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Let $(\Omega, \mathcal{F}, \mathbb{P})$ be the unit interval with Lebesgue measure on the Borel subsets. Then we can find independent random variables $X_1, X_2, X_3, \dots$ defined on $(\Omega, \mathcal{F}, \mathbb{P})$, each normal mean zero, variance $1$.

This is a special case of the Borel isomorphism theorem, but I was wondering if anyone could supply/refer me to a proof which does not invoke it, since it seems like overkill here. Thanks!

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If we can find independent random variables $U_j$ uniform on $[0,1]$ we can transform them to $\mathcal N(0,1)$. If $Y$ is uniform on $[0,1]$, let $D_j$ be its $j$'th decimal place, i.e. $Y = \sum_{j=1}^\infty 10^{1-j} D_j$ with $D_j \in \{0,1,\ldots, 9\}$. If $p_j$ is the $j$'th prime, let $U_j = \sum_{k=1}^\infty 10^{1-k} D_{p_j^k}$. Then $U_j$ are iid uniform on $[0,1]$.

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  • $\begingroup$ As to how to transform them to $\mathcal{N}(0,1)$, let $\Phi(x) = \int_{-\infty}^x \frac{1}{\sqrt{2\pi}} e^{-x^2/2}\,dx$ be the normal cdf, and set $X_j = \Phi^{-1}(U_j)$. $\endgroup$
    – Nate Eldredge
    Commented Oct 1, 2015 at 15:58
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    $\begingroup$ @NateEldredge and of course there's nothing special about the normal distribution here: For any desired output distribution with cdf $F$, we can set $X = F^{-1}(U)$. (Hope this comment is not overly pedantic...) $\endgroup$
    – usul
    Commented Oct 1, 2015 at 16:29
  • $\begingroup$ @usul: Indeed. Though a little more work is needed if $F$ is not strictly increasing, i.e. not 1-1. $\endgroup$
    – Nate Eldredge
    Commented Oct 1, 2015 at 16:31

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