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For a nonempty finite list of real numbers $x_1,\dots ,x_n$, the arithmetic mean isIt is the weighted mean with equal weights, giving each observation equal influence. For $(2,4,9)$, the sum is $15$ and the arithmetic mean is $5$. Replacing every value by $5$ retains the same total. The residuals $-3,-1,4$ sum to zero, so the mean is also the balancing point of these equally weighted observations.

The arithmetic mean minimizes the sum of squared deviations by the weighted mean minimizes squared deviations identity with all weights one. A large input can consequently have a substantial effect: replacing $9$ by $90$ changes the mean to $32$, while the median remains $4$. The choice of summary depends on what aspect of the data matters.

Under loss $w(\theta )|\theta -a|$, assume $0<\mathbb{E}[w(\theta )\mid x]<\infty$. The posterior expected loss is a positive constant times ordinary absolute-error loss under the probability density proportional to $w(\theta )\pi (\theta \mid x)$. Its minimizers are the medians of that weighted posterior: slopes of the loss change sign where the weighted probability below $a$ crosses one half. In particular, relative absolute loss has $w(\theta )=1/\theta$ for positive parameters. For a Gamma distribution posterior of shape two and rate $b$, the weighted posterior is an exponential distribution of rate $b$, so the estimate is $(\log 2)/b$.

A box plot compresses a statistical sample into a box spanning its lower and upper quartiles $Q_1,Q_3$, a line at its median $Q_2$, and two whiskers. It makes location and central spread easy to compare across groups on a common value axis. The box represents the central half in terms of quantile levels; with ties or interpolated quartiles, it need not contain exactly half the observations. Whisker endpoints require a stated convention: they can be the observed extremes, or they can stop short of separately marked observations. ^[ref]

Here is one explicitly toy dataset, reused under histogram: twelve invented, unitless values, with repeated values counted separately, already arranged as order statistics:For this box plot we choose the interpolated quartiles defined under quartile, using R's type-7 rule. This givesThe interquartile range is $I=Q_3-Q_1=3.5$. Our whisker convention uses the fencesThe whiskers end at the smallest and largest observations in the closed interval $[L,U]$: here $1$ and $8$. Values strictly outside the fences receive separate outlier markers: here only $20$. An observation exactly on a fence is retained within the whisker under this convention. The fences are thresholds, not observed whisker endpoints; the upper whisker is at $8$, not $11.5$. R's documented whisker rule likewise includes observations no more than $1.5$ box lengths away, although its box uses hinges, which can differ from our type-7 quartiles. ^[ref]

A separate outlier marker is a prompt to investigate, not a finding that the value is bad data and not an instruction to delete it. A distant observation can be a recording error, a legitimate unusual outcome, or evidence that the assumed statistical model misses some behavior. Check its provenance and context before deciding how to analyze it. No measurement-error claim is possible for our invented $20$: we deliberately put it there to demonstrate the rule. ^[ref]

The beauty of a box plot is its compact comparison of ordered positions rather than every observation. That compression also loses information: many different statistical samples have the same quartiles and whiskers, so the box alone cannot reveal every cluster or gap that a histogram might show. For example, replacing the two invented $6$ values by $5.5$ and $6$ leaves the type-7 quartiles, whiskers and separate marker unchanged. Displaying the individual values alongside the box plot is especially helpful for such a small statistical sample. Always state both the quartile rule and the whisker rule; changing either can change which values are marked.

A concentration inequality bounds the probability that a random variable differs substantially from a typical value such as its expected value or median.

Let $m$ be an integer median for the longest convex chain length of independent uniform points in a triangle. If $L_n(x)\ge a$, fix a certificate of $a$ coordinates. Every configuration with $L_n\le b$ must disagree with at least $a-b$ of those coordinates; equal weights $1/\sqrt{a}$ prove Talagrand convex distance separation at least $(a-b)/\sqrt{a}$. Talagrand's convex distance inequality then gives $\mathbb{P}(L_n\ge a)\mathbb{P}(L_n\le b)\le e^{-(a-b{)}^2/(4a)}$. Set $b=m$, then $a=m$, in the two respective applications to obtain the displayed tails for positive integer $t$. The exact chain-counting bound gives $m\le C{n}^{1/3}$. Hence any interval centred at $m$ with length $\omega (n)n^{1/6}$, where $\omega (n)\to \infty$, contains $L_n$ with high probability. No upper restriction on the growth of $\omega$ is required.

At a smooth point with nonzero gradient, rotate coordinates so $\nabla u=G{e}_2$. Write $u=u_0+Gy+\frac{1}{2}(A{x}^2+2Bxy+C{y}^2)+O(|(x,y){|}^3)$. A candidate median $u_0+d{h}^2$ has level boundary $y=(d{h}^2-A{x}^2/2)/G+O(h^3)$ in the disk. The area imbalance is $(2d-A/3)h^3/G+o(h^3)$; equal half-areas force $d=A/6$. Here $A=\Delta u-\nabla u^T(D^{2}u)\nabla u/|\nabla u{|}^2$ is the tangential second derivative, giving the formula. The disk uses uniform area measure, not its boundary measure.

Extreme value theory studies limiting probability distributions of normalized sample maxima and related extremes. The standard limits for maxima include the Fréchet distribution, negative Weibull distribution, and Gumbel distribution.

The question is often practical: if a batch finishes only when its slowest task finishes, its completion time is a sample maximum. A typical individual delay or a sample mean does not answer that question. The central limit theorem describes suitably normalized sums and sample means under its assumptions; it does not give a limit law for sample maxima. Extreme value theory instead describes how the largest observation changes as the number of opportunities for a large observation grows.

For independent and identically distributed random variables $X_1,X_2,\dots$ with distribution function $F$, define the sample maximumThe event $M_n\le x$ means that every observation is at most $x$, so independence givesExtreme value theory asks whether there are constants $a_n>0$ and $c_n\in \mathbb{R}$ for which $(M_n-c_n)/a_n$ has convergence in distribution to a non-degenerate probability distribution. Equivalently, $F(c_n+a_{n}z{)}^n$ must tend to a limiting distribution function at its continuity points. The extremal types theorem says that any such limit belongs, up to a change of location and positive scale, to the Gumbel distribution, Fréchet distribution, or negative Weibull distribution. It does not guarantee that a limit exists for every $F$. ^[ref].

The maximum of independent exponential random variables gives a particularly transparent worked example: its center grows as $(\log n)/\lambda$, and its fluctuations have a Gumbel distribution limit on the scale $1/\lambda$. Here and below $\log$ denotes the natural logarithm. This normalization depends on the exponential distribution. Other survival functions can lead to the Fréchet distribution or negative Weibull distribution, with different constants $a_n,c_n$; the Gumbel distribution and the exponential example's normalization are not universal.

Independence is essential to the product calculation. If instead $X_1=X_2=\cdots =X$ for one rate-$\lambda$ exponential random variable, then $M_n=X$ for every $n$. Its distribution function stays $1-e^{-\lambda x}$ for $x\ge 0$ and zero below zero, and its median stays $(\log 2)/\lambda$. Perfect dependence creates no logarithmic growth: repeating the same random variable supplies no new chance of a larger value.

A historical starting point is R. A. Fisher and L. H. C. Tippett's 1928 paper on limiting laws of the largest or smallest sample member, published in volume 24, issue 2, pages 180–190 of the Proceedings of the Cambridge Philosophical Society. Its abstract studies the possible limiting forms and emphasizes that approach to a limit can be slow. ^[ref].

Consider the population statistical model $F_{\epsilon ,x}=(1-\epsilon )N(0,1)+\epsilon {\delta }_x$. The following are deterministic statistical functional calculations, not observed contamination experiments. Compare the actual location bias with the linear approximation $\epsilon \mathrm{IF}(x;T,F)$. The mean bias is exactly $\epsilon x$. For positive $x$ exceeding the shifted median and $0\le \epsilon <1/2$, the generalized-inverse median bias is exactlyThis formula applies to both $x=1$ and $x=8$ throughout the plotted interval $0\le \epsilon \le 0.1$. If the contaminating atom lies inside the quantile jump containing probability $1/2$, the generalized-inverse median instead equals that atom.

For the Huber location estimator with $s=1$ and $k=1.345$, solve its actual population equation, rather than substituting the influence function:Writing $a=t-k$ and $b=t+k$, integration of the normal distribution giveswhere $\varphi$ is the probability density function of $N(0,1)$. Its derivative is $-[\Phi (t+k)-\Phi (t-k)]<0$, so the mixture score is strictly decreasing for $\epsilon <1$. For $\epsilon <1/2$, its limits at negative and positive infinity have opposite signs even with arbitrary distant point contamination. The displayed examples use the unique root, evaluated by the bisection method to numerical precision.

At $x=8$ and $\epsilon =0.01$, the exact mean, median, and Huber location estimator biases are $0.080000$, $0.012660$, and $0.016541$; the influence function predictions are $0.080000$, $0.012533$, and $0.016375$. At $\epsilon =0.05$, they become $0.400000$, $0.066012$, and $0.086241$, versus predictions $0.400000$, $0.062666$, and $0.081875$. The bounded alternatives control the effect of the distant atom, but their actual response curves bend away from the tangent.

The nearby atom $x=1$ gives a useful comparison: at $\epsilon =0.05$, the Huber location estimator bias is $0.060237$, slightly below its linear approximation $0.060874$, whereas the median bias is still $0.066012$. The direction and size of approximation error depend on the contaminating location. Even accurate local influence calculations must specify $F$, the scale, the cutoff, and the actual contamination fraction.

Take $N_0$ independently decaying atomic nuclei, each with lifetime $T_i$ drawn from an exponential distribution of rate $\lambda$. With no replenishment, the survivor count isEach indicator variable has success probability $p(t)$, so the binomial distribution givesThe smooth population law is the expected value $\overline{N}(t)=\mathbb{E}[N(t)]$, satisfying ${\overline{N}}^{\prime }=-\lambda \overline{N}$. One realization instead loses whole atomic nuclei at discrete times. An observed count above the smooth curve need not contradict the mathematical model.

For an original chosen example, let $N_0=1000$ and half-life $3\mathrm{h}$. Then $\lambda =\log 2/(3\mathrm{h})$. After $9\mathrm{h}$, the survival probability is $1/8$, givingThe standard deviation describes variation between repeated populations governed by these assumptions. It is not an uncertainty interval for a noiseless deterministic curve.

The half-life and the mean lifetime answer different questions:The first is the median of one lifetime, or the time when half the population survives on average. The second is the lifetime's expected value. In this example they are $3\mathrm{h}$ and approximately $4.33\mathrm{h}$.

Here is an original reproducible finite set with 40 points and 512 coordinates. Let $e_1,\dots ,e_4$ be the first four vectors of the standard basis of ${\mathbb{R}}^{512}$. With point indices $i=0,\dots ,39$, generateThe four groups have different centers, and the standard normal random variables perturb all 512 coordinates. Generate the coordinates in row order using NumPy's PCG64 pseudorandom number generator with seed 20261013. Then keep this finite set fixed. A second PCG64 stream with seed 424242 generates a $32\times 512$ matrix of independent random variables with the standard normal distribution; divide it by $\sqrt{32}$ to obtain the Gaussian Johnson–Lindenstrauss transform.

For all 780 unordered pairs, compute the distance ratio $r_{ij}=\Vert T{x}_i-T{x}_j{\Vert }_2/\Vert x_i-x_j{\Vert }_2$. In this draw, its minimum is approximately $0.733010$ and its maximum $1.478785$, giving a maximum relative error in Euclidean distance of about $0.478785$. The squared-distance ratios $r_{ij}^2$ instead range from approximately $0.537304$ to $2.186806$: their maximum relative error is about $1.186806$. Thus this particular draw fails a 20% distance tolerance. The Johnson–Lindenstrauss lemma does not assert that every draw into 32 dimensions must succeed.

The two bands of original Euclidean distances distinguish pairs within a group from pairs between groups. Their spread after the linear map shows why checking an average distance is insufficient: the Johnson–Lindenstrauss lemma controls every pair. The dashed 20% distance band is an illustrative tolerance, not a confidence interval or a squared-distance band. The affine hull of 40 points has dimension at most 39; 512 is the ambient coordinate count, and no full-dimensionality assumption is required. For this draw the bi-Lipschitz distortion is approximately $1.478785/0.733010=2.017414$.

Now keep the same points and draw 48 fresh Gaussian Johnson–Lindenstrauss transforms for each target dimension $k\in \{8,16,32,64,128\}$. NumPy's SeedSequence(314159).spawn(240) supplies distinct PCG64 streams, ordered first by target dimension, then by draw. Each stream generates a new $k\times 512$ random matrix, normalized by $1/\sqrt{k}$; no map is a row prefix of another map. Each plotted value is

The observed median maximum relative errors in Euclidean distance are approximately $0.735$, $0.520$, $0.381$, $0.267$, and $0.184$, respectively. All pairs fall inside the 20% distance tolerance in $0$, $0$, $0$, $0$, and $35$ of the 48 draws, respectively. These are computations on one synthetic finite set, rather than optimal dimension bounds or measured physical data. The displayed quantiles describe the sampled maps; they do not certify a success probability for other sets. Even at 128 dimensions, 13 of the sampled maps fail this tolerance.

The practical benefit is that each point acquires fewer coordinates while its pairwise Euclidean distances can remain useful. The choice of target dimension balances compression, accuracy, and failure probability. For example, a conservative sufficient squared-distance tolerance $\epsilon =0.2$ with $n=40$ and $\delta =0.05$ gives $k\ge 1195$ by the explicit bound above. Here that exceeds the ambient dimension, so the identity map is better for a certified guarantee. The much smaller dimensions in the plots illustrate observed behavior without claiming that this sufficient bound is sharp.

Francis Galton's paper Regression towards mediocrity in hereditary stature by Francis Galton analyzes adult offspring heights against the average of their parents' adjusted heights. He multiplied female heights by $1.08$ before making that parental average. Table I contains 928 adult offspring from 205 parental pairs; the opening prose instead states 930. Its grouped readout is a median, and the explanation of Plate IX describes a smoothed offspring deviation about two-thirds of the parental deviation.^[ref]

These were parent-offspring records, not repeat readings of an unchanged signal. The paper's group tendency does not say that every child must be closer to the population mean. Galton also reports uncertainty over whether some recorded heights included shoes, a concrete measurement error limitation.^[ref] The $0.6$ Pearson correlation coefficient and all generated values above are our chosen model, not a reconstruction of his data.

For a positive response, the model $\log Y=\eta +\epsilon$ with $\epsilon \sim N(0,{\sigma }^2)$ gives conditional median $e^{\eta }$ and conditional mean $e^{\eta +{\sigma }^2/2}$. The latter follows from the moment-generating function of a normal distribution. Additive factor effects in $\eta$ become multiplicative effects on these response summaries. Its constant log-scale variance corresponds to original-scale variance proportional to the square of the mean; direct exponentiation of fitted log means incurs retransformation bias.

The standard Gumbel distribution for maxima has distribution function $\Lambda (z)=e^{-e^{-z}}$ for every real $z$. Its probability density function is the derivative of its distribution function:This article uses the maxima convention throughout. Reflecting a standard Gumbel distribution random variable $Z$ to $-Z$ gives the minima convention, with distribution function $1-\exp (-e^z)$. The two conventions have opposite directions of asymmetry, so their formulas must not be interchanged. ^[ref].

For a location $\mu \in \mathbb{R}$ and positive scale $b>0$, let $Y=\mu +bZ$. This location-scale family has cumulative distribution function and probability density functionSolving $\Lambda (z)=p$ gives the standard quantile function $q(p)=-\log (-\log p)$, $0<p<1$, and hence the standard medianThe median of $Y$ is therefore $\mu -b\log (\log 2)$. Location $\mu$ is not the median: at $x=\mu$ the cumulative distribution function is $e^{-1}\approx 0.367879$.

To derive the Gumbel distribution as an extreme-value limit, take the maximum of independent exponential random variables with rate $\lambda >0$ and setThe exact distribution function isThe finite-$n$ lower boundary is required because $M_n\ge 0$. For each fixed real $z$, eventually $z>-\log n$, and the natural logarithm of the first branch satisfiesExponentiating proves convergence in distribution to the standard Gumbel distribution. It is the familiar limit $(1-u/n{)}^n\to e^{-u}$ with $u=e^{-z}$: a simple calculation reveals a whole limiting shape. The name honors Emil Gumbel; Siegrist gives this exponential-maxima derivation and the standard Gumbel distribution formulas. ^[ref].

For the toy delay model $\lambda =1{\mathrm{s}}^{-1}$, $n=100$, the limiting Gumbel distribution approximation to $M_{100}$ has $\mu =(\log 100)\mathrm{s}$ and $b=1\mathrm{s}$. Its predicted median iscompared with the exact median $-\log (1-2^{-1/100})\mathrm{s}\approx 4.975147\mathrm{s}$. The approximation is about $0.003464$ seconds lower in this example. It is not an exact finite-sample probability distribution: the approximating Gumbel distribution assigns positive probability to negative delays, while the exact sample maximum cannot be negative.

The maximum domain of attraction of the Gumbel distribution is characterized by an exponential tail ratio on the scale of a Gumbel auxiliary function. This allows many original probability distributions to share the limiting shape, but it does not make the centering $\log n$ appropriate to all of them.

The Huber loss is quadratic near zero and linear beyond a fixed cutoff; its derivative is the clipped Huber score.

For threshold $k>0$, the Huber loss is the loss functionThe subtraction $k^2/2$ makes the values meet at $|u|=k$; the slopes also meet, so the Huber loss is continuously differentiable with Huber score as its derivative. It is a convex function, since that derivative is a monotone function. These quadratic and linear pieces define the original clipped location criterion.^[ref]

For $k=2$, ${\rho }_2(1)=1/2$, ${\rho }_2(2)=2$, and ${\rho }_2(5)=8$, whereas the squared criterion $u^2/2$ at $u=5$ is $12.5$. Increasing a standardized difference from $5$ to $6$ raises the Huber loss by $2$, but raises the squared criterion by $5.5$. The compromise preserves quadratic sensitivity to small differences while limiting the slope for large ones. This makes the Huber loss useful when ordinary observations should retain their detailed distances yet isolated outliers should exert limited pull.

For a fixed finite dataset and scale, taking $k$ large enough that every difference from the sample mean lies within $ks$ recovers that sample mean. Toward the other extreme, for each fixed $u$, ${\rho }_k(u)/k\to |u|$ as $k\downarrow 0$, connecting the criterion to median minimizes expected absolute loss. The normalization is essential: putting $k=0$ into the unscaled Huber loss gives zero everywhere, not a usable median criterion.

For the longest increasing subsequence length $L$ on a product space of independent coordinates, a witness of length $\ell$ uses only $\ell$ coordinates. Any point with subsequence length at most $a$ must differ on at least $\ell -a$ witness positions. Weights $1/\sqrt{\ell }$ on those positions show $d_T(x,\{L\le a\})\ge (\ell -a)/\sqrt{\ell }$. Thus for $0\le a<b$,At a median $m$, this gives upper tail $2e^{-t^2/[4(m+t)]}$ and lower tail $2e^{-t^2/(4m)}$. Integrating the tails places the mean within $O(\sqrt{m}+1)$ of the median, so the natural concentration scale near the mean is $\sqrt{\mathbb{E}L}$. A random permutation is represented by the ranks of independent continuous variables; arbitrary dependent sequences need not obey this conclusion.

An influence function asks how much a population answer moves when an infinitesimal fraction of its probability distribution is relocated to a specified point. It distinguishes the unbounded response of the mean from the bounded response of the median and the Huber location estimator. The normal-location influence curves and finite contamination bias of normal location below make that difference quantitative.

Let $T$ be a real-valued statistical functional, $F$ a reference probability measure, and ${\delta }_x$ the Dirac measure concentrated at $x$. Along the epsilon-contamination neighborhood path $F_{\epsilon ,x}=(1-\epsilon )F+\epsilon {\delta }_x$, define the right derivativeThe name influence curve refers to the same object. The derivative is in the signed direction ${\delta }_x-F$, so the calculation removes baseline mass as well as adding a Dirac measure. The definition is a limit for each fixed $x$; it does not automatically give a remainder uniform over arbitrarily distant $x$.^[ref]

The statistical functional must be defined on these contaminated probability measures, and the right derivative must exist and be finite. For a mean, a finite first absolute moment suffices. For a median, require a unique baseline median and a probability density function continuous and strictly positive there. For a smooth M-estimator, require an isolated population root, justified differentiation of its expected value, and a nonzero derivative with respect to the fitted parameter. These are specific conditions for the formulas below. Existence of the pointwise influence function alone proves neither an asymptotic linear representation nor pathwise differentiability of a statistical functional; each needs control of its own remainder and admissible statistical paths.

For the statistical functional $T(F)=\mu ={\mathbb{E}}_{F}X$, assuming ${\mathbb{E}}_F|X|<\infty$, the contaminated mean is exactlyThus the influence function of the mean is $x-\mu$, and its gross-error sensitivity is infinite on an unbounded sample space. The linear approximation is exact here: any fixed positive contamination fraction allows an arbitrarily large change by moving $x$ far enough. No finite second moment is needed for this derivative, although a finite variance is needed for the usual central limit theorem for the sample mean.

For comparison, the influence function of the sample median iswhen the median $m$ is unique and the probability density function $f$ is positive and continuous at $m$. The sign function records which side of $m$ receives the contaminating mass. The influence function of a quantile supplies the short derivation. With the generalized-inverse quantile convention, contamination exactly at $x=m$ leaves the median at $m$, giving derivative zero there. If $f(m)$ is very small, this bounded influence function can still be large; if $f(m)=0$, the displayed formula is unavailable.

The jackknife variance estimator estimates the sampling variance of the full-data estimator, using the deletion estimates and their mean from jackknife resampling:The square root is the estimated standard error. The factor $n-1$ multiplies the divisor-$n$ variance of the deletion estimates: those estimates vary less than estimates from new independent experiments because almost all their observations are shared. Ryan Tibshirani's ^[ref] give the same scaling in terms of the divisor-$(n-1)$ sample variance of the deletion estimates.

For the sample mean, this formula has an exact algebraic simplification. With $s_n^2$ denoting the usual sample variance,and thereforeUnder the independent and identically distributed assumption and a finite second moment, $s_n^2$ is an unbiased estimator of the population variance ${\sigma }^2$. Consequently ${\hat{V}}_J$ is an unbiased estimator of the sample mean's sampling variance ${\sigma }^2/n$. The equality ${\hat{V}}_J=s_n^2/n$ itself holds for any observed data with $n>1$; its sampling interpretation uses the stated assumptions.

For $(1,2,3,4,10)$, the sum of squared deviations from the sample mean is $50$, giving sample variance $s_5^2=50/4=12.5$. The deletion sample means have squared deviations totaling $3.125$, soHere $12.5$ estimates the variance of individual observations, while $2.5$ estimates the sampling variance of their sample mean. These answer different questions.

For more general statistics, validity requires regularity beyond the deletion formula, such as a sufficiently smooth function of sample means with suitable moments. Ordinary delete-one jackknife resampling can fail for nonsmooth quantiles, including the median; see Larry Wasserman's ^[ref]. With statistical dependence, deleting individual observations need not reproduce the relevant sampling variance; the deletion scheme must reflect that statistical dependence. Shalizi's ^[ref] distinguish the basic independent and identically distributed case from extensions.

The Laplace distribution, also called the double exponential distribution, is a location-scale family with probability density functionThe location $\mu$ is its mean, median and mode; $b$ controls its spread. In the centred rate convention, the same law has $\mu =0$ and $\theta =1/b>0$, giving $f(x)=\theta e^{-\theta |x|}/2$. Keeping track of whether a parameter is a scale or its reciprocal prevents reversing the effect of increasing it. ^[ref]

The two halves of the probability density function are reflected exponential decay curves, each with total probability $1/2$. Their meeting point has a sharp corner: the left and right derivatives are $1/(2b^2)$ and $-1/(2b^2)$. This finite peak is not a positive probability mass at $\mu$. Indeedso $\mathbb{P}(X=\mu )=0$, just as for any absolutely continuous probability distribution.

The figure compares a Laplace distribution with $\mu =0$, $b=1$ against a normal distribution with mean zero and standard deviation $\sigma =\sqrt{2}$: both have variance 2. The right panel uses a logarithmic scale for the two-sided tail probability $\mathbb{P}(|X|>r)$. Its straight Laplace distribution curve comes from $e^{-r}$; the normal distribution tail falls faster. Thus values far from the center can be more plausible under a Laplace distribution even after matching the variance. These are theoretical curves, not measured frequencies. The tails still have exponential decay; this is not a heavy-tailed distribution in the convention requiring every positive exponential moment to diverge.

Integrating either half of the probability density function gives the cumulative distribution functionIn particular $F(\mu )=1/2$. The basic moments areThus the scale is the expected value of the absolute distance from the center, and is not the standard deviation. ^[ref] A short way to obtain the moments is to note that $R=|X-\mu |$ has an exponential distribution with rate $1/b$: $\mathbb{P}(R>r)=e^{-r/b}$ for $r\ge 0$. Integrating gives $\mathbb{E}R=b$ and $\mathbb{E}{R}^2=2b^2$; the symmetric distribution gives the centered first moment zero.

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.