Question about multidimensional iid random variable 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?
 A: 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 
$F(x_1,x_2;[0,1]^2)=\int_{0 }^{x_1}\frac{\frac1{\sigma_1} \cdot f_{X_1}(s)}{F_{X_1}(1)-F_{X_1}(0)} \, ds\cdot \int_{0 }^{x_2}\frac{\frac1{\sigma_2} \cdot f_{X_2}(t)}{F_{X_2}(1)-F_{X_2}(0)} \, dt$
