1.7. Continuous Random Variables 55a simple formula for the pdf ofYin terms of the pdf ofX, which we record in the
next theorem.
Theorem 1.7.1.LetXbe a continuous random variable with pdffX(x)and support
SX.LetY=g(X),whereg(x)is a one-to-one differentiable function, on the sup-
port ofX,SX. Denote the inverse ofgbyx=g−^1 (y)and letdx/dy=d[g−^1 (y)]/dy.
Then the pdf ofYis given by
fY(y)=fX(g−^1 (y))∣
∣
∣
∣dx
dy∣
∣
∣
∣, fory∈SY, (1.7.11)where the support ofY is the setSY={y=g(x):x∈SX}.
Proof: Sinceg(x) is one-to-one and continuous, it is either strictly monotonically
increasing or decreasing. Assume that it is strictly monotonically increasing, for
now. The cdf ofYis given by
FY(y)=P[Y≤y]=P[g(X)≤y]=P[X≤g−^1 (y)] =FX(g−^1 (y)). (1.7.12)Hence, the pdf ofYis
fY(y)=d
dy
FY(y)=fX(g−^1 (y))dx
dy
, (1.7.13)wheredx/dyis the derivative of the functionx=g−^1 (y). In this case, becausegis
increasing,dx/dy >0. Hence, we can writedx/dy=|dx/dy|.
Supposeg(x) is strictly monotonically decreasing. Then (1.7.12) becomesFY(y)=
1 −FX(g−^1 (y)). Hence, the pdf ofYisfY(y)=fX(g−^1 (y))(−dx/dy). But sinceg
is decreasing,dx/dy <0 and, hence,−dx/dy=|dx/dy|. Thus Equation (1.7.11) is
true in both cases.^5.
Henceforth, we refer todx/dy=(d/dy)g−^1 (y)astheJacobian(denoted byJ)
of the transformation. In most mathematical areas,J=dx/dyis referred to as the
Jacobian of the inverse transformationx=g−^1 (y), but in this book it is called the
Jacobian of the transformation, simply for convenience.
We summarize Theorem 1.7.1 in a simple algorithm which we illustrate in the
next example. Assuming that the transformationY =g(X) is one-to-one, the
following steps lead to the pdf ofY:
- Find the support ofY.
- Solve for the inverse of the transfomation; i.e., solve forxin terms ofyin
y=g(x), thereby obtainingx=g−^1 (y). - Obtaindxdy.
- The pdf ofYisfY(y)=fX(g−^1 (y))
∣
∣
∣dxdy∣
∣
∣.(^5) The proof of Theorem 1.7.1 can also be obtained by using the change-of-variable technique as
discussed in Chapter 4 ofMathematical Comments.