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For linear regression with fixed predictor vectors $X_i$, assume conditionally independent responses with a common scale $b>0$:Each conditional median and mean is $X_i^T\beta$. The log-likelihood function is $-n\log (2b)-S(\beta )/b$, so maximum likelihood estimation minimizes the sum of absolute regression residuals:This is least absolute deviations. With only an intercept, it reduces to the Laplace location and scale maximum likelihood problem, so a sample median is a fitted intercept. Provided the minimum $S(\hat{\beta })>0$, the scale estimate is $\hat{b}=S(\hat{\beta })/n$; the conditional standard deviation is $\sqrt{2}b$. If the scale is denoted $\sigma$ instead of $b$, the conditional standard deviation is $\sqrt{2}\sigma$.

Minimizing a sum of absolute losses does not automatically give a unique coefficient vector. Even an intercept-only fit to an even sample size can have an interval of solutions, and redundant predictor columns can introduce further nonuniqueness. If a fit makes every regression residual zero, the joint likelihood function is unbounded as $b\downarrow 0$, so there is no finite joint maximum-likelihood estimate with positive scale.

The linear growth of absolute loss limits sensitivity to an extreme response at fixed predictors compared with squared-error loss. It does not protect against every outlier: an extreme predictor can have large regression leverage. The numerical example above changes a response in an intercept-only problem, and makes no general claim of immunity to such predictor changes.

A finite contour based at an integrable branch point, or a closed contour around an integer-order pole, gives a solution analytic at $z=0$. It is therefore a constant multiple of the unique Frobenius power-series solution there.

Let $f_r(m)$ be the lower-shadow size prescribed by the binomial representation in the Kruskal-Katona theorem. ThenInduction using Pascal's identity proves the inequality and drives the standard ground-set induction for the theorem.

Every intuitionistically underivable proposition has a finite rooted Kripke countermodel. Filtration and witness pruning bound the model in terms of the number of subformulas.

The unravelling of a rooted Kripke model has finite increasing paths from the root as worlds, ordered by extension and labelled by their endpoints. It is tree-like and preserves forcing.

Filtration identifies worlds that force the same formulas in a fixed finite subformula-closed set. With order induced by inclusion of these finite theories, the quotient preserves forcing of every retained formula and has at most $2^n$ worlds for $n$ formulas.

A proposition is derivable in intuitionistic propositional logic exactly when it is forced at every world of every intuitionistic Kripke model.

If a decipherable binary code has $N$ codewords of lengths $l_1,\dots ,l_N$, thenThis follows by applying the entropy lower bound with the uniform source distribution.

Let $X_1,\dots ,X_n$ be real independent and identically distributed random variables with unknown probability density function $f$, let $n\ge 1$, and choose a nonnegative, unit-integral kernel for density estimation $K$ and a smoothing bandwidth $h>0$. The kernel density estimator isPlace a bump at every observation and add their contributions. Each bump $K_h(x-X_i)/n$ has area $1/n$, so for every realized statistical sampleThe change of variables is $u=(x-X_i)/h$. Summing the unweighted bumps $K_h(x-X_i)$ would give area $n$; summing just $K((x-X_i)/h)$ would give area $nh$. Those raw sums do not have the normalization of a probability density function.

Unlike a histogram, the construction does not require choosing bin origins, and a Gaussian density kernel gives a smooth curve. Its appeal is that identical local contributions can combine into a probability density function with several peaks without first choosing their number. This makes kernel density estimation useful for exploring an unknown probability distribution and estimating interval probabilities, while the smoothing bandwidth controls how much detail is visible.

Before the statistical sample is observed, ${\hat{f}}_{n,h,K}$ is a random function; afterwards it is a fitted probability density function. It is an estimator of the unknown $f$, not the true probability distribution. For fixed $K,h$, its expectation is the convolution $K_h\ast f$, generally different from $f$, and repeating the sampling changes the fitted curve. The exact mean integrated squared error of a kernel density estimator separates this smoothing bias of an estimator from sampling variance. Allowing signed kernels of order ell retains the formula but can lose nonnegativity.

Every Boolean function of degree at most $k$ is a $k{2}^{k-1}$-junta.

If $f:\{-1,1{\}}^n\to \{-1,1\}$ satisfies $\Vert f^{\le k}{\Vert }_2^2\ge 1-\epsilon$, then it has a real-valued $J$-junta approximation $g$ with $\Vert f-g{\Vert }_2^2\le 2\epsilon$ and

If the jump chain visits a state of finite holding rate infinitely often, the sum of its independent positive holding times at that state diverges almost surely, preventing explosion.

For holding rates $q_i$, a measure $\pi$ satisfies $\pi Q=0$ exactly when ${\mu }_i=q_i{\pi }_i$ is invariant for the jump chain. Thus ${\pi }_i={\mu }_i/q_i$ when normalization is possible.

On $\mathbb{N}$, let the chain jump from every $i\ge 1$ to $i+1$ with probability $p$ and to $i-1$ with probability $1-p$, and jump from zero to one. It is recurrent exactly when $p\le 1/2$; at $p=1/2$ it is null recurrent.

A nilpotent Jordan block has zeros on its diagonal and ones on its superdiagonal. Its $k$th power has ones on the $k$th superdiagonal, and a block of size $d$ satisfies $J_d(0{)}^d=0$.

For exobase number density $n_e$ and $v_{\mathrm{th}}=\sqrt{2k_{B}T/m}$, the collisionless Jeans number flux is

The Jacobian determinant is the determinant of the square Jacobian matrix. Its absolute value is the local volume-scaling factor in the change of variables formula.

The Jacobian matrix and determinant describe respectively the linear derivative and its signed volume factor.

If $A=D+L+L^T$ is symmetric positive definite and tridiagonal, set $S=\mathrm{diag}(1,-1,1,-1,\dots )$. Thenis positive definite. Applying the Householder-John theorem to $M=D$ and $N=-(L+L^T)$ proves that the Jacobi iteration converges.

For Jacobi coordinates $\mathbf{a},\mathbf{b},\mathbf{c}$ and their corresponding reduced masses $\alpha ,\beta ,\gamma$, total angular momentum is