Stat 5101 Lecture Notes - School of Statistics

10 Stat 5101 (Geyer) Course NotesOne-To-One TransformationsA transformation (change of variable)Sg−→ Tis one-to-one if g maps each point x to a different value g(x) from all otherpoints, that is,g(x 1 ) ≠ g(x 2 ), whenever x 1 ≠ x 2 .A way to say this with fewer symbols is to consider the equationy = g(x).If for each fixed y, considered as an equation to solve for x, there is a uniquesolution, then g is one-to-one. If for any y there are multiple solutions, it isn’t.Whether a function is one-to-one or not may depend on the domain. Soif you are sloppy and don’t distinguish between a function and an expressiongiving the value of the function, you can’t tell whether it is one-to-one or not.Example 1.2.2 (x 2 ).The function g : R → R defined by g(x) =x 2 is not one-to-one becauseg(x) =g(−x), x ∈ R.So it is in fact two-to-one, except at zero.But the function g :(0,∞)→Rdefined by the very same formula g(x) =x 2is one-to-one, because there do not exist distinct positive real numbers x 1 andx 2 such that x 2 1 = x 2 2. (Every positive real number has a unique positive squareroot.)This example seems simple, and it is, but every year some students getconfused about this issue on tests. If you don’t know whether you are dealingwith a one-to-one transformation or not, you’ll be in trouble. And you can’t tellwithout considering the domain of the transformation as well as the expressiongiving its values.Inverse TransformationsA function is invertible if it is one-to-one and onto, the latter meaning thatits codomain is the same as its range.Neither of the functions considered in Example 1.2.2 are invertible. Thesecond is one-to-one, but it is not onto, because the g defined in the examplemaps positive real numbers to positive real numbers. To obtain a function thatis invertible, we need to restrict the codomain to be the same as the range,defining the functiong :(0,∞)→(0, ∞)byg(x) =x 2 .