Affine Transformations of Uniform Variables

Why Affine Transformations Matter

Uniform distributions model outcomes that are equally likely across an interval. In applications, however, a quantity may be recorded in a different unit or shifted to a new origin. If \(X\) is uniform on an interval and \(Y=c+dX\), the transformation shifts and rescales the interval. When \(d\) is negative, it also reverses the order of values. Getting the transformed support and density right requires accounting for that reversal.

Assume throughout that \(X\sim\operatorname{Uniform}(a,b)\), where \(a<b\), and that \(c,d\) are fixed real constants. The density of \(X\) is \(1/(b-a)\) on \([a,b]\) and zero outside it; endpoint values of a density do not affect probabilities.

Support and Distribution When \(d\ne0\)

The transformed endpoints are found by applying \(x\mapsto c+dx\) to \(a\) and \(b\). If \(d>0\), this map preserves order, so \(c+da<c+db\). If \(d<0\), it reverses order, so \(c+db<c+da\). In either case, the interval width is \(|d|(b-a)\).

The following result gives both the CDF and density. It is useful to state the cases explicitly: the negative-\(d\) CDF comes from reversing the inequality when solving for \(X\).

Proof. By definition, \(F_Y(y)=P(Y\le y)\). First suppose \(d>0\). Then \(c+dX\le y\) is equivalent to \(X\le (y-c)/d\). For \(c+da\le y\le c+db\), the threshold \((y-c)/d\) lies in \([a,b]\). The uniform CDF of \(X\) therefore gives

Below this interval the event has probability zero, and above it the event has probability one. This proves the stated CDF for \(d>0\).

Now suppose \(d<0\). Dividing the inequality \(c+dX\le y\) by \(d\) reverses its direction, so it is equivalent to \(X\ge (y-c)/d\). For \(c+db\le y\le c+da\), this threshold lies in \([a,b]\). Since \(X\) is continuous, the probability of equality at the threshold is zero, and

The final equality uses \(d=-|d|\). Below the transformed interval the event is impossible; above it the event is certain. Thus the CDF has the stated negative-\(d\) form. In either case, the CDF increases linearly across an interval of width \(|d|(b-a)\), with slope \(1/(|d|(b-a))\). Differentiating on the interval's interior gives the stated density. It integrates to one because its height times the interval width is \(\{1/(|d|(b-a))\}|d|(b-a)=1\). The CDF is continuous at both endpoints, so there are no endpoint point masses. \(\square\)

Worked Transformations

The Zero-Scale Case and a Common Pitfall

If \(d=0\), then \(Y=c\) with probability one, regardless of the distribution of \(X\). This is a point mass at \(c\), not a continuous uniform distribution on an interval of positive width. Its CDF is zero for \(y<c\) and one for \(y\ge c\); it has no density with respect to ordinary length on the real line.

A common mistake is to transform the endpoints and then use them in their original order even when \(d<0\). For example, with \(Y=10-3X\) and \(X\) uniform on \([1,5]\), the image of \(a=1\) is 7 and the image of \(b=5\) is \(-5\). Using 7 as the lower endpoint would produce a decreasing CDF across the interval, which is impossible: if \(y_1\le y_2\), then \(\{Y\le y_1\}\subseteq\{Y\le y_2\}\), so \(F_Y(y_1)\le F_Y(y_2)\). The correction is to sort the two image endpoints and, for the CDF, use \((y-\text{lower endpoint})/(\text{interval width})\) within the support.

The absolute value in the density is another essential safeguard. The density height is the reciprocal of the transformed interval's width, \(|d|(b-a)\), so it must be positive for either sign of \(d\). The total probability is then the height times that width, equal to one.