Characteristic function of Cantor distribution

Question

In Wiki, it provided the Characteristic function of Cantor distribution. That is, $e^{\mathrm{i}\,t/2}\prod_{i= 1}^{\infty} \cos{\left(\frac{t}{3^{i}} \right)}$. How to show this?

Answer 1

Suppose $X$ is Cantor distributed and we know that it is symmetric around $\frac{1}{2}$ and we have to rewrite as :

$$X - \frac{1}{2} = \sum_{k=1}^\infty \frac{X_k}{3^k}$$ where $\Pr[X_k = -1] = \Pr[X_k = 1] = 1/2$ and $X_k$ are iid for all $k$.

Remark: It is the same as what John Dawkins write, noted that $\frac{1}{2} = \sum_{k=1}^\infty \frac{1}{3^k}$.

For each $X_k$, $$\begin{align}\varphi_{X_k}(t) &= \sum\Pr[X_k=l]e^{itl}\\ &=\frac{1}{2}e^{-it} + \frac{1}{2}e^{it} = \cos(t)\\ \therefore \varphi_{X_k/3^k}(t)&=\varphi_{X_k}(t/3^k)\\ &=\cos\left(\frac{t}{3^k}\right)\end{align} $$

By independence, the sum of $X_k/3^k$ will be the product in the characteristic function,

$$\begin{align} \varphi_{X-1/2} (t) &= \prod_{k=1}^\infty \cos\left(\frac{t}{3^k}\right) \\ \varphi_{X} (t) &= e^{it/2} \prod_{k=1}^\infty \cos\left(\frac{t}{3^k}\right) \end{align}$$

For most mathematical beauty, a little bit of location-scale transformation is always needed.

References: Lukacs, Eugene (1970). Characteristic Functions, 2nd edition, Griffin, London.