Likelihood-Ratio Order Under Increasing Transformations

Why a Change of Scale Matters

Likelihood-ratio order compares two distributions through the way their density ratio changes across the sample space. A common change of scale changes each density: probability that was spread over a small interval is redistributed according to the transformation’s derivative. Yet when the same smooth, increasing transformation is applied to both distributions, the Jacobian factor is identical in the two transformed densities. It therefore cancels in their ratio.

The cancellation alone is not the whole argument. Likelihood-ratio order requires the ratio to be nondecreasing in the argument, and an increasing transformation preserves the ordering of arguments as well. This tutorial states the regularity conditions for that reasoning, proves the result, and checks it in several transformed models.

Definition and Transformation Formula

Definition: Let \(X\) and \(Y\) have densities \(f_X\) and \(f_Y\) that are strictly positive on the same interval \(I\) and zero outside \(I\). We say \(X\) is smaller than \(Y\) in likelihood-ratio order, written \(X\le_{\mathrm{lr}}Y\), if the ratio \(f_Y(x)/f_X(x)\) is nondecreasing in \(x\) on \(I\). Equivalently, for all \(x_1<x_2\) in \(I\), \(f_Y(x_1)/f_X(x_1)\le f_Y(x_2)/f_X(x_2)\).

The shared-support assumption keeps the ratio well defined throughout the interval. Likelihood-ratio comparisons with different supports require additional support conditions; these are not needed for the theorem below.

Let \(T:I\to J\) be a continuously differentiable, strictly increasing bijection between intervals, with \(T'(x)>0\) for every \(x\in I\). Its inverse \(T^{-1}:J\to I\) is differentiable, and the change-of-variables formula gives the density of \(T(X)\):

f_{T(X)}(y)=\frac{f_X(T^{-1}(y))}{T'(T^{-1}(y))}, \qquad y\in J.

The same formula applies to \(T(Y)\), with \(f_Y\) in place of \(f_X\). Both transformed distributions have support \(J\).

The Invariance Theorem

Theorem: Suppose \(X\) and \(Y\) have strictly positive densities on the same interval \(I\), and \(X\le_{\mathrm{lr}}Y\). If \(T:I\to J\) is a continuously differentiable, strictly increasing bijection with \(T'(x)>0\) throughout \(I\), then \(T(X)\le_{\mathrm{lr}}T(Y)\).

The proof tracks both the transformed ratio and the order of its arguments. In particular, it does not assume that the transformed densities equal the original densities.

Transform both densities.
Apply the change-of-variables formula using the same \(T\) and the same Jacobian factor.

Form the ratio.
Divide the transformed density of \(T(Y)\) by that of \(T(X)\); the Jacobian cancels.

Check monotonicity.
The inverse \(T^{-1}\) is increasing, so composing the original nondecreasing ratio with it preserves nondecreasingness.

Proof: Write \(R(x)=f_Y(x)/f_X(x)\), which is nondecreasing on \(I\) by the assumed likelihood-ratio order. For \(y\in J\), the transformed density ratio is

\frac{f_{T(Y)}(y)}{f_{T(X)}(y)} = \frac{f_Y(T^{-1}(y))/T'(T^{-1}(y))} {f_X(T^{-1}(y))/T'(T^{-1}(y))} = R(T^{-1}(y)).

If \(y_1<y_2\), then \(T^{-1}(y_1)<T^{-1}(y_2)\), because \(T^{-1}\) is increasing. Since \(R\) is nondecreasing, \(R(T^{-1}(y_1))\le R(T^{-1}(y_2))\). Thus the transformed density ratio is nondecreasing on \(J\), which is exactly \(T(X)\le_{\mathrm{lr}}T(Y)\). \(\square\)

The same calculation also proves the converse under these conditions. If the transformed ratio is nondecreasing on \(J\), then \(R(x)=f_{T(Y)}(T(x))/f_{T(X)}(T(x))\). Since \(T\) is increasing, this ratio is nondecreasing in \(x\). Consequently, for a common increasing bijection satisfying the stated regularity conditions, likelihood-ratio order holds before transformation if and only if it holds after transformation.

Worked Applications

Worked Example: Equal-Variance Normals Transformed to Lognormals

Let \(X\sim N(0,1)\) and \(Y\sim N(1,1)\), and apply \(T(x)=e^x\). First check the original ratio. The two normal densities are \(f_X(x)=(2\pi)^{-1/2}e^{-x^2/2}\) and \(f_Y(x)=(2\pi)^{-1/2}e^{-(x-1)^2/2}\). Hence

\frac{f_Y(x)}{f_X(x)} =\exp\left(-\frac{(x-1)^2}{2}+\frac{x^2}{2}\right) =\exp\left(x-\frac12\right).

Its derivative is \(\exp(x-\tfrac12)>0\), so \(X\le_{\mathrm{lr}}Y\). The map \(T\) is increasing from \(\mathbb{R}\) onto \((0,\infty)\), with \(T'(x)=e^x>0\). For \(y>0\), \(T^{-1}(y)=\log y\) and \(T'(T^{-1}(y))=y\). Therefore the transformed ratio is

\frac{f_{T(Y)}(y)}{f_{T(X)}(y)} =\frac{f_Y(\log y)/y}{f_X(\log y)/y} =\exp\left(\log y-\frac12\right) =\frac{y}{\sqrt{e}}.

This is increasing on \((0,\infty)\), confirming \(e^X\le_{\mathrm{lr}}e^Y\). The common factor \(1/y\), not an unchanged density, is what cancels.

Worked Example: Exponential Variables After a Log Transformation

Let \(X\) be exponential with rate \(2\), and \(Y\) exponential with rate \(1\). Their densities on \((0,\infty)\) are \(f_X(x)=2e^{-2x}\) and \(f_Y(x)=e^{-x}\). The ratio is

\frac{f_Y(x)}{f_X(x)} =\frac12 e^x,

which is increasing, so \(X\le_{\mathrm{lr}}Y\). Now take \(T(x)=\log x\), a strictly increasing bijection from \((0,\infty)\) to \(\mathbb{R}\), with derivative \(1/x>0\). For \(z\in\mathbb{R}\), the inverse is \(e^z\), and the transformed densities are

f_{\log X}(z)=f_X(e^z)e^z=2e^z\exp(-2e^z), \qquad f_{\log Y}(z)=f_Y(e^z)e^z=e^z\exp(-e^z).

Thus

\frac{f_{\log Y}(z)}{f_{\log X}(z)} =\frac{e^z\exp(-e^z)}{2e^z\exp(-2e^z)} =\frac12\exp(e^z).

Its derivative with respect to \(z\) is \(\tfrac12 e^z\exp(e^z)>0\). Therefore \(\log X\le_{\mathrm{lr}}\log Y\), as the theorem predicts.

Worked Example: Beta Variables Under the Odds Transformation

Let \(X\) have a \(\operatorname{Beta}(2,1)\) distribution and \(Y\) a \(\operatorname{Beta}(3,1)\) distribution. Their densities on \((0,1)\) are \(f_X(x)=2x\) and \(f_Y(x)=3x^2\). The ratio \(f_Y(x)/f_X(x)=3x/2\) is increasing, so \(X\le_{\mathrm{lr}}Y\).

Use the odds transformation \(T(x)=x/(1-x)\). It is a strictly increasing bijection from \((0,1)\) to \((0,\infty)\), and \(T'(x)=1/(1-x)^2>0\). Its inverse is \(T^{-1}(t)=t/(1+t)\). The transformed ratio, with the common Jacobian canceled, is

\frac{f_{T(Y)}(t)}{f_{T(X)}(t)} =\frac{f_Y(t/(1+t))}{f_X(t/(1+t))} =\frac{3}{2}\frac{t}{1+t},\qquad t>0.

Differentiating gives \(\frac{3}{2(1+t)^2}>0\). Hence the odds-transformed variables retain the likelihood-ratio order. The ratio increases toward \(3/2\) as \(t\to\infty\), which is consistent with being nondecreasing; it need not increase without bound.

When the Direction Changes

Increasingness is essential to the monotonicity step. Under a smooth, strictly decreasing bijection \(T\) with \(T'(x)<0\), the change-of-variables formula uses \(|T'(T^{-1}(y))|\). This absolute Jacobian still cancels in the density ratio, leaving \(R(T^{-1}(y))\). But now \(T^{-1}\) is decreasing: as \(y\) increases, \(T^{-1}(y)\) decreases. Therefore a nondecreasing original ratio becomes nonincreasing in the transformed argument. A decreasing transformation generally reverses the likelihood-ratio comparison rather than preserving it in the same direction.

Common mistake: “The Jacobians cancel, so likelihood-ratio order is preserved under any one-to-one transformation.” The cancellation only identifies the transformed ratio as \(R(T^{-1}(y))\). Its monotonicity also depends on whether \(T^{-1}\) is increasing or decreasing. For the theorem above, the transformation must be increasing.

There is a further regularity point. A strictly increasing differentiable function need not have a strictly positive derivative everywhere; for example, its derivative can vanish at some points. The density formula using an ordinary differentiable inverse is directly justified by the stated condition \(T'(x)>0\). If a transformation has vanishing derivative, preservation may still be established under broader change-of-variables conditions, but the proof must address those conditions rather than assume the inverse formula automatically applies.

Why the Result Is Useful

The theorem lets a comparison be made on whichever scale makes the densities easiest to understand. A ratio can be simple on the log scale, an odds scale, or another monotone scale, while the corresponding distributions on the original scale have less familiar density formulas. Once a likelihood-ratio comparison is established on one such scale, a common smooth increasing bijection carries it to the other without a new monotonicity calculation from scratch.

The key is to separate the argument into two checks: the Jacobian cancels because both distributions use the same transformation, and monotonicity is retained because the transformation preserves the order of points. If either condition is missing—for example, if the transformations differ or the common map is decreasing—the conclusion does not follow from this theorem.

Check Your Understanding

For each question, distinguish the density-ratio calculation from the monotonicity argument.

Suppose \(f_Y(x)/f_X(x)=e^{2x}\) on a common interval and \(T(x)=3x+1\). Write the transformed density ratio as a function of \(y\), and state why it is nondecreasing.
For the normal example, identify the Jacobian factor in each transformed density and show why it disappears from their ratio.
If \(f_Y(x)/f_X(x)\) is strictly increasing and \(T\) is strictly decreasing and differentiable with nonzero derivative, what monotonicity does the transformed ratio have?
Why does the theorem specify \(T'(x)>0\), rather than merely saying that \(T\) is differentiable and strictly increasing?
Does a nondecreasing transformed ratio have to be strictly increasing or unbounded? Explain using the beta odds example.