Density and Support Under Location-Scale Changes

From a Base Density to a Location-Scale Density

A location-scale transformation changes the measurement scale in two ways: it shifts values by $\mu$ and stretches them by $\sigma$. The density must account for both changes. In particular, stretching spreads probability over a wider range, so the transformed density is not just the original density evaluated at a new argument; it also has a Jacobian factor.

Let $Z$ have a density $f_Z$, and define $X=\mu+\sigma Z$, where $\mu\in\mathbb{R}$ and $\sigma>0$. The inverse transformation is $z=(x-\mu)/\sigma$. The change-of-variables formula gives a factor $1/\sigma$, while the support is shifted and stretched along with the values of $Z$.

Definition: For a real-valued random variable $Z$, its support is the set of points $z$ such that every open interval containing $z$ has positive probability under the distribution of $Z$. Write this set as $S_Z$. If $Z$ has a density, the support is determined by the probability distribution, not by the value assigned to the density at an individual point.

Theorem: Suppose $Z$ has density $f_Z$, and let $X=\mu+\sigma Z$ with $\mu\in\mathbb{R}$ and $\sigma>0$. A density of $X$ is $$ f_X(x)=\frac{1}{\sigma}f_Z\!\left(\frac{x-\mu}{\sigma}\right),\qquad x\in\mathbb{R}. $$ The support of $X$ is $$ S_X=\{\mu+\sigma z:z\in S_Z\}. $$

When a density is given as zero outside a stated region, its transformed formula should include the corresponding transformed region. For example, if $f_Z(z)=0$ outside an interval $(a,b)$, then $f_X(x)=0$ outside $(\mu+\sigma a,\mu+\sigma b)$. The support, in the sense defined above, includes the interval’s boundary points when every neighborhood of each boundary point has positive probability. Whether the density is set to zero or assigned some other value at a boundary does not change the distribution.

Why the Jacobian Appears

The density formula follows by computing probabilities on measurable sets. The same change of variables also verifies that the resulting function integrates to one.

Proof: For any Borel set $A\subseteq\mathbb{R}$, the event $X\in A$ is the event $\mu+\sigma Z\in A$. Since $Z$ has density $f_Z$, $$ P(X\in A)=\int_{\{z:\,\mu+\sigma z\in A\}} f_Z(z)\,dz. $$ Make the substitution $x=\mu+\sigma z$. Since $\sigma>0$, $dz=dx/\sigma$, and the transformed region of integration is $A$. Therefore $$ P(X\in A)=\int_A \frac{1}{\sigma}f_Z\!\left(\frac{x-\mu}{\sigma}\right)\,dx. $$ This establishes the claimed density. In particular, taking $A=\mathbb{R}$ gives $$ \int_{-\infty}^{\infty}\frac{1}{\sigma}f_Z\!\left(\frac{x-\mu}{\sigma}\right)\,dx =\int_{-\infty}^{\infty}f_Z(z)\,dz=1. $$ To establish the support statement, fix $x=\mu+\sigma z$. The inverse map sends the interval $(x-\varepsilon,x+\varepsilon)$ to $(z-\varepsilon/\sigma,z+\varepsilon/\sigma)$. Consequently, every neighborhood of $x$ has positive probability under $X$ exactly when every neighborhood of $z$ has positive probability under $Z$. The affine map is a bijection, so this correspondence gives $S_X=\{\mu+\sigma z:z\in S_Z\}$. $\square$

The theorem applies even when the support is not a single interval: the entire set is mapped by $z\mapsto\mu+\sigma z$. For the usual positive scale, its order is preserved. Thus a finite interval $[a,b]$ becomes $[\mu+\sigma a,\mu+\sigma b]$, and a half-line $[a,\infty)$ becomes $[\mu+\sigma a,\infty)$.

Worked Transformations

Worked Example: Stretching a Uniform Distribution

Let $Z\sim\operatorname{Uniform}(0,1)$, so its density is $f_Z(z)=1$ for $0<z<1$ and zero elsewhere. Set $X=2+3Z$.

Here $\mu=2$ and $\sigma=3$. Substitute $z=(x-2)/3$ into the base density and multiply by $1/3$: $$ f_X(x)=\frac{1}{3}f_Z\!\left(\frac{x-2}{3}\right) =\begin{cases} \frac{1}{3},&2<x<5,\\ 0,&\text{otherwise}. \end{cases} $$ The transformed support is $\{2+3z:0\leq z\leq1\}=[2,5]$. Thus $X$ is uniform on $(2,5)$, with support including the endpoints.

As a check, the density integrates to $(5-2)(1/3)=1$. The interval’s length tripled, from 1 to 3, and the density height was divided by 3.

Worked Example: Shifting and Scaling an Exponential Distribution

Let $Z\sim\operatorname{Exponential}(1)$, with density $f_Z(z)=e^{-z}$ for $z>0$ and zero otherwise. Define $X=1+2Z$.

The inverse transformation is $z=(x-1)/2$, and the Jacobian factor is $1/2$. The condition $z>0$ becomes $x>1$, so $$ f_X(x)= \begin{cases} \frac{1}{2}e^{-(x-1)/2},&x>1,\\ 0,&x\leq1. \end{cases} $$ The support is $[1,\infty)$: the transformed values satisfy $X>1$, and every neighborhood of 1 has positive probability.

The normalization check is $$ \int_1^\infty \frac{1}{2}e^{-(x-1)/2}\,dx =\int_0^\infty e^{-u}\,du=1, $$ where $u=(x-1)/2$, so $dx=2\,du$. The location shifts the lower endpoint from 0 to 1, and the positive scale stretches the unbounded range without changing its direction.

Worked Example: Transforming a Beta Density and Its Bounded Support

Suppose $Z\sim\operatorname{Beta}(2,3)$, with density $f_Z(z)=12z(1-z)^2$ for $0<z<1$ and zero elsewhere. Let $X=-1+4Z$.

Here $\mu=-1$, $\sigma=4$, and $z=(x+1)/4$. The condition $0<z<1$ is equivalent to $-1<x<3$. Applying the formula, $$ f_X(x)=\frac{1}{4}\,12\left(\frac{x+1}{4}\right) \left(1-\frac{x+1}{4}\right)^2 =\frac{3(x+1)(3-x)^2}{64},\qquad -1<x<3. $$ Outside this interval the density is zero. The support is $[-1,3]$, since the base support $[0,1]$ is mapped by $z\mapsto-1+4z$.

To verify normalization, substitute $z=(x+1)/4$, so $dx=4\,dz$: $$ \int_{-1}^{3}\frac{3(x+1)(3-x)^2}{64}\,dx =\int_0^1 12z(1-z)^2\,dz =12\left(\frac{1}{2}-\frac{2}{3}+\frac{1}{4}\right)=1. $$ The calculation confirms both the Jacobian factor and the transformed support.

Support, Endpoints, and Parameter Dependence

The density formula records where probability can occur, but support is a property of the distribution rather than of one particular density expression. Densities that differ on a set of Lebesgue measure zero represent the same distribution. In particular, changing a density’s value at an endpoint does not add or remove probability there. For a uniform distribution on $(a,b)$, for instance, the density can be specified on the open interval while the support is the closed interval $[a,b]$.

The support can depend on the parameters. In the uniform example, changing $\sigma$ changes the right endpoint $\mu+\sigma$; in the exponential example, changing $\mu$ changes the lower endpoint. Such a parameter-dependent region matters when writing a density: its indicator or piecewise conditions must move with the parameters. This is a basic feature of parameter-dependent support, rather than a cosmetic choice of how to write the formula.

A useful way to keep the calculation organized is to transform the support separately from the density values:

Write the inverse map.
For $x=\mu+\sigma z$, solve for $z=(x-\mu)/\sigma$.

Transform the density height.
Evaluate the base density at the inverse value and multiply by $1/\sigma$.

Transform the support.
Map each point in the base support through $z\mapsto\mu+\sigma z$, or translate the support conditions using the inverse map.

Check the result.
Confirm that the density is nonnegative and integrates to one over its stated support.

Common Mistake: Forgetting the Scale Factor

A frequent error is to write $f_X(x)=f_Z((x-\mu)/\sigma)$ and stop. This uses the correct inverse argument but omits the Jacobian. For the transformation $X=2+3Z$ with $Z\sim\operatorname{Uniform}(0,1)$, that incorrect expression assigns density 1 throughout $(2,5)$, whose integral is 3, not 1. The correct density is $1/3$ there, because the factor $1/\sigma$ compensates for the tripling of interval length.

Another error is to keep the base support unchanged. In the same example, the support of $X$ is $[2,5]$, not $[0,1]$. The density’s argument is transformed by the inverse map, but the values and support are described on the $x$-scale.

Key takeaway: For $X=\mu+\sigma Z$ with $\sigma>0$, the density is $f_Z((x-\mu)/\sigma)/\sigma$, and the support is the image of the base support under $z\mapsto\mu+\sigma z$. Keep the Jacobian and the transformed support together.

Check Your Understanding

Use the change-of-variables formula and track the support as well as the density.

If $Z\sim\operatorname{Uniform}(0,1)$ and $X=5+2Z$, what is the density and support of $X$?
If $Z\sim\operatorname{Exponential}(1)$ and $X=-3+4Z$, what is the lower endpoint of the support, and what is the density on that support?
Why does a density’s value at an endpoint not determine whether the endpoint belongs to the support?
A proposed density for $X=1+5Z$, where $Z\sim\operatorname{Uniform}(0,1)$, is $1$ on $(1,6)$. Identify and correct the error.
If the base support is $[a,b]$, express the transformed support when $\sigma>0$.