std/MAD
Two ways to measure “spread”:
- std $=\sqrt{\mathbb E[X^2]}$ — squares everything, so one big outlier dominates. Measures the extremes.
-
MAD $=\text{median} X $ — half the values are above, half below. One outlier changes nothing. Measures the typical.
std/MAD tells us: how far out are the extremes, relative to pure Gaussian? For a Gaussian, $\text{median} |
X | = 0.6745\sigma$, so |
Any $r > 1.4826$ means heavier-tailed than Gaussian. Real data at $r = 23.2$ is 15.7× more tail-heavy than Gaussian.
What is ACF?
Autocorrelation function. For a pixel and its neighbour $k$ steps away:
\[\rho(k) = \frac{\mathbb E[X_i X_{i+k}]}{\mathbb E[X_i^2]}\]$\rho(0)=1$ always. $\rho(k)\to 0$ as $k$ grows. The shape of that decay is the texture of the noise.
- $\rho(k)=0$ for all $k\neq0$ → pure salt-and-pepper grain
- $\rho$ decays slowly → soft blobby mottling
The key theorem: if you build a field by blurring white noise with a kernel $h$, the field’s ACF is the kernel’s own autocorrelation:
\[\rho(k) = \frac{\sum_j h[j],h[j+k]}{\sum_j h[j]^2}\]For our Gaussian kernel with $\sigma=0.82$, the autocorrelation of a Gaussian is another Gaussian of width $\sigma\sqrt2$:
\[\rho(k) = e^{-k^2/(4\sigma^2)}\]Theory matches our measurement to three decimals. So we understand our own field exactly — and the problem is unambiguous: $k^2$ in the exponent kills it. Real decays like $a^k$ (linear in $k$), we decay like $e^{-k^2}$. By lag 4 we’re 30× too small. The fix, and why it’s exact. Use a one-sided exponential kernel $h[j]=a^j$ for $j\ge0$: \(\sum_{j\ge0} a^j a^{j+k} = a^k\sum_{j\ge0}a^{2j} = \frac{a^k}{1-a^2} \quad\Longrightarrow\quad \rho(k) = a^k\)
Exponential ACF by construction, no fitting. Set $a_{\text{row}}=0.651$, $a_{\text{col}}=0.489$. Honest limit: look at the last column. Exponential overshoots at high lag (0.276 vs 0.213 at lag 3). It’s ~3× better than Gaussian, not perfect. Real sits between the two. Worth taking the improvement and re-measuring.
Gotcha: $h$ is one-sided, so it’s a causal filter. padding=K//2 assumes a symmetric kernel and will shift the whole field by $K/2$ pixels. It needs explicit left-padding only.
