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This is on the same theme as in this post, where the Fourier transform was derived using simple function.

Let $f:[0,1] \to [0,1]$ be the Cantor function.
Then $f$ is the cumulative distribution of a Cantor distributed random variable $$ X=\sum_{n=1}^\infty 3^{-n} Y_n $$ where the $Y_n$ are i.i.d. and takes values $0$ and $2$ with equal probability.

In this MO post, it is stated that $$ E(e^{itY_n})=e^{it/2}\cos(3^{-n} t). $$

How do we get that? I have $$ E(e^{itY_n})=\frac{e^{it2/3^n}+1}{2}. $$

Also, it is stated in the post that $$ \hat f(t)=\frac{1}{it} -\frac{1}{it}\hat {f'}(t). $$

How do we get this one? I thought $$ \hat f(t)=\frac{1}{it}\hat {f'}(t) $$ only.

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It makes sense to link to closely related questions that you asked before, since it provides more context and may allow people to build on what's already been done. –  joriki Oct 25 '12 at 12:00
    
@joriki I though this was done automatically somehow. Thanks for notifying me that I should do it explicitly. –  Nicolas Essis-Breton Oct 25 '12 at 13:04
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up vote 3 down vote accepted

We have \begin{align} E[e^{itX}]&=\prod_{n=1}^{+\infty}E[e^{it3^{-n}Y_n}]\\ &=\prod_{n=1}^{+\infty}\frac{e^{it2\cdot 3^{-n}}+1}2\\ &=\prod_{n=1}^{+\infty}e^{it \cdot 3^{-n}}\cos(t3^{—n})\\ &=\prod_{n=1}^{+\infty}e^{it \cdot 3^{-n}}\prod_{n=1}^{+\infty}\cos(t3^{-n})\\ &=\exp\left(it\sum_{n\geq 1}3^{—n}\right)\prod_{n=1}^{+\infty}\cos(t3^{-n})\\ &=\exp(it/2)\prod_{n=1}^{+\infty}\cos(t3^{-n}). \end{align}

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