Why the Parameter Must Be Tracked in Three Places
When a transformation \(Y=g_\theta(X)\) depends on a parameter \(\theta\), it is tempting to make a substitution in a familiar formula and treat \(\theta\) as a fixed multiplier. That can miss an important feature: changing \(\theta\) may alter the inverse map, the derivative used in the density formula, the range of possible values of \(Y\), or some combination of the three. The support must be derived from the transformation, not guessed from the appearance of the resulting density.
Throughout, \(\theta\) is fixed when deriving the distribution of \(Y\). If the distribution of \(X\) also depends on \(\theta\), its density must be indexed accordingly and included in the calculation. The results below apply to a continuously distributed \(X\) and a one-to-one monotone transformation on an interval containing \(X\) with probability one.
CDFs and Densities for a Fixed Parameter Value
The CDF calculation begins with the event \(g_\theta(X)\leq y\). If \(g_\theta\) is increasing, this event is equivalent to \(X\leq g_\theta^{-1}(y)\). If \(g_\theta\) is decreasing, the inequality reverses: it is equivalent to \(X\geq g_\theta^{-1}(y)\). These statements apply when \(y\) is in the range of the transformation; outside that range, the CDF is determined by the fact that \(Y\) cannot take values there.
if \(g_\theta\) is increasing, \(F_Y(y)=F_X(h_\theta(y))\); if \(g_\theta\) is decreasing, \(F_Y(y)=1-F_X(h_\theta(y))\).
To prove the increasing case, for \(y\in J_\theta\), strict monotonicity gives \(P(Y\leq y)=P(g_\theta(X)\leq y)=P(X\leq h_\theta(y))=F_X(h_\theta(y))\). For the decreasing case, the inequality reverses, so \(P(Y\leq y)=P(X\geq h_\theta(y))\). Absolute continuity implies \(P(X=h_\theta(y))=0\), hence \(P(X\geq h_\theta(y))=1-F_X(h_\theta(y))\). The CDF is 0 or 1 outside the range as appropriate to the lower or upper end of that range, since \(P(X\in I)=1\). This proves both formulas. \(\square\)
\(f_Y(y)=f_X(h_\theta(y))\left|h_\theta'(y)\right|=\dfrac{f_X(h_\theta(y))}{|g_\theta'(h_\theta(y))|}\) for \(y\in J_\theta\), and \(f_Y(y)=0\) for \(y\notin J_\theta\).
For any Borel set \(A\subseteq J_\theta\), the event \(Y\in A\) is equivalent to \(X\in h_\theta(A)\). The one-dimensional change-of-variables formula, applied to the inverse map, gives
The inverse function theorem applies because \(g_\theta'\) is continuous and nowhere zero on \(I\), so \(h_\theta'(y)=1/g_\theta'(h_\theta(y))\). Taking its absolute value gives the stated density. Outside \(J_\theta\), \(Y\) cannot lie because \(X\in I\) with probability one. In particular, the proposed density has total integral \(\int_{J_\theta}f_Y(y)\,dy=\int_I f_X(x)\,dx=P(X\in I)=1\). The support condition is essential: monotonicity on \(I\) alone does not justify setting the transformed density to zero outside \(g_\theta(I)\) if \(X\) can also fall outside \(I\). \(\square\)
Worked Transformations
Worked Example: A Parameter Changes the Inverse, Jacobian, and Support
Let \(X\sim\operatorname{Uniform}(0,1)\) and, for \(\theta>0\), set \(Y=\theta X\). The map \(g_\theta(x)=\theta x\) is increasing on \(I=(0,1)\), and \(P(X\in I)=1\). Its range is \(J_\theta=(0,\theta)\), its inverse is \(h_\theta(y)=y/\theta\), and \(h_\theta'(y)=1/\theta\).
For \(0<y<\theta\), the CDF formula gives \(F_Y(y)=F_X(y/\theta)=y/\theta\), since \(F_X(x)=x\) on \((0,1)\). Completing the CDF outside the range,
Differentiating on the range, or using the density formula directly, gives \(f_Y(y)=1/\theta\) for \(0<y<\theta\), and zero elsewhere. The parameter changes the inverse \(y/\theta\), the absolute inverse derivative \(1/\theta\), and the upper support endpoint \(\theta\). As a check, \(\int_0^\theta (1/\theta)\,dy=1\).
Worked Example: A Decreasing Transformation With a Moving Interval
Let \(X\sim\operatorname{Uniform}(0,1)\) and let \(Y=\theta-X\), where \(\theta\) can be any real number. The map is decreasing, its range is \((\theta-1,\theta)\), and its inverse is \(h_\theta(y)=\theta-y\). The inverse derivative is \(-1\), so the absolute Jacobian factor is 1.
For \(\theta-1<y<\theta\), the decreasing CDF formula yields \(F_Y(y)=1-F_X(\theta-y)=1-(\theta-y)=y-\theta+1\). Thus
The density is \(f_Y(y)=1\) on \((\theta-1,\theta)\), zero elsewhere. Here \(\theta\) changes the inverse and shifts both support endpoints, but it does not change the absolute Jacobian factor. The interval has length 1, so \(\int_{\theta-1}^{\theta}1\,dy=1\), as required.
Worked Example: A Bounded Range From an Unbounded Input
Let \(X\sim\operatorname{Exponential}(\lambda)\), with \(\lambda>0\), and define \(Y=X/(1+\theta X)\) for \(\theta>0\). Here \(I=(0,\infty)\), and \(P(X\in I)=1\). The function \(g_\theta(x)=x/(1+\theta x)\) is increasing because \(g_\theta'(x)=1/(1+\theta x)^2>0\). As \(x\) increases from 0 to infinity, \(g_\theta(x)\) increases from 0 toward \(1/\theta\), so \(J_\theta=(0,1/\theta)\).
Solving \(y=x/(1+\theta x)\) for \(x\) gives \(y+\theta xy=x\), hence \(x=y/(1-\theta y)\). This inverse is valid precisely for \(0<y<1/\theta\), and \(h_\theta'(y)=1/(1-\theta y)^2\). Since \(f_X(x)=\lambda e^{-\lambda x}\) for \(x>0\), the density is
The CDF on the range is \(F_Y(y)=F_X(y/(1-\theta y))=1-\exp(-\lambda y/(1-\theta y))\). It tends to 0 as \(y\downarrow0\) and to 1 as \(y\uparrow1/\theta\), confirming the endpoint behavior. Differentiating this CDF gives the displayed density: the derivative of \(y/(1-\theta y)\) is \(1/(1-\theta y)^2\), so the chain rule gives exactly \(\lambda\exp(-\lambda y/(1-\theta y))/(1-\theta y)^2\). Thus the density integrates to the difference of the CDF limits, 1. In this example, the parameter changes the inverse, the Jacobian, and the finite upper endpoint, even though \(X\) has an unbounded range.
How to Check a Parameter-Dependent Calculation
The parameter should be carried through the derivation rather than inserted only in the final expression. A reliable order of work is:
State an interval \(I\) on which the map is one-to-one, and verify \(P(X\in I)=1\).
Determine \(J_\theta=g_\theta(I)\) and solve \(y=g_\theta(x)\) for \(x=h_\theta(y)\). Check that the inverse is defined on exactly the claimed range.
Keep the inequality direction: it is preserved for an increasing map and reversed for a decreasing map.
Evaluate \(|h_\theta'(y)|\), set the density to zero off \(J_\theta\), and verify its integral is one or that its CDF has the correct endpoint limits.
A common mistake is to find a monotone interval \(I\), apply the density formula there, and declare the result zero outside \(g_\theta(I)\) without checking where \(X\) lies. For example, the identity map is monotone on \(I=(0,1)\), but if \(X\sim N(0,1)\), then \(P(X\in I)<1\). A formula using only \(x\in(0,1)\) accounts for only part of the law of \(Y=X\); it cannot be the full density. The correction is to choose an interval containing \(X\) almost surely, or to account separately for all parts of the distribution not included in \(I\). For the one-interval formula proved here, the required condition is \(P(X\in I)=1\).
Check Your Understanding
For each question, pay attention to both the transformed range and the parameter dependence.
- If \(X\sim\operatorname{Uniform}(0,1)\) and \(Y=X/\theta\) for \(\theta>0\), what are the inverse map, transformed support, and density?
- Why does the CDF calculation for a decreasing transformation use \(1-F_X(h_\theta(y))\) when \(X\) is absolutely continuous?
- For the transformation \(Y=X/(1+\theta X)\) with \(X>0\) and \(\theta>0\), why must the transformed support have a finite upper endpoint?
- What condition on \(P(X\in I)\) is needed before a density derived on \(g_\theta(I)\) can be set to zero everywhere outside that range?
- In \(Y=\theta-X\) with \(X\sim\operatorname{Uniform}(0,1)\), which parts of the transformation calculation change with \(\theta\), and which do not?