I need to find the probability density of some distribution with characteristic function given by:

$$\frac{1}{9} + \frac{4}{9} e^{iw} + \frac{4}{9} e^{2iw}$$

I know the formula for inverting a characteristic function is:

$$f_X(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \phi(\omega) e^{-i\omega x} \mathop{d\omega}$$

But obviously putting this function inside the formula, will make the integral diverge. So my question is how does one invert a characteristic function, when this integral diverges? Or is this supposed to always converge and there is something wrong with my characteristic function?

  • 2
    $\begingroup$ In this particular case note that a constant characteristic corresponds to a multiple of the point mass (Dirac $\delta$) centered at $0$. Note also that multiplying by $e^{i\omega}$ is a "phase shift" which corresponds to translation in physical space. So your probability distribution is one with three points masses sitting at $x = 0,1,2$. $\endgroup$ Commented Jun 6, 2013 at 15:24

2 Answers 2


The inversion formula you cite is restricted to integrable characteristic functions. The case in your question is $\varphi_X(\omega)=\sum\limits_{k=1}^np_k\mathrm e^{\mathrm i \omega a_k}$ with $p_k\gt0$ and $\sum\limits_{k=1}^np_k=1$, which is never integrable.

Assume that $n=1$, that is, that $\varphi_X(\omega)=\mathrm e^{\mathrm i \omega a}$. Can you identify the distribution of $X$ in this case? Hint: there is no density. Then the general case might be straightforward.


This is an old question, but I still like to write an answer.

First note that for your c.f. it holds that $\phi_x(\omega)=\phi_x(\omega+2\pi)$, indicating that your random variable takes only value in $\mathbb{Z}$ and hence is of lattice type for which a simple inversion formula is $f_X(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi} e^{-i \omega x}\phi_x(\omega)d\omega$.

An evaluation for $\phi_x(\omega)=\frac{1}{9}+\frac{4}{9}e^{i \omega}+\frac{4}{9}e^{2 i \omega}$ shows that $f_X(x)=\dfrac{2(2+x+x^2)}{9x(2-3x+x^2)}\dfrac{\sin(\pi x)}{2\pi}$.

Now note that $f_X(x)=0$ for $x\in\mathbb{Z}-\{0,1,2\}$, and that $\lim \limits_{x\rightarrow 0}f_X(x)=\frac{1}{9}$, $\lim\limits_{x\rightarrow 1}f_X(x)=\frac{4}{9}$ and $\lim\limits_{x\rightarrow 2}f_X(x)=\frac{4}{9}$. Therefore $f_X(x)=\frac{1}{9}$ for $x=0$, $f_X(x)=\frac{4}{9}$ for $x=1$, $f_X(x)=\frac{4}{9}$ for $x=2$, and $f_X(x)=0$ everywhere else.

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    $\begingroup$ Actualy, it seems that $f_X(x)\ne0$ for (almost) every noninteger $x$... $\endgroup$
    – Did
    Commented Nov 25, 2014 at 12:11
  • $\begingroup$ But the inversion formula used applies to lattice characteristic functions only, i.e. we know beforehand that we are after a function that takes only integers. I mean that find a function $f_X(x)$ with $\mathbb{Z}$ as its domain not $\mathbb{R}$. $\endgroup$
    – Math-fun
    Commented Nov 25, 2014 at 14:05

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