Four Familiar Location-Scale Families
A location-scale family starts with a fixed standardized random variable \(Z\) and forms \(X=\mu+sZ\), where \(\mu\) is a location parameter and \(s>0\) is a scale parameter. Changing \(\mu\) shifts the distribution; changing \(s\) stretches or contracts it. The earlier material on location-scale transformations gives the general rules for the resulting distribution, density, support, moments, and quantiles. Here the emphasis is on recognizing those rules in four common scalar families and interpreting their different scale conventions.
The same numerical value of \(s\) need not mean the same standard deviation across families. In particular, a scale parameter is not automatically a standard deviation. The standardized base law determines how spread relates to \(s\), as well as the shape and support.
Standard Bases and Parameter Conventions
The following table fixes a base distribution and the resulting parameters for each family. For the normal family, the notation \(N(\mu,\sigma^2)\) uses variance as its second argument. For the other three families, \(s\) denotes their conventional scale parameter; the formulas in the table show how it relates to standard deviation.
| Family | Standardized base \(Z\) | Location-scale form | Support and standard deviation |
|---|---|---|---|
| Normal | \(Z\sim N(0,1)\) | \(X=\mu+\sigma Z\), so \(X\sim N(\mu,\sigma^2)\) | \(\mathbb{R}\); standard deviation \(\sigma\) |
| Logistic | \(g(z)=e^{-z}/(1+e^{-z})^2\) | \(X=\mu+sZ\) | \(\mathbb{R}\); standard deviation \(s\pi/\sqrt{3}\) |
| Laplace | \(g(z)=\tfrac12e^{-|z|}\) | \(X=\mu+bZ\) | \(\mathbb{R}\); standard deviation \(b\sqrt{2}\) |
| Uniform | \(Z\sim \operatorname{Unif}(-\tfrac12,\tfrac12)\) | \(X=\mu+sZ\) | \([\mu-s/2,\mu+s/2]\); standard deviation \(s/\sqrt{12}\) |
The letters \(s\) and \(b\) are used here to emphasize that logistic and Laplace scales are family-specific conventions; neither is a standard deviation. The uniform scale \(s\) in this parameterization is the full width of its support. Another common uniform description uses a center \(\mu\) and half-width \(a\), so its support is \([\mu-a,\mu+a]\) and \(s=2a\).
The corresponding densities make the differences in shape explicit. The normal density is \(f(x)=(\sigma\sqrt{2\pi})^{-1}\exp(-(x-\mu)^2/(2\sigma^2))\). The logistic density is \(f(x)=s^{-1}e^{-(x-\mu)/s}/(1+e^{-(x-\mu)/s})^2\). The Laplace density is \(f(x)=(2b)^{-1}\exp(-|x-\mu|/b)\). The uniform density equals \(1/s\) on its support and zero outside it. These formulas use the density transformation stated in the definition.
Quantiles and Shape
All four families are symmetric about their location \(\mu\), and each has median \(\mu\). Their densities also attain their maximum at \(\mu\). Symmetry does not make their shapes interchangeable: the normal density has tails that decay at a Gaussian rate, while the logistic and Laplace tails decay exponentially. The Laplace density has a sharp peak at its center; the logistic density is smoother there. The uniform density is constant over a bounded interval and has no tails outside that interval.
For the normal distribution, quantiles are written using the standard normal quantile function: \(Q_X(p)=\mu+\sigma\Phi^{-1}(p)\), for \(0<p<1\). For the logistic family, the standard base CDF is \(F_Z(z)=1/(1+e^{-z})\). Its inverse is particularly simple:
To verify the logistic formula, solve \(p=1/(1+e^{-z})\). Rearranging gives \(1/p=1+e^{-z}\), hence \(e^{-z}=(1-p)/p\), so \(z=\log(p/(1-p))\). Since \(X=\mu+sZ\) with \(s>0\), its \(p\)-quantile is \(\mu+s\) times the base quantile in the affine sense: \(Q_X(p)=\mu+sQ_Z(p)\). Substitution gives the stated expression. For the uniform base, \(F_Z(z)=z+1/2\) on \([-1/2,1/2]\), so \(Q_Z(p)=p-1/2\), yielding the displayed uniform quantile. \(\square\)
Worked Example: A Logistic Location and Scale
Let \(X=10+2Z\), where \(Z\) has the standard logistic density in the table. Find the median, first and third quartiles, interquartile range, and standard deviation.
The location is \(\mu=10\), and the logistic scale is \(s=2\). The median is the \(p=1/2\) quantile. The quantile formula gives \(Q_X(1/2)=10+2\log(1)=10\).
For the first quartile, \(Q_X(1/4)=10+2\log((1/4)/(3/4))=10+2\log(1/3)=10-2\log 3\approx7.803\). For the third quartile, \(Q_X(3/4)=10+2\log((3/4)/(1/4))=10+2\log 3\approx12.197\). Thus the interquartile range is \((10+2\log 3)-(10-2\log 3)=4\log 3\approx4.394\).
The logistic standard deviation is \(s\pi/\sqrt{3}\), not \(s\). Here it is \(2\pi/\sqrt{3}\approx3.628\). The distinction matters: the scale is 2, while the standard deviation is approximately 3.628.
Moments from the Base Distributions
The moment formulas in the table can be checked directly for the uniform and Laplace bases. The derivations also illustrate why the standardized base determines how scale translates into spread.
For the Laplace law, symmetry about \(\mu\) gives \(E[X]=\mu\). With \(Z=(X-\mu)/b\), the standardized density is \(\tfrac12e^{-|z|}\). Its mean is zero, and \(E[Z^2]=\int_0^\infty z^2e^{-z}\,dz=2\). The last integral follows by integration by parts twice: \(\int_0^\infty z^2e^{-z}\,dz=2\int_0^\infty ze^{-z}\,dz=2\int_0^\infty e^{-z}\,dz=2\); the boundary terms vanish because \(z^ke^{-z}\to0\) as \(z\to\infty\) for \(k=1,2\). Therefore \(\operatorname{Var}(X)=b^2\operatorname{Var}(Z)=2b^2\), and the standard deviation is \(b\sqrt{2}\). The logistic variance \(s^2\pi^2/3\) is the standard logistic moment formula; its scale-to-standard-deviation conversion is shown in the table.
Worked Example: Uniform Support Width Versus Standard Deviation
Suppose \(X\) is uniform on \([4,16]\). Express it using the center-and-width parameterization, then find its mean and standard deviation.
The center is \(\mu=(4+16)/2=10\), and the full support width is \(s=16-4=12\). Equivalently, \(X=10+12Z\) for \(Z\sim\operatorname{Unif}(-1/2,1/2)\). The mean is 10. The standard deviation is \(\sqrt{s^2/12}=\sqrt{12^2/12}=\sqrt{12}=2\sqrt{3}\approx3.464\).
In the alternative half-width convention, the same support is \([10-a,10+a]\) with \(a=6\). Its standard deviation is \(a/\sqrt{3}=6/\sqrt{3}=2\sqrt{3}\), the same result. This is one reason to state the support or parameter convention when reporting a uniform distribution.
Worked Example: A Laplace Model and Its Scale
Let \(X\) have a Laplace distribution with location \(\mu=4\) and scale \(b=3\). Find its density, median, variance, and probability of lying within one scale unit of its location.
The density is \(f(x)=\frac{1}{6}\exp(-|x-4|/3)\), for \(x\in\mathbb{R}\). Symmetry gives median 4. The variance is \(2b^2=2(3^2)=18\), so the standard deviation is \(\sqrt{18}=3\sqrt{2}\approx4.243\).
For the probability, the Laplace CDF is \(F(x)=\tfrac12e^{(x-\mu)/b}\) when \(x<\mu\), and \(F(x)=1-\tfrac12e^{-(x-\mu)/b}\) when \(x\geq\mu\). Consequently, \(P(1\leq X\leq7)=F(7)-F(1)\). Here \(F(7)=1-\tfrac12e^{-1}\) and \(F(1)=\tfrac12e^{-1}\), so the probability is \(1-e^{-1}\approx0.6321\).
Worked Example: Converting a Normal Model
Let \(Z\sim N(0,1)\) and set \(X=5+1.5Z\). Identify the distribution of \(X\), its standard deviation, and the probability \(P(X\leq8)\).
An affine transformation of a standard normal gives \(X\sim N(5,1.5^2)=N(5,2.25)\), using variance as the second parameter. Its standard deviation is 1.5. For the probability, standardize the cutoff: \(P(X\leq8)=P((X-5)/1.5\leq(8-5)/1.5)=P(Z\leq2)=\Phi(2)\approx0.9772\).
Unlike the logistic, Laplace, and uniform conventions above, the normal parameter \(\sigma=1.5\) is already the standard deviation. Its variance, which appears in \(N(5,2.25)\), is \(1.5^2=2.25\).
How to Choose and Report a Parameterization
These families are useful as compact descriptions of different distributional shapes. A normal model is a familiar symmetric, unbounded benchmark. A logistic model has heavier tails than the normal and a convenient quantile formula. A Laplace model has a sharper central peak and heavier tails than the normal. A uniform model describes constant density over a bounded interval. Selecting one is a modeling decision: the location-scale representation does not itself establish that a particular family fits data.
When reporting a model, give both the family and the parameter convention. For example, “uniform with center 10 and full width 12” identifies support \([4,16]\); “uniform on \([4,16]\)” states the support directly. For logistic and Laplace models, specify whether the reported spread is the family scale or the standard deviation. For a normal model, state whether the second argument is variance, as in \(N(\mu,\sigma^2)\).
Common Mistakes
- Calling every scale a standard deviation. For example, Laplace scale \(b=3\) gives standard deviation \(3\sqrt{2}\), not 3. Use the family-specific conversion.
- Confusing uniform width and half-width. In this tutorial \(s\) is full support width, so the endpoints are \(\mu\pm s/2\). A half-width parameter \(a\) instead gives endpoints \(\mu\pm a\).
- Reading the normal parameter incorrectly. In \(N(\mu,\sigma^2)\), the second argument is variance, while \(\sigma\) is the standard deviation.
- Assuming equal location and scale imply equal shapes. Normal, logistic, and Laplace distributions all have support \(\mathbb{R}\), but their peak and tail behavior differ; the uniform law even has bounded support.
Check Your Understanding
For each question, attend to both the base distribution and the parameter convention.
- If \(X=\mu+\sigma Z\) with \(Z\sim N(0,1)\), what are the mean, variance, and standard deviation of \(X\)?
- A logistic variable has location 0 and scale 1. What is its median, and what is its standard deviation?
- A uniform variable has center 3 and full width 8. What is its support and standard deviation?
- For a Laplace variable with location 2 and scale 1, give its standard deviation and explain why it is not equal to its scale.
- Which of these four families has bounded support? Describe one shape difference between the logistic and Laplace densities.