# Distribution of $e$ if $g=\tilde{g}+e$, $g=|f|$, $f \sim \mathcal{N}(0,1)$ and $\tilde{g}$ is the quantization of $g$

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 ?

• 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 stats.stackexchange.com/a/117711.
– whuber
Feb 11, 2016 at 21:30
• @whuber thank you for your valuable comment. Question edited.. I have no idea what is "binning"
– din
Feb 11, 2016 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 $g\geq \tilde{g}_{i+1}$...." Feb 11, 2016 at 21:56
• @Glen_b I edited the question. Now we use $g=|f|$. Thank you!
– din
Feb 11, 2016 at 22:03
• @dln In respect of your earlier comment to whuber, a simple google search on binning gives me these as the top two hits: one, two. We also have a binning tag. Feb 11, 2016 at 23:32

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, $$f_e(y) = \sum_{i=0}^n 2\phi(\tilde{g}_{i} + y) \ 1(0\leqslant y < \tilde{g}_{i+1} - \tilde{g}_i)\,.$$
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.