The Ten Distributions You Need, and the Geometry of the Multivariate Gaussian

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TL;DR: Pick a distribution by asking what the data is: a yes/no outcome is Bernoulli, a count of independent rare events is Poisson, a waiting time is Exponential, a probability is Beta, a probability vector is Dirichlet. The multivariate Gaussian is the one worth knowing in depth — its covariance matrix eigendecomposes into the axes and widths of its elliptical level sets, and its marginals and conditionals stay Gaussian, which is why it is everywhere.

The reference table

$\operatorname{Var}$ is the variance of a single draw; parameters are on the left.

DistributionModelsParametersMeanVariance
Bernoulli$(p)$one binary trial$p\in[0,1]$$p$$p(1-p)$
Binomial$(n,p)$successes in $n$ trials$n\in\mathbb{N},\,p$$np$$np(1-p)$
Categorical$(\boldsymbol{\pi})$one draw from $K$ classes$\boldsymbol{\pi}$, $\sum_k\pi_k=1$$\boldsymbol{\pi}$$\operatorname{diag}(\boldsymbol{\pi})-\boldsymbol{\pi}\boldsymbol{\pi}^{\top}$
Poisson$(\lambda)$counts of rare independent events$\lambda>0$$\lambda$$\lambda$
Uniform$(a,b)$no preference on an interval$a \lt b$$\frac{a+b}{2}$$\frac{(b-a)^2}{12}$
Gaussian$(\mu,\sigma^2)$sums of many small effects$\mu,\ \sigma^2>0$$\mu$$\sigma^2$
Exponential$(\lambda)$memoryless waiting time$\lambda>0$$1/\lambda$$1/\lambda^2$
Beta$(\alpha,\beta)$an unknown probability$\alpha,\beta>0$$\frac{\alpha}{\alpha+\beta}$$\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}$
Dirichlet$(\boldsymbol{\alpha})$an unknown probability vector$\alpha_k>0$, $\alpha_0=\sum_k\alpha_k$$\alpha_k/\alpha_0$$\frac{\alpha_k(\alpha_0-\alpha_k)}{\alpha_0^2(\alpha_0+1)}$
Laplace$(\mu,b)$sharply peaked, heavy-tailed noise$\mu,\ b>0$$\mu$$2b^2$

Three relationships tie the table together. Binomial is a sum of independent Bernoullis. Poisson is the limit of Binomial$(n,\lambda/n)$ as $n\to\infty$, which is why it models rare events among many opportunities. Beta is the one-dimensional case of Dirichlet.

The three that matter most in ML

Categorical. Every classifier’s output layer. A softmax over $K$ logits is a categorical parameter vector, and the training loss is the negative log-likelihood of the observed class under it. Its covariance $\operatorname{diag}(\boldsymbol{\pi})-\boldsymbol{\pi}\boldsymbol{\pi}^{\top}$ is exactly the Jacobian of the softmax, which is why the softmax–cross-entropy gradient collapses to $\hat{\mathbf{y}} - \mathbf{y}$.

Beta and Dirichlet. These are the conjugate priors for Bernoulli and Categorical: start with Beta$(\alpha,\beta)$, observe $s$ successes and $f$ failures, and the posterior is Beta$(\alpha+s,\beta+f)$. The parameters act as pseudo-counts, which is the cleanest way to read Laplace smoothing — adding 1 to every count is a Dirichlet$(\mathbf{1})$ prior. Dirichlet is also what latent Dirichlet allocation puts over topic proportions.

Laplace. Its log-density is $-x-\mu/b$ up to a constant, so a Laplace prior on weights gives an $\ell_1$ penalty and a Gaussian prior gives $\ell_2$. Lasso versus ridge is a choice of noise model, not just a choice of penalty.

The multivariate Gaussian

For $\mathbf{x}\in\mathbb{R}^d$ with mean $\boldsymbol{\mu}$ and positive-definite covariance $\boldsymbol{\Sigma}$:

\[ p(\mathbf{x}) = (2\pi)^{-d/2}\,|\boldsymbol{\Sigma}|^{-1/2}\exp\!\left(-\tfrac12 (\mathbf{x}-\boldsymbol{\mu})^{\top}\boldsymbol{\Sigma}^{-1}(\mathbf{x}-\boldsymbol{\mu})\right). \]

Everything about the shape is in the quadratic form. The quantity $\Delta^2 = (\mathbf{x}-\boldsymbol{\mu})^{\top}\boldsymbol{\Sigma}^{-1}(\mathbf{x}-\boldsymbol{\mu})$ is the squared Mahalanobis distance, and the density is constant wherever $\Delta$ is constant — so the level sets are ellipsoids.

Diagonalise, $\boldsymbol{\Sigma} = \mathbf{U}\boldsymbol{\Lambda}\mathbf{U}^{\top}$ with orthonormal eigenvectors \(\mathbf{u}_i\) and eigenvalues $\lambda_i > 0$. The ellipsoid’s axes point along the \(\mathbf{u}_i\), and its semi-axis in direction $i$ has length proportional to $\sqrt{\lambda_i}$. Take

\[ \boldsymbol{\Sigma} = \begin{pmatrix} 1 & 0.8 \\ 0.8 & 1\end{pmatrix} \ \Rightarrow\ \lambda_1 = 1.8 \ \text{along}\ \tfrac{1}{\sqrt2}(1,1), \quad \lambda_2 = 0.2 \ \text{along}\ \tfrac{1}{\sqrt2}(1,-1). \]

The contours are ellipses tilted at 45°, elongated by $\sqrt{1.8/0.2} = 3$ — three times wider along the diagonal than across it. Strong positive correlation is that elongation.

Level sets of a correlated bivariate GaussianThree concentric ellipses centred at the origin, tilted 45 degrees, for a covariance matrix with ones on the diagonal and 0.8 off-diagonal. The long axis lies along the direction (1,1) with eigenvalue 1.8 and the short axis along (1,-1) with eigenvalue 0.2, so the ellipse is three times longer than it is wide. u₁, λ₁ = 1.8u₂, λ₂ = 0.2 x₁x₂ Σ = [[1, 0.8],[0.8, 1]] axis ratio 3
Notice that the ellipse is not aligned with the coordinate axes: the marginal variances are both 1, yet almost all the probability mass lies near the \(x_1 = x_2\) diagonal. Correlation is a rotation of the level sets, not a change in their marginal widths.

Two structural facts make the family tractable. Marginals and conditionals of a Gaussian are Gaussian, with the conditional mean \(\boldsymbol{\mu}_1 + \boldsymbol{\Sigma}_{12}\boldsymbol{\Sigma}_{22}^{-1}(\mathbf{x}_2-\boldsymbol{\mu}_2)\) — linear in what you conditioned on, which is where linear regression and Kalman filters come from. And any affine map of a Gaussian is Gaussian: with $\boldsymbol{\Sigma}=\mathbf{L}\mathbf{L}^{\top}$, sampling is $\mathbf{x} = \boldsymbol{\mu}+\mathbf{L}\boldsymbol{\epsilon}$ with $\boldsymbol{\epsilon}\sim\mathcal{N}(\mathbf{0},\mathbf{I})$ — the reparameterisation trick that makes VAEs differentiable.

Interview trap — "Gaussian marginals plus zero correlation means independent": only if the pair is jointly Gaussian. Let \(X\sim\mathcal{N}(0,1)\) and let \(S=\pm1\) be an independent fair sign; put \(Y = SX\). Then \(Y\sim\mathcal{N}(0,1)\) too, and \(\operatorname{Cov}(X,Y)=\mathbb{E}[S]\,\mathbb{E}[X^2]=0\). But \(|Y|=|X|\) always, so they are strongly dependent — the joint distribution is not Gaussian, it is mass on two lines. Diagonal covariance implies independence only inside the joint-Gaussian family.
Key Insight — the Gaussian is the maximum-entropy distribution for a fixed covariance: among all densities on \(\mathbb{R}^d\) with covariance \(\boldsymbol{\Sigma}\), the Gaussian has the largest entropy. So assuming Gaussian noise is the least committal assumption you can make once you have decided on second moments — a real justification rather than convenience, and one that fails the moment the tails matter.

Where this goes next

Why sums of almost anything drift towards the Gaussian is limit theorems; fitting these families to data is the subject of statistics basics.

Recap

  • Choose by data type: binary → Bernoulli, counts → Poisson, waiting times → Exponential, probabilities → Beta/Dirichlet.
  • Poisson has mean equal to variance, both \(\lambda\); Binomial\((n,\lambda/n)\) converges to it.
  • Beta and Dirichlet are conjugate to Bernoulli and Categorical, with parameters acting as pseudo-counts.
  • Gaussian level sets are ellipsoids with axes \(\mathbf{u}_i\) and widths \(\propto\sqrt{\lambda_i}\); \(\rho=0.8\) gives an axis ratio of 3.
  • Zero correlation implies independence only under joint Gaussianity.

References

  1. Bishop, C. M. Pattern Recognition and Machine Learning, ch. 2. Springer, 2006.
  2. Murphy, K. P. Probabilistic Machine Learning: An Introduction. MIT Press, 2022.
  3. Kingma, D. P. & Welling, M. Auto-Encoding Variational Bayes. ICLR 2014.
  4. Blei, D. M., Ng, A. Y. & Jordan, M. I. Latent Dirichlet Allocation. JMLR 3, 993–1022, 2003.