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I came across this fact that if we flip an unbiased coin infinite times and represent the outcome as $X(\omega)=(X_1(\omega),X_2(\omega),\ldots )$ then the new random variable $U$ defined by

$$ U(w)=\sum_{i=1}^{\infty} \frac{X_i(\omega)}{2^i} $$

would have a uniform distribution on $[0,1]$.

I want to prove this above fact in addition to it's converse that given a uniform random variable, it's binary representation is equivalent to infinite i.i.d coin tosses. Though converse is fairly easy, the if part is turning out to be difficult.

In essence, the experiment of infinite fair coin tosses serves as a building block for generating any other random variable.

pikachuchameleon
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    I got halfway through answering this question when I realized that I had answered it before. – heropup Sep 15 '15 at 04:44
  • It seems like you're basically just creating a random binary number, with equal chances for 0/1 on each digit, that is, an equal chance for any value in [0, 1]. – Daniel R. Collins Sep 15 '15 at 05:48
  • Not all values in [0,1] are equally likely.Using Chebychev's theorem (Law of large numbers) we can show that $n^{-1} \sum_{i \leq n} X_i(w)$ converges to $1/2$ with probability $1$. – DanielWainfleet Sep 15 '15 at 18:41
  • @user254665 Is this fact related to the question? – Did Sep 16 '15 at 06:54
  • @DanielR.Collins "an equal chance for any value in [0, 1]" Not sure about the meaning of this? Or rather, the meanings I can imagine are much better expressed differently. – Did Sep 16 '15 at 06:55

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A sum of random variables is a convolution and a coin flip is a sum of two dirac distributions. Division by 2 "shrinks" the filter by a factor of two. So we can view it as a kind of fractal, successively refining the interval. It also happens to be the Haar Inverse Discrete Wavelet Transform for a coefficient of the lowest frequency. All of these concepts come together nicely in the book Analysis and Probability - Wavelets, Signals, Fractals by Palle Jorgensen.

mathreadler
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