2
$\begingroup$

Consider $\Omega = \{0,1\}^\mathbb{N} $. Let $$U(\omega) =\sum_{i=1}^\infty \frac{\omega_i}{2^i} $$ where all $\omega_i \in \{0,1\}$. We can show, that $2U+1$ is uniformly distributed in $[-1,1]$.Therefore we know, that its characteristic function is $$\frac{\sin(t)}{t}.$$

We now want to establish, that sinus is an infinte product. Therefore we have to calculate $$E[e^{it (2U-1)}]$$ in a different way as a infinite product $$E[e^{it (2U-1)}] = \prod_{i=1}^\infty...$$ How can one do that? Does anyone know the ansatz for that?

  • 5
    Well, use the fact that the characteristic function of a sum with independent increments is the product of the characteristic functions. After that, you might need the characteristic function of the random variable $X_n$ defined by $X_n(\omega)=2\omega_n-1$ hence you may begin by computing this.2011-10-25
  • 0
    Eugene Lukacs wrote a book called _Characteristic Functions_ in which, if I recall correctly, you will find this. But I think Didier Piau's comment pretty much covers it.2011-10-25
  • 0
    Note, incidentally, that the English name for the $\sin$ function is "(the) sine" rather than "sinus".2011-10-25
  • 0
    Thanks a lot! With Didier Piau's post it follows by easy computations.2011-10-25
  • 1
    @Didier: Doesn't that get you [Viète's infinite product](http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Vi.C3.A8te.27s_infinite_product)? Not sure which infinite product the OP is asking for.2011-10-25
  • 0
    @Raskolnikov: yes it does!2011-10-25

1 Answers 1

5

Use the fact that the characteristic function of a sum with independent increments is the product of the characteristic functions. After that, you might need the characteristic function of the random variable $X_n$ defined by $X_n(\omega)=2\omega_n-1$ hence you may begin by computing this.

As mentioned by @Raskolnikov, the result involves Viète's infinite product.