The normal distribution shows up everywhere in scientific and statistical research. An obvious question to ask is, why? Why would this distribution be the one to manifest all over the place? One could simply quote the central limit theorem, but I find that unsatisfying. A better answer to that question is "because it is a maximizer of entropy". This can be shown using methods from calculus of variations.

The Lagrangian of entropy, $L$, can be constructed as follows: $$ \begin{align} L[\rho] &= \int_{-\infty}^{\infty}S[\rho]\,dx + \lambda_{0}\left(\int_{-\infty}^{\infty}\rho\,dx - 1 \right)\\ &\quad+ \lambda_{1}\left(\int_{-\infty}^{\infty}x^{2}\rho\,dx - \sigma^{2} \right)\,dx \end{align} $$ where the terms with $\lambda_{0},\,\lambda_{1}$ coefficients represent the normalization and finite variance constraints of the solution $\rho = \rho(x)$.

Similarly to the rules of standard calculus, we can differentiate $L$ and set the derivative equal to zero in order to determine an optimum functional argument. Taking the functional derivative gives $$ \begin{align} \frac{\delta L}{\delta \epsilon}[\rho] &= \lim_{\epsilon \rightarrow 0}\frac{L[\rho + \epsilon\eta] - L[\rho]}{\epsilon} \\ &= \int_{-\infty}^{\infty}\eta\left( \log(\rho) - 1 + \lambda_{1}x^{2} + \lambda_{0}\right)\,dx \end{align} $$ The function $\eta = \eta(x)$ is arbitrary and satisfies the same boundary conditions as $\rho$. Thus, for $\delta L / \delta\epsilon$ to be equal to 0 when $\rho$ is an optimal function, we must have: $$ \begin{align} &\log(\rho) - 1 + \lambda_{1}x^{2} + \lambda_{0} = 0 \\ \Rightarrow \quad &\rho = \hat{\lambda}_{0}\exp(\hat{\lambda}_{1}x^{2}). \end{align} $$ Applying the normalization and finite variance constraints to solve for the constants $\hat{\lambda}_{0}$, $\hat{\lambda}_{1}$ gives: $$ \rho = \frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\left( -\frac{x^{2}}{2\sigma^{2}}\right). $$ Voilà! We see that $\rho$ is the density function for the normal distribution with variance $\sigma^{2}$. (Naturally, one should confirm that this optimal function gives a maximum for $L$ but I won't do that here. Too much math. Just take my word for it)

Isn't that cool?!

Stochastic differential equations (SDEs) are used in many disciplines to model the behaviour of continuously changing quantities that are subject to random (unpredictable) forces, such as option prices, the electrical potential across the membranes of neurons, and the dynamics of microscopic particles.

A generic SDE process, $x_{t}$, can be modelled by the following equation: $$ dx_{t} = f(x_{t},t)\,dt + \sigma(x_{t},t)\,dW_{t}, \quad t \geq 0. $$ This equation describes the change in $x_{t}$ , denoted $dx_{t}$, over a miniscule time interval $dt \to 0^{+}$ as the sum of a deterministic component $f\,dt$ and a random component $\sigma\,dW_{t}$. The term $dW_{t}$ typically represents a Gaussian white noise forcing, which essentially means that, over time change $dt$, the trajectory of the system is perturbed by a Gaussian random variable with mean zero and variance $dt$. The effect of this perturbation is scaled by the quantity $\sigma$ which can be state-dependent.

My interest in these processes is driven by my interest in differential equations in general, and seeing what you can learn about time-varying processes in the presence of unpredictable factors.

In analogy to why the normal distribution is an "attracting distribution" for processes with finite second moments, there is a broader class of stochastic process that are the attractors for stochastic processes having non-finite second moments and non-finite first moments. These processes are called $\alpha$-stable processes, where the $0 < \alpha \leq 2$ refers to the stability index of the distribution (essentially the power-law behaviour of the tails of the distribution)

Analysis of these distributions tends to be much easier via their characteristic functions, since the associated distributions do not have neat closed form expressions in real space (except in a handful of cases). The characteristic function of a distribution is the Fourier transform of its density function and for a symmetric $\alpha$-stable distribution it is equal to $$ \psi(k) = \exp\left(-\sigma^\alpha |k|^{\alpha}\right). $$ It is interesting to note that the characteristic function of the normal distribution with variance $\nu^{2}$ is consistent with this form when $\alpha = 2$ and $\sigma = \frac{\nu}{\sqrt{2}}$: $$ \exp\left( -\frac{\nu^{2}}{2}k^{2}\right) $$ Indeed, the normal distribution is a special case of the $\alpha$-stable distribution and the most useful to statisticians dealing with finite variance problems. However, $\alpha$-stable processes have been shown to have use modelling the behaviour of option pricing and atmospheric processes subject to advection.