Laplace Distribution Calculator

Why the median falls out of the likelihood

The normal distribution makes the mean the best estimate of its centre. The Laplace makes the median. Both follow from the same one-line argument.

The log-likelihood is a sum of absolute deviations. The density has |x − μ| in the exponent, so maximising the likelihood over μ means minimising Σ|xᵢ − μ|.

The median is what minimises that sum. Exactly as the mean minimises Σ(xᵢ − μ)². Two famous estimators, one argument, two different error models.

The tool shows it rather than stating it. On the worked sample the log-likelihood at the median is −29.169 and at the mean −32.364. The median wins by 3.2 log units, and the verification suite confirms it wins on 400 of 400 random samples.

The scale estimate follows too: b is the mean absolute deviation from that median, which is the Laplace analogue of the standard deviation being the root mean squared deviation from the mean.

So “use the median, it is robust” has a precise version. The median is not a compromise or a safety measure here — it is the maximum-likelihood answer, given an explicit claim about how the errors are distributed.

Which is where least-absolute-deviations regression comes from

The same substitution, applied to a regression line rather than a single centre.

Ordinary least squares assumes normal errors. Minimising squared residuals IS maximising a normal likelihood, which is why OLS and the normal distribution always appear together.

Minimising absolute residuals assumes Laplace errors. Least-absolute-deviations regression is not a robustness patch bolted onto least squares; it is the maximum-likelihood fit under a different, explicit error model.

Which reframes the usual advice. “Use LAD when there are outliers” is really “use LAD when the errors have heavier tails than a normal” — a claim about the process, testable rather than tactical.

And it generalises to quantile regression. Weighting the absolute residuals asymmetrically fits a conditional quantile instead of a conditional median, which is the same machinery with one parameter added.

The cost is that there is no closed form. Least squares has normal equations; LAD needs linear programming, and the solution can be non-unique. That is the practical reason least squares dominates, not a statistical one.

Sharper peak, heavier tails — both at once

Matched on standard deviation, a Laplace is not simply “wider” than a normal. It is different in two directions simultaneously.

Inside about one standard deviation the Laplace tail is LIGHTER. The sharp peak concentrates probability near the centre, so there is less of it in the shoulders. The tool’s tail ratio plot starts below 1.

Past about two standard deviations it overtakes and keeps climbing. At four standard deviations the Laplace puts many times as much probability beyond the point as the normal does, and the ratio grows without limit.

Its excess kurtosis is exactly 3, against 0 for a normal — the standard summary of “peaked centre, heavy tails”, and one of the few distributions where the figure is a whole number.

The cusp at the centre is real. The density is not differentiable at μ, because it is built from an absolute value rather than a square. It is not a plotting artefact, and it is the visual signature of the whole family.

Which is why forecast errors are usually Laplace-shaped. Most forecasts are nearly right, a few are badly wrong, and almost none are moderately wrong — exactly the shape a normal cannot produce.

Where it comes from, and where it is used now

Three constructions and one modern application, all of which give the same distribution.

It is the difference of two independent exponentials. Which is why it appears wherever two independent waiting times are compared — arrival-time differences, clock offsets, timing jitter.

It is the prior behind L1 regularisation. Putting a Laplace prior on a regression coefficient and taking the maximum-a-posteriori estimate gives the lasso. The sharp peak at zero is what drives coefficients exactly to zero rather than merely shrinking them.

It is the noise mechanism in differential privacy. Adding Laplace noise scaled to a query’s sensitivity gives ε-differential privacy exactly, and the proof depends on the exponential decay: the ratio of densities at two nearby points is bounded, which a normal does not deliver.

And it predates the normal distribution’s dominance. Laplace proposed it as a law of errors before the least-squares tradition settled on the Gaussian, which is a reminder that the usual assumption was a choice.

Fitting one, and checking whether it belongs

The fit is the easiest in statistics. Deciding whether it is the right family takes more care.

Both estimates are one pass. Sort for the median, then average the absolute deviations from it. No iteration, no optimiser, and it is exactly the maximum-likelihood answer.

The diagnostic is the shape of the residuals, not their spread. A Laplace fit is indicated by a sharp central spike with occasional large values, and contraindicated by a smooth shoulder. A normality test rejecting is not by itself evidence for a Laplace — many alternatives reject too.

Compare the two log-likelihoods directly. Fit a normal and a Laplace to the same data and compare, penalising for parameters if the counts differ. Both have two parameters here, so the comparison is immediate.

And watch for a mixture pretending to be a Laplace. Two normals with different variances also give a peaked centre and heavy tails, and the distinction matters if you are going to extrapolate into the tail.

Reporting a Laplace fit

Four things, and the second is the one that trips readers up.

Name the parameters. Location and scale, not mean and standard deviation — even though the mean does exist here, because the scale b is NOT the standard deviation.

Say that the standard deviation is b√2. A scale of 1 gives a standard deviation of 1.414. Comparing a Laplace b directly against a normal sigma understates the spread by 29%.

Report the mean absolute deviation rather than the SD if you can. For this distribution it is exactly b, which makes it the natural summary and removes the conversion entirely.

And say why you chose the family. A Laplace fit is a claim that the errors have heavier tails than a normal, which changes every tail probability you go on to quote.

Sources and methodology

References for the Laplace distribution and absolute-error methods.

• NIST/SEMATECH e-Handbook, 1.3.6.6.12 — the double exponential distribution
• Dwork, McSherry, Nissim & Smith, “Calibrating Noise to Sensitivity in Private Data Analysis”, TCC 2006 — the Laplace mechanism
• Tibshirani, “Regression Shrinkage and Selection via the Lasso”, JRSS-B 58(1), 1996 — where the Laplace prior produces exact zeros
• Mean, Median and Mode Calculator — the estimator this distribution makes optimal
• Normal Distribution Calculator — the squared-error alternative
• How we calculate — methodology

Method. The fit is the exact maximum-likelihood solution rather than a numerical optimisation: the location is the sample median and the scale the mean absolute deviation from it. The suite verifies the underlying claim as a property rather than on one example — the median beats the mean on Laplace log-likelihood in 400 of 400 randomly generated heavy-tailed samples — and pins the worked case at −29.169226. The density is confirmed to integrate to its own CDF range by Simpson’s rule over a wide window, the CDF is checked monotone, and a single-value input correctly returns no fit. That engine is verified on every change against 75 assertions. The count and the per-case breakdown are published on the formula verification page.

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