Question

Our proof of Theorem \(1.8 .1\) was for the discrete case. The proof for the continuous case requires some advanced results in in analysis. If, in addition, though, the function \(g(x)\) is one-to-one, show that the result is true for the continuous case. Hint: First assume that \(y=g(x)\) is strictly increasing. Then use the change-ofvariable technique with Jacobian \(d x / d y\) on the integral \(\int_{x \in \mathcal{S}_{X}} g(x) f_{X}(x) d x\)

Short answer

By changing variables in the integral using the Jacobian \(dx/dy\), the integral in the continuous case for \(y=g(x)\) where \(g(x)\) is strictly increasing can be derived as \(\int_{y \in \mathcal{S}_{Y}} y f_{Y}(y) dy\).

Step-by-step solution

Identifying the Problem

Consider the given integral \(\int_{x \in \mathcal{S}_{X}} g(x) f_{X}(x) d x\). It involves a one-to-one function \(g(x)\). It is given that \(g(x)\) is strictly increasing. As the problem hints, we first treat \(y\) as \(g(x)\), as this suggests we can perform a change of variables from \(x\) to \(y\).

Changing Variables

Performing a change of variables from \(x\) to \(y\), we have \(y=g(x)\) that implies \(x=g^{-1}(y)\). Then replace \(x\) with \(g^{-1}(y)\) and \(dx\) with \(\frac{dy}{g'(g^{-1}(y))}\) in the integral.

Calculating the Integral

The integral now becomes \(\int_{y \in \mathcal{S}_{Y}} y f_{Y}(y) dy\), as the function \(fY(y)\) represents the probability density function of \(Y\). Thus, by changing variables, the integral for \(y = g(x)\) has been successfully derived.