Does logarithm of Gaussian image still gaussian distribution?

Question

I have an image 2D that pixel intensity follows multi Gaussian distribution such as

$$p \left( I(x) \in \Omega_i \mid (I(x)\right)=\frac{1}{2\pi \sigma_i}\exp\left(-\frac {(I(x)-\mu_i)^2}{2\sigma_i^2} \right )$$

where $I(x):\Omega \to R$, $\sigma_i,\mu_i$ are std, and mean of region $\Omega_i$. (For example: Image in $\Omega$ domain can spearate into three region $\Omega_1,\Omega_2,\Omega_3$)

I will take the logarithm of image I such as $$I_{\log}=\log(I)$$ Then, could I said that the pixel intensity in logarithm space still follows multi Gaussian distribution such as? $$p \left( I(x) \in \Omega_i \mid (I(x)\right)=\frac{1}{2\pi \sigma_i}\exp\left(-\frac {(I_{\log}-\mu_i)^2}{2\sigma_i^2} \right )$$

Update: This is image distribution

Answer 1

Since any linear transformation of a Gaussian variable is Gaussian, you get that "locally" (if your Gaussian distribution for $I$ is concentrated strongly enough around a small enough region) then $\log I$, or any differnentiable function $f(I)$ for that matter, will look Gaussian in most of the probability mass for $I$ and you can easily estimate the mean and variance for the new Gaussian distribution of $f(I)$.

Answer 2

If $I(x)$ has a Gaussian distribution, then $\Pr(I(x)<0)>0$, and you cannot take the logarithm of a negative number, so the logarithm cannot have exactly a Gaussian distribution.

For values of $I$ that are very close to $\mu$, the curvature of the graph of $y=\log I$ is too small to matter, and the graph is approximated by the tangent line, which is the graph of the function $$ I\mapsto \frac 1 \mu (I-\mu) + \log \mu. \tag 1 $$

The slope $\dfrac 1 \mu$ is $\left.\dfrac d {dI} \log I \right|_{I=\mu}$ , i.e. the slope of the graph of the logarithm function at that point.

If $\sigma$ is very small compared to $\mu$, then with high probability $I$ is in the small area in which $\log I$ can reasonably be approximated by the expression $(1)$ above. And that expression has a Gaussian distribution with expected value $0$ and variance $\dfrac 1 {\mu^2} \cdot \sigma^2\vphantom{\frac{\displaystyle\int}{\displaystyle\int}}$ (multiplying a random variable by $1/\mu$ has the effect of multiplying the variance by $1/\mu^2$).

Summary: $I(x)$ has approximately a Gaussian distribution if $\sigma$ is small compared to $\mu$.