Logistic Location-Scale Distribution

From a Standard Logistic Variable to a Location-Scale Family

A location-scale family separates two features of a distribution: where it is centered and how spread out it is. For the logistic family, both features can be read directly from its parameters. Starting with the standard logistic law also makes the derivations transparent: a location shift moves the distribution, while a positive scale change stretches it.

We first define the standard variable through its cumulative distribution function. The general family will then follow by transforming that variable. This approach gives the CDF, density, median, quantiles, and symmetry from a single starting point.

Definition: A standard logistic random variable $Z$ has CDF $$ F_Z(z)=\frac{1}{1+e^{-z}}, \qquad z\in\mathbb{R}. $$ For $\mu\in\mathbb{R}$ and $s>0$, the logistic location-scale distribution with location $\mu$ and scale $s$ is the distribution of $$ X=\mu+sZ. $$

The requirement $s>0$ matters: it makes the transformation increasing, so inequalities involving $X$ can be translated into inequalities involving $Z$ without reversing their direction. The parameter $\mu$ sets the center, and $s$ controls the horizontal scale.

Deriving the CDF and Density

Because $s>0$, the event $X\le x$ is equivalent to $Z\le (x-\mu)/s$. Substitution into the standard CDF gives the general CDF.

Theorem: If $X=\mu+sZ$, where $Z$ is standard logistic, $\mu\in\mathbb{R}$, and $s>0$, then $$ F_X(x)=P(X\le x) =\frac{1}{1+\exp\!\left(-\frac{x-\mu}{s}\right)}, \qquad x\in\mathbb{R}, $$ and the density is $$ f_X(x)= \frac{\exp\!\left(-\frac{x-\mu}{s}\right)} {s\left[1+\exp\!\left(-\frac{x-\mu}{s}\right)\right]^2}, \qquad x\in\mathbb{R}. $$

To obtain the density, write $z=(x-\mu)/s$. Differentiating the CDF by the chain rule gives

\begin{aligned} f_X(x) &=\frac{d}{dx}\left[1+\exp\!\left(-\frac{x-\mu}{s}\right)\right]^{-1}\\ &=\left[1+\exp\!\left(-\frac{x-\mu}{s}\right)\right]^{-2} \cdot \exp\!\left(-\frac{x-\mu}{s}\right)\frac{1}{s}\\ &=\frac{\exp\!\left(-\frac{x-\mu}{s}\right)} {s\left[1+\exp\!\left(-\frac{x-\mu}{s}\right)\right]^2}. \end{aligned}

The factor $1/s$ is essential: changing the horizontal scale changes the density height as well as its width. The CDF tends to 0 as $x\to-\infty$ and to 1 as $x\to\infty$. Since it is differentiable with the nonnegative derivative displayed above, this derivative is a density whose integral over the real line is $1-0=1$.

Worked Example: Evaluating a CDF and Density

Let $X$ have a logistic distribution with $\mu=2$ and $s=3$. Find $F_X(5)$ and $f_X(5)$.

At $x=5$, the standardized value is $(5-2)/3=1$. Therefore,

F_X(5)=\frac{1}{1+e^{-1}} \approx \frac{1}{1+0.367879} \approx 0.7311.

For the density, retain the scale factor $1/3$:

f_X(5)=\frac{e^{-1}}{3(1+e^{-1})^2} \approx\frac{0.367879}{3(1.367879)^2} \approx 0.0655.

Thus about $73.11\%$ of the distribution lies at or below 5, and the density at 5 is approximately $0.0655$ per unit on the $X$-scale. A density is not itself a point probability.

Location, Scale, and the Median

At $x=\mu$, the standardized value is zero. The CDF is therefore $1/2$, and the density is $1/(4s)$. The center of the distribution is at $\mu$; increasing $s$ makes the distribution wider and lowers its density at that center.

F_X(\mu)=\frac12, \qquad f_X(\mu)=\frac{1}{4s}.

The median is the value $m$ for which $P(X\le m)\ge 1/2$ and $P(X\ge m)\ge 1/2$. Here the CDF is continuous and strictly increasing, since its density is positive for every real $x$. Consequently, the point where the CDF equals $1/2$ is the unique median.

Result: The logistic location-scale distribution has median $\mu$. Its CDF is strictly increasing, so this median is unique.

Indeed, substituting $x=\mu$ gives $F_X(\mu)=1/2$. Strict increase follows because the density formula is positive for all $x$, so no other value can have CDF equal to $1/2$.

Worked Example: Reading the Center and Scale

Suppose $X$ has a logistic distribution with $\mu=10$ and $s=2$. Find its median, its CDF at 10, and its density at 10.

The median is the location parameter, so it is 10. Direct substitution confirms this:

F_X(10)=\frac{1}{1+e^{-(10-10)/2}}=\frac12.

The density at the center is

f_X(10)=\frac{e^0}{2(1+e^0)^2} =\frac{1}{2\cdot 4} =\frac18=0.125.

If the location stayed at 10 but the scale increased, the median would remain 10, while the density at the center would decrease according to $1/(4s)$.

Quantile Function

For $0<p<1$, define the $p$-quantile $Q_X(p)$ as the unique value $x$ satisfying $F_X(x)=p$. This definition is unambiguous here because the CDF is continuous and strictly increasing. Solving the CDF equation gives the quantile function.

Result: For $0<p<1$, the logistic location-scale quantile function is $$ Q_X(p)=\mu+s\log\!\left(\frac{p}{1-p}\right). $$

To derive it, set the CDF equal to $p$, then isolate $x$:

\begin{aligned} p &=\frac{1}{1+\exp\!\left(-\frac{x-\mu}{s}\right)},\\ \frac{1-p}{p} &=\exp\!\left(-\frac{x-\mu}{s}\right),\\ \log\!\left(\frac{p}{1-p}\right) &=\frac{x-\mu}{s},\\ x &=\mu+s\log\!\left(\frac{p}{1-p}\right). \end{aligned}

The odds $p/(1-p)$ appear because the CDF has the form $1/(1+e^{-z})$. For $p=1/2$, the logarithm is $\log(1)=0$, recovering $Q_X(1/2)=\mu$, the median.

Worked Example: Finding a Logistic Quantile

Let $X$ have location $\mu=4$ and scale $s=2$. Find its 75th percentile and verify the result using the CDF.

For $p=0.75$, the odds are $0.75/0.25=3$. Thus,

Q_X(0.75)=4+2\log(3)\approx 4+2(1.098612)=6.1972.

To check, standardize this value:

\frac{6.1972-4}{2}\approx 1.0986=\log(3).

Therefore,

F_X(6.1972) \approx\frac{1}{1+e^{-\log(3)}} =\frac{1}{1+1/3} =\frac34.

The computed value is indeed the 75th percentile, up to rounding.

Symmetry About the Location

The logistic distribution is symmetric around its location parameter. One way to verify this is to compare the CDF at points equally far from $\mu$. For every $t\in\mathbb{R}$,

\begin{aligned} F_X(\mu-t) &=\frac{1}{1+e^{t/s}},\\ 1-F_X(\mu+t) &=1-\frac{1}{1+e^{-t/s}} =\frac{1}{1+e^{t/s}}. \end{aligned}

Thus $F_X(\mu-t)=1-F_X(\mu+t)$. Because the distribution is continuous, this identity says that the probability to the left of $\mu-t$ equals the probability to the right of $\mu+t$. Equivalently, $X-\mu$ and $-(X-\mu)$ have the same distribution.

Theorem: If $X$ has a logistic location-scale distribution with location $\mu$ and scale $s>0$, then $X$ is symmetric about $\mu$. In particular, for every $t\ge 0$, $$ P(X\le \mu-t)=P(X\ge \mu+t). $$

The density gives the same conclusion directly. Replacing $x$ by $2\mu-x$ changes $(x-\mu)/s$ to its negative. The two density values are equal because

f_X(\mu+t) =\frac{e^{-t/s}}{s(1+e^{-t/s})^2} =\frac{e^{t/s}}{s(1+e^{t/s})^2} =f_X(\mu-t).

The middle equality follows by multiplying the numerator and denominator of the first fraction by $e^{2t/s}$. Symmetry also appears in the quantile function: replacing $p$ by $1-p$ changes the sign of the logarithm, so

Q_X(1-p)=2\mu-Q_X(p), \qquad 0<p<1.

For example, the 25th and 75th percentiles lie equally far below and above $\mu$. This reflection property is a consequence of symmetry, not an additional parameter assumption.

Common Mistake and Practical Use

A frequent error is to use the scale parameter as though it were a standard deviation. The parameter $s$ controls the horizontal stretch in the transformation $X=\mu+sZ$; it is not defined as the standard deviation. Another common error is to omit $1/s$ when differentiating the CDF. The chain rule requires this factor, and without it the resulting expression would not have the correct total integral.

A quick check helps catch both errors: at the center, the density must be $1/(4s)$, while the CDF must equal $1/2$. The location $\mu$ is both the median and the center of symmetry. These facts are useful when interpreting a fitted logistic distribution or checking calculations, without confusing location and scale with other summaries.

Check Your Understanding

Use the definitions and derivations above to answer the following questions.

For $X=\mu+sZ$ with $s>0$, derive $F_X(x)$ by expressing the event $X\le x$ in terms of $Z$.
Why does the density contain a factor $1/s$, and what is the density at $x=\mu$?
Show from the quantile formula that the median is $\mu$.
For a logistic distribution with $\mu=1$ and $s=2$, calculate $Q_X(0.25)$ and $Q_X(0.75)$, and verify their symmetry about $\mu$.
State the CDF identity that expresses symmetry about $\mu$, and explain how it compares probabilities on opposite sides of the center.