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Let $X=(x_1,\ldots,x_d)^\top\in[0,1]^d$ be the row-wise representation of an $n\times n$ image ($d=n\times n$). Each element of $X$ is the value of a pixel, which we assume it belongs to $[0,1]$.

If we assume that $X$ is distributed normally with covariance matrix $\Sigma=\sigma^2I_d$, i.e., $X\sim\mathcal{N}(\mathbf{0},\Sigma)$, what can we say about the various sub-regions of $X$?

Starting from a single variable, say $x_i$ for some $i\in\{1,\ldots,d\}$, is it true that $x_i\sim\mathcal{N}(0,\sigma^2)$? If so, is it also true that a random selection of $x_i$'s, for instance the vector $Y=(x_3, x_7, x_{11})^\top$, is also distributed normally with the same variance? That is, in the case of the $3$-dimensional $Y$ for instance, is it true that $Y\sim\mathcal{N}(\mathbf{0},\sigma^2I_3)$?.

Finally, if we assume that the original $X$ is independent and identically distributed (iid), would that be a sufficient condition?

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    $\begingroup$ Independency is a sufficient condition. If the $X_i´s$ are independent then they are uncorrelated as well. $\endgroup$
    – callculus42
    Jul 19, 2016 at 13:31
  • $\begingroup$ If each pixel value $X_i$ is in the unit interval $[0,1]$, then it cannot be Gaussian (i.e., "normally distributed"). $\endgroup$
    – Michael
    Jul 19, 2016 at 13:43
  • $\begingroup$ @Michael, what do you mean? Why they can't be normally distributed? Say that some univariate variable $x\in[0,1]$ is distributed normally with mean 0.5 and variance, say 0.01, where is the problem? $\endgroup$
    – nullgeppetto
    Jul 19, 2016 at 13:47
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    $\begingroup$ @nullgeppetto: $X_i$ can not be normally distributed by definition of the normal distribution which requires the density function be positive at all values. Your present distribution is truncated at $0$ and $1$. However, this does not affect the conclusion so long as the pre-truncation distribution is independent. However, the dependency or covariance will be affected by truncation if the pre-trancation distribution is not independent. $\endgroup$
    – Hans
    Jul 19, 2016 at 19:30
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    $\begingroup$ @nullgeppetto : Hans gives the explanation. Indeed the Gaussian density for mean $m$ and variance $\sigma^2$ is $$f_X(x) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-m)^2}{2\sigma^2}} \quad , \forall x \in \mathbb{R}$$ That is, $f_X(x)$ is defined over $x \in (-\infty, \infty)$, not just over $x \in [0,1]$. Values of a Gaussian random variable $X$ can be arbitrarily large or arbitrarily small. $\endgroup$
    – Michael
    Jul 19, 2016 at 19:41

1 Answer 1

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I consider the two variable case and take into account that the means of $X_1$ and $X_2$ are 0.

The pdf of the univariate untruncated distribution of $X_1$ is

$f_{X_1}(x_1)=\frac{1}{\sqrt{2\cdot \pi \sigma_1^2}}\cdot e^{-\frac12 \left[\frac{x_1^2}{\sigma_1 ^2}\right]} $

The cdf of the univariate untruncated distribution of $X_1$ is

$F_{X_1}(x_1)=\int_{-\infty}^{x_1}\frac{1}{\sqrt{2\cdot \pi \sigma_1^2}}\cdot e^{-\frac12 \left[\frac{s^2}{\sigma_1 ^2}\right]} \, ds$

The pdf of the univariate truncated distribution of $X_1$ on $[0,1 ] $is

$f(x_1;0,1)=\frac{\frac1{\sigma_1} \cdot f_{X_1}(x_1)}{F_{X_1}(1)-F_{X_1}(0)} $

The cdf of the univariate truncated distribution of $X_1$ on $[0,1 ] $is

$F(x_1;0,1)=\int_{0 }^{x_1}\frac{\frac1{\sigma_1} \cdot f_{X_1}(s)}{F_{X_1}(1)-F_{X_1}(0)} \, ds$

Since $X_1$ and $X_2$ are independent the joint truncated cumulative distribution function on $[0,1 ]^2 $ is

$F(x_1,x_2;[0,1]^2)=\int_{0 }^{x_1}\frac{\frac1{\sigma_1} \cdot f_{X_1}(s)}{F_{X_1}(1)-F_{X_1}(0)} \, ds\cdot \int_{0 }^{x_2}\frac{\frac1{\sigma_2} \cdot f_{X_2}(t)}{F_{X_2}(1)-F_{X_2}(0)} \, dt$

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