# Gibbs measure and normal distribution

On Wikipedia, the Gibbs measure defines the probability as:

$$P(X=x) = \frac{1}{Z(\beta)}\exp(-\beta E(x))$$

Now, the familiar form of the normal distribution is:

$$P(x) = \frac{1}{\sqrt{2\pi}\sigma}\exp-{\frac{(x-\mu)^{2}}{2\sigma^2}}$$

Now, it seems that the normal distribution is a Gibbs measure and I was wondering whether I can equate some of the terms. So, for example what would be the $\beta$ term and more importantly, what is the corresponding energy term in the normal distribution. Can I say $\beta$ is $\frac{1}{2}$ and the energy term is $\frac{(x-\mu)^{2}}{\sigma^2}$ but then I do not understand how the partition function $Z$ is a function of $\beta$.

• Luca: the $x$'s in your formulas have very different meanings. In the Gibbs measure, one possibility is that $x$ is a point in the configuration space of a system of particles, that is, in this case $x$ is a $3N$-dimensional vector containing the cartesian components of the positions and velocities of $n$ particles. Another possibility is that $x$ is a vector representing the discrete values of the spins in a network. In the normal distribution, the $x$ is a real number: one of the possible values of a random variable whose support is the real line. – Zen Jan 23 '14 at 19:10
• ahhhh…so, they are not exactly comparable. How does one then use energy functions (like used for regularisation) in a probabilistic framework? – Luca Jan 23 '14 at 19:12
• Take a look at this math.ntnu.no/~joeid/TMA4250/image_ana.pdf – Zen Jan 23 '14 at 19:35
• – Zen Jan 23 '14 at 19:35
• The reason for the appearence of the Gibbs measure in these problems is a deep mathematical result known as the Hammersley-Clifford Theorem. – Zen Jan 23 '14 at 19:36

1. If you consider a mechanical system, a term like $\frac{1}{2}k {(x-\mu)^2}$ can occur as measure of elastic energy: For instance when you have a linear spring with spring constant $k$, which can only move in x-direction and is in equilibrium when $X=\mu$. We choose $\sigma$ such that $k=\frac{1}{\sigma^2}$. Larger values of $k$ represent stiffer springs.
2. Remember that the gibbs measure comes from statistical mechanics, describing an ensemble of particles and where $\beta$ is the inverse temperature: The smaller the value of $\beta$, the faster is the average speed of the particles, i.e. the more likely are states which are "far away" from the equilibrium (in the sense of states with a high energy function $E(x)$). In this way, also $\beta$ represents the "stiffness" of the ensemble of particles.
3. The terms $\frac{1}{\sqrt{2 \pi \sigma^2}}$ and $\frac{1}{Z(\beta)}$ play a very similar role: They are normalization constants which ensure that the resulting functions are probability densities. These constants naturally depend on $\beta$ and $\sigma^2$.
4. In conclusion, we can reasonably define $E(X=x)=\frac{1}{2} {(x-\mu)^2}$ as an energy function and set $\beta = \frac{1}{\sigma^2}$ as inverse temperature or measure of stiffness, where smaller values of $\beta$ make high-energy states more likely.