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I would like to calculate the distribution function from the characteristic function. There is a formula given as

$$F_X(x)=\frac{1}{2}+\frac{1}{2\pi}\int_{0}^\infty \frac{e^{i w x}\phi(-w)-e^{-i w x}\phi(w)}{i w}\mbox{d}w$$ Actually, I want to evaluate $F_X(x)$ at some real number $b<0$. I wrote some codes in Mathematica with unfortunately no success. The codes give some results but they seem to be incorrect.

Here are my codes:

opts = {Method -> {Automatic, "SymbolicProcessing" -> None}, AccuracyGoal -> 8}
b = -2;
f1[x_] := PDF[NormalDistribution[2, 2], x]
ff1[\[Omega]_] := InverseFourierTransform[f1[x], x, \[Omega]]
qn1 =1/2 + 1/(2*Pi) NIntegrate[(Exp[I \[Omega] b]*ff1[-\[Omega]] -Exp[-I \[Omega] b]*ff1[\[Omega]])/(I \[Omega]), {\[Omega], 0,Infinity}, Evaluate@opts]
NIntegrate[f1[x], {x, -Infinity, b}]

What I expect is $qn1$ to be equal to the last line of my code. But it just $0$. I cannot see the problem. Perhaps someone can?

Thanks in advance

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2 Answers 2

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This is already bult into Mathematica

$Assumptions = {s > 0};

dist = NormalDistribution[m, s];

pdf[x_] = PDF[dist, x];

cf[w_] = CharacteristicFunction[dist, w];

pdf[x] == InverseFourierTransform[cf[w], w, x,
   FourierParameters -> {1, 1}] // Simplify

True

cf[w] == FourierTransform[pdf[x], x, w,
   FourierParameters -> {1, 1}] == 
  Expectation[Exp[I w x], x \[Distributed] dist] //
 Simplify

True

EDITED:

(cdf[x_] = Integrate[pdf[t], {t, -Infinity, x}]) ==
  CDF[dist, x] // Simplify

True

For a specific example of deriving the PDF from the characteristic function

dist = NormalDistribution[2, 2];

cf[w_] = CharacteristicFunction[dist, w];

(pdf[x_] = 
    InverseFourierTransform[cf[w], w, x, FourierParameters -> {1, 1}]) == 
  PDF[dist, x] //
 Simplify

True

(cdf[x_] = Integrate[pdf[t], {t, -Infinity, x}]) ==
  CDF[dist, x] // FullSimplify

True

pdf /@ Range[-4., 8., 2.]

{0.00221592, 0.0269955, 0.120985, 0.199471, 0.120985, 0.0269955, 0.00221592}

Plot[pdf[x], {x, -4, 8}]

enter image description here

cdf /@ Range[-4., 8., 2.]

{0.0013499, 0.0227501, 0.158655, 0.5, 0.841345, 0.97725, 0.99865}

Plot[cdf[x], {x, -4, 8}]

enter image description here

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3
  • $\begingroup$ It seems I get $1/2$ for all $x$ not only for $x=-2$. I tried to draw $F_X$ for the given formula and mathematica gave me a constant line on $y=0.5$. $\endgroup$
    – Seyhmus Güngören
    Commented Jun 21, 2014 at 21:52
  • $\begingroup$ Specific example added above to demonstrate clearly that this works. $\endgroup$
    – Bob Hanlon
    Commented Jun 22, 2014 at 5:25
  • $\begingroup$ Thank you. The problem was about the Fourier parameters, which should have been chosen as $(1,1)$. $\endgroup$
    – Seyhmus Güngören
    Commented Jun 22, 2014 at 11:11
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Clear["Global`*"]
b = -2;
f1[x_] := Evaluate@PDF[NormalDistribution[2, 2], x];
ff1[\[Omega]_] := Evaluate@InverseFourierTransform[f1[x], x, \[Omega]];
f1[x]
ff1[x]
(Exp[I \[Omega] b]*ff1[-\[Omega]] - 
   Exp[-I \[Omega] b]*ff1[\[Omega]])/(I \[Omega]) // Simplify

enter image description here

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1
  • $\begingroup$ Thank you very much for the answer. I used "Evaluate@" so that I can draw a figure out of this formulation. $\endgroup$
    – Seyhmus Güngören
    Commented Jun 22, 2014 at 11:11

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