to understand why Mathematica is not finding the integral.

Here's the code:

expr1[x_] := PDF[JohnsonDistribution["SU", γ, δ, ξ, λ], 

Integrate[expr1[x]*Exp[I*t*x], {x, -Infinity, Infinity}]
  • 1
    $\begingroup$ I'm voting to close this question as off-topic because the OP is asking for functionality that is not supported given the constraints the OP is putting on the solution. $\endgroup$
    – m_goldberg
    Jun 19, 2019 at 15:28
  • 4
    $\begingroup$ @m_goldberg Closing a question from a first timer abruptly may alienate the person from this forum and worse: from the Wolfram community. $\endgroup$ Jun 20, 2019 at 7:55

2 Answers 2


Please look at the characteristic function of the lognormal distribution.
In some cases there is no closed form.

One way to address this issue is using empirical' s one.

emCF = With[{dist = #}, 
    RandomVariate[dist, 100] // Exp[I t #] & /@ # & // Mean // 
     Set[approx, #] &;
    ReIm[approx]] &;

approximated characteristic function of NormalDistribution[0,1]

Plot[Evaluate[emCF[NormalDistribution[0, 1]]], {t, -1, 1}]

Mathematica graphics

exact characteristic function of NormalDistribution[0,1]

Plot[ReIm@CharacteristicFunction[NormalDistribution[0, 1], t], {t, -1,

Mathematica graphics

approximated characteristic function of JohnsonDistribution["SU", a, b, c, d]

 Plot[Evaluate[emCF[JohnsonDistribution["SU", a, b, c, d]]], {t, -1, 
   1}], {a, -5, 5}, {b, 0.01, 5}, {c, -5, 5}, {d, 0.01, 5}]

Mathematica graphics


The first assignment is incorrect, resp. I don't know what it should mean from a Wolfram Mathematica syntax perspective. It can mean

expression: a correct form is

 expr1 = PDF[JohnsonDistribution["SU", γ, δ, ξ, λ], x]

or function definition: a correct form is

 expr1[x_] := PDF[JohnsonDistribution["SU", γ, δ, ξ, λ], x]

However, I think that this is not a major issue. It is that apparently the characteristic function for Johnson's distribution cannot be expressed in the form of a closed expression. Neither wikipedia nor other source know the formula for general characteristic function. The Mathematica answer is "I'm not able to calculate it", i.e.

 CharacteristicFunction[JohnsonDistribution["SU", γ, δ, ξ, λ], t]

CharacteristicFunction[JohnsonDistribution["SU", γ, δ, ξ, λ], t]

  • 2
    $\begingroup$ There's nothing wrong with the function definition as supplied. It simply uses immediate evaluation instead of delayed. The real answer is your second portion to this. $\endgroup$
    – b3m2a1
    Jun 20, 2019 at 7:58
  • 1
    $\begingroup$ @b3m2a1 Exactly right. Function assignments of the form f[x_] = ... are simply equivalent to f[x_] := Evaluate[...]. $\endgroup$ Jun 20, 2019 at 8:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.