Stat 5101 Lecture Notes - School of Statistics

8 Stat 5101 (Geyer) Course Noteswhere S is the sample space of the probability model describing X. We couldhave written g ( X(s) ) in place of g(s) in (1.12), but since X is the identityrandom variable for the P X model, these are the same. Putting this all together,we get the following theorem.Theorem 1.1. If X ∼ P X and Y = g(X), then Y ∼ P YwhereP Y (A) =P X (B),the relation between A and B being given by (1.12).This theorem is too abstract for everyday use. In practice, we will use at lotof other theorems that handle special cases more easily. But it should not beforgotten that this theorem exists and allows, at least in theory, the calculationof the distribution of any random variable.Example 1.2.1 (Constant Random Variable).Although the theorem is hard to apply to complicated random variables, it isnot too hard for simple ones. The simplest random variable is a constant one.Say the function g in the theorem is the constant function defined by g(s) =cfor all s ∈ S.To apply the theorem, we have to find, for any set A in the sample of Y ,which is the codomain of the function g, the set B defined by (1.12). Thissounds complicated, and in general it is, but here is it fairly easy. There areactually only two cases.Case I:Suppose c ∈ A. Thenbecause g(s) =c∈Afor all s in S.B = { s ∈ S : g(s) ∈ A } = SCase II:Conversely, suppose c/∈A. ThenB = { s ∈ S : g(s) ∈ A } = ∅because g(s) =c/∈Afor all s in S, that is there is no s such that the conditionholds, so the set of s satisfying the condition is empty.Combining the Cases: Now for any probability distribution the empty sethas probability zero and the sample space has probability one, so P X (∅) =0and P X (S) = 1. Thus the theorem says{1, c ∈ AP Y (A) =0, c /∈ A