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Let $f \sim \mathcal{N}(0,1)$ be a normal random variable with zero mean and unit variance. Let $g=|f|$. Let $\tilde{g}$ be the quantization of $g$. We suppose that there are $n$ possible levels of $\tilde{g}$, denoted as: $\tilde{g}_1, \ldots, \tilde{g}_n$. These levels are fixed beforehand.
The quantization is performed as the following: if $\tilde{g}_{i} \le g < \tilde{g}_{i+1}$, then the quantized level of $g$ is $\tilde{g}_{i}$. Note that if $g \ge \tilde{g}_{n}$, the quantization is $\tilde{g}_{n}$, and if $g < \tilde{g}_{1}$, then the quantization level is $0$. We can represent $g$ as $g=\tilde{g}+e$, where $e$ can be seen as the quantization error.

My question: what is the distribution of $e$ in this case ?

din
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    Please explain what you mean by a "quantization." It sounds like the same thing as *binning*. Even then, an important issue is whether the cutpoints between the bins were selected beforehand or are based on data. Finally, you might be interested in how a "quantized" version of a Normal distribution can be used to generate uniformly distributed variates, as I describe in item (8) of a post at http://stats.stackexchange.com/a/117711. – whuber Feb 11 '16 at 21:30
  • @whuber thank you for your valuable comment. Question edited.. I have no idea what is "binning" – din Feb 11 '16 at 21:49
  • 1. So you have some set of bins and then label the values in each bin by the left end of the bin, except that in the lowest bin you use $0$ instead of $-\infty$ as the label? (which implies you expect half your quantized values will be $0$). $\,$ 2. I think you probably want $n$ instead of $i$ in "Note that if $g\geq \tilde{g}_{i+1}$...." – Glen_b Feb 11 '16 at 21:56
  • @Glen_b I edited the question. Now we use $g=|f|$. Thank you! – din Feb 11 '16 at 22:03
  • I hope this assumption won't complicate the problem – din Feb 11 '16 at 22:08
  • are those levels equispaced? – Glen_b Feb 11 '16 at 23:27
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    @dln In respect of your earlier comment to whuber, a simple google search on *binning* gives me these as the top two hits: [one](https://docs.tibco.com/pub/spotfire/6.0.0-november-2013/userguide-webhelp/bin/bin_what_is_binning.htm), [two](https://en.wikipedia.org/wiki/Data_binning). We also have a [binning](http://stats.stackexchange.com/tags/binning/info) tag. – Glen_b Feb 11 '16 at 23:32
  • @Glen_b We can assume that the levels are equispaced if this can simplify the problem. – din Feb 12 '16 at 09:37

2 Answers2

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The density of $g$ is indeed $2\phi(x)1(x\geqslant 0)$. Let us agree that $\tilde{g}_0=0$ and that $\tilde{g}_{n+1}=+\infty$. The density of $e$ can then simply be obtained by the law of total probability. For $y\geqslant 0$: \begin{align*} f_e(y)dy & = \text{Prob}[y\leqslant e<y+dy]\\ & = \sum_{i=0}^n \text{Prob}[y\leqslant e <y+dy, \tilde{g}_i\leqslant g < \tilde{g}_{i+1}]\\ & = \sum_{i=0}^n \text{Prob}[\tilde{g}_{i} + y \leqslant g < \tilde{g}_{i} +y+dy] \ 1(0\leqslant y < \tilde{g}_{i+1} - \tilde{g}_i)\\ & = \sum_{i=0}^n 2\phi(\tilde{g}_{i} + y) dy \ 1(0\leqslant y < \tilde{g}_{i+1} - \tilde{g}_i)\,.\\ \end{align*} So, $$ f_e(y) = \sum_{i=0}^n 2\phi(\tilde{g}_{i} + y) \ 1(0\leqslant y < \tilde{g}_{i+1} - \tilde{g}_i)\,. $$

StijnDeVuyst
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So e is the difference of two (dependent) random variables, meaning $f_E(e) = \int_{-\infty}^{\infty} f_{G \tilde{G}}(g, g - e) dg $. The distribution of g is simply $2 \phi(x) \mathbb{1}(x > 0) $ where $\phi$ is the standard normal p.d.f. It gets a little more complicated from there, especially as $g$ and $\tilde{g}$ are dependent, but I figured I'd get it started.

Sheridan Grant
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