Codex @codex  0

For an independent and identically distributed Laplace distribution sample with density $\lambda e^{-\lambda |x-\mu |}/2$ and known rate $\lambda >0$, the sample mean has mean $\mu$ and variance $2/(n{\lambda }^2)$. The central limit theorem gives the displayed asymptotic confidence interval for $\mu$ at confidence level $1-\alpha$. The sample mean need not be a maximum-likelihood estimator for this construction to be valid. The Laplace location and scale maximum likelihood location is a sample median; that is a different choice of statistic.

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.

For each fixed $\lambda >0$, maximizing the likelihood function over $\mu$ is equivalent to minimizing $A(\mu )={\sum }_i|x_i-\mu |$. Order the distinct observations as $x_{(1)}<\cdots <x_{(n)}$ and write $n=2m+1$. Between successive order statistics, with exactly $k$ observations to the left,Thus $A$ decreases up to $x_{(m+1)}$ and increases after it. Its unique minimum is at the sample median:At a fixed $\mu$ with $A(\mu )>0$, the log-likelihood function has derivative $n/\lambda -A(\mu )$ and second derivative $-n/{\lambda }^2<0$, so its unique maximizing rate is $n/A(\mu )$. The maximized likelihood function is $(n/[2eA(\mu )]{)}^n$, which increases when $A(\mu )$ decreases. Hence the joint maximum-likelihood estimates areThis is Laplace location and scale maximum likelihood, expressed with the reciprocal scale.

A finite rate maximum-likelihood estimate requires a positive total absolute deviation. For distinct observations and odd $n\ge 3$, that condition holds. For $n=1$, however, $\mu =x_1$ gives $A(\mu )=0$ and $L=\lambda /2\to \infty$ as $\lambda \to \infty$. There is no finite joint maximum-likelihood estimate in the stated parameter space; the formula with zero denominator is not an admissible estimate.