The characteristic function for a Bernoulli distribution is

$$\phi(t) = (q+pe^{it}) \text{ where } p+q=1$$

I also know that the relationship between $\phi(t)$ and the pdf $f(k)$ is the Fourier Transform

$$f(k) = \frac{1}{2\pi}\int\limits_{-\infty}^{\infty}e^{-itk}(q+pe^{it})dt$$

However I do not see how to evaluate this integral to arrive at the expected pdf

$$f(k) = \begin{cases} p & k=1 \\ q & k=0 \end{cases}$$

as far as I got was

$$f(k) = \frac{1}{2\pi}\lim\limits_{T \rightarrow \infty}\int\limits_{-T}^{T}e^{-itk}(q+pe^{it})dt = \frac{-i}{\pi}\lim\limits_{T \rightarrow \infty}[q\sin(Tk) + p\sin(T(1-k))]$$

So if $k=0$ then the first term vanishes and the second term is undefined. Vice versa for $k=1$. And for $k \notin \{0,1\}$ both terms are undefined (although this last part is okay I think)

  • $\begingroup$ The Bernoulli distribution does not have a PDF: you are confusing the PDF with the probability function. If you want to approach the problem this way, you must think of the probability function as being a linear combination of the generalized functions $\delta_0$ and $\delta_1$. $\endgroup$
    – whuber
    Dec 17, 2016 at 21:41

1 Answer 1


I think the identity you're using for this problem isn't ideal. If $X$ is integer-valued, then there's another easier identity you can use: $$P(X=k) = \dfrac 1 {2\pi} \int\limits_{-\pi}^{\pi} e^{ikt} \phi_X(t) dt $$ Evaluating this integral leaves you with $$\dfrac {(2 k p - k + 1 - p) \sin(π k)} {π (1-k) k}$$

Now, you can't directly substitute $0$ or $1$ for $k$. Instead, you'll have to take the limit as $k$ approaches these values. If you do that, you end up with the correct limits of $p$ and $1-p$.

  • 1
    $\begingroup$ I take it that you mean characteristic function for CF and probability mass function for PDF. You should avoid acronyms as there can be ambiguities. In fact PDF usual stands for probability density function which doesn't exist for discrete distributions. $\endgroup$ Dec 18, 2016 at 2:27
  • $\begingroup$ I know it's been awhile, but can anyone provide a reference for that inversion formula? $\endgroup$ Feb 4, 2017 at 21:02
  • $\begingroup$ @GeoMatt22 tagging so you get a nofication $\endgroup$ Feb 6, 2017 at 15:43
  • $\begingroup$ I used corollary 6.10 in Probability by Alan Karr. $\endgroup$
    – jjet
    Feb 8, 2017 at 13:42

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