Why Transform Utility?
A utility function assigns numbers to bundles so that the numbers represent a consumer’s preferences. But those numbers are not unique: different numerical scales can describe the same rankings. This matters whenever a problem asks whether a proposed change to a utility function preserves the preferences it represents. The essential question is whether the transformation preserves the order of every pair of utility values.
The earlier material on ordinal and cardinal utility distinguishes the ranking information in utility from the numerical scale used to represent it. Here we make that distinction precise. The result is useful for checking whether a transformed function is an equally valid representation, including transformations such as positive rescaling, translation, and logarithms. This topic is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus.
“Strictly increasing” means that whenever \(r_1<r_2\) are utility values in the relevant range, \(f(r_1)<f(r_2)\). The condition concerns the transformation’s behaviour over the utility values the consumer can actually attain. A transformation need not be defined outside that range, but it must be defined throughout it.
If \(u(x)>u(y)\), the consumer strictly prefers \(x\) to \(y\), and a strictly increasing transformation preserves that strict ranking. If \(u(x)=u(y)\), the consumer is indifferent between the bundles, and the transformed values remain equal. Thus the same bundles are tied with one another before and after transformation. The numbers assigned to those bundles may change substantially; the preference comparisons do not.
Why Strictly Increasing Transformations Preserve Rankings
The result follows from the way a strictly increasing function preserves order. If \(u(x)>u(y)\), strict increase gives \(f(u(x))>f(u(y))\). If \(u(x)=u(y)\), applying the same function to both sides gives \(f(u(x))=f(u(y))\). These facts show that every original strict ranking and tie remains after transformation.
The reverse comparison also matters: the transformation must not create a new ranking or tie that contradicts the original one. Since a strictly increasing function is one-to-one on its domain, equal transformed values can only come from equal original values. More generally, if \(u(x)<u(y)\), then \(f(u(x))<f(u(y))\), so a transformed comparison cannot reverse the original order. In compact form:
Let their utility values be \(u(x)\) and \(u(y)\).
If \(u(x)>u(y)\), then \(f(u(x))>f(u(y))\); if \(u(x)=u(y)\), their transformed values are equal.
Every strict preference and every indifference between bundles is preserved. Therefore the transformed function represents the same preferences.
Positive Rescaling
Multiplying every utility value by the same positive number is a strictly increasing transformation. If \(c>0\), then \(f(r)=cr\) is strictly increasing: for \(r_1<r_2\), \(cr_1<cr_2\). Multiplication changes the numerical gaps between utility values but does not change which value is larger or whether two values are equal.
Worked Example: Rescaling a Utility Function
A consumer’s utility function is \(u(x,y)=x+2y\), where \(x\) and \(y\) are quantities of two goods. Compare bundles \(A=(2,3)\), \(B=(4,1)\), and \(C=(1,2)\). Then consider the transformed function \(v(x,y)=4u(x,y)\). Determine the rankings and ties under both functions.
Method. Calculate the original utility for each bundle, then multiply each result by the same positive factor. Compare the resulting values rather than comparing the bundle coordinates separately.
For \(A=(2,3)\), \(u(A)=2+2(3)=2+6=8\). For \(B=(4,1)\), \(u(B)=4+2(1)=4+2=6\). For \(C=(1,2)\), \(u(C)=1+2(2)=1+4=5\). Thus \(8>6>5\), so \(A\) is preferred to \(B\), and \(B\) to \(C\).
Now rescale each utility value: \(v(A)=4(8)=32\), \(v(B)=4(6)=24\), and \(v(C)=4(5)=20\). The transformed values satisfy \(32>24>20\), so the ranking remains \(A\) above \(B\) above \(C\). There are no ties in either set of values.
Check. The multiplying factor is \(4>0\), and each transformed value is exactly four times its original value. The ranking is unchanged, as required. The common mistake to avoid is thinking that the larger numerical gaps after rescaling represent changed preferences; the comparisons, not the size of the gaps, determine the ranking.
Translation
Adding the same constant to every utility value also preserves preferences. For any real constant \(b\), the transformation \(f(r)=r+b\) is strictly increasing because \(r_1<r_2\) implies \(r_1+b<r_2+b\). The constant may be positive or negative. Consequently, utility values can become negative without changing the preference representation.
Worked Example: Translating Utility Values
For three bundles \(D\), \(E\), and \(F\), a utility function assigns \(u(D)=5\), \(u(E)=5\), and \(u(F)=2\). A second function assigns each bundle \(v=u-8\). Find the transformed values and state all the preference comparisons.
Method. Subtract \(8\) from each utility value, then compare the results using the usual order on numbers.
For \(D\), \(v(D)=5-8=-3\). For \(E\), \(v(E)=5-8=-3\). For \(F\), \(v(F)=2-8=-6\). Since \(-3=-3\), \(D\) and \(E\) remain tied. Since \(-3>-6\), both \(D\) and \(E\) are strictly preferred to \(F\).
Check. The transformation adds the same constant, \(-8\), to each original value. It keeps the tie between \(D\) and \(E\) and preserves their advantage over \(F\). A negative utility value is not, by itself, evidence of a worse preference representation; it is the ordering of values that matters.
Logarithms and Their Domain
The natural logarithm is strictly increasing for positive inputs. Therefore, if \(u(x)>0\) for every bundle under consideration, transforming utility to \(v(x)=\ln(u(x))\) preserves the rankings and ties represented by \(u\). The positivity condition is essential: the real-valued natural logarithm is not defined for zero or negative inputs.
Worked Example: Applying a Logarithm to Positive Utility
Suppose \(u(x,y)=xy\) for bundles with positive quantities of both goods. Compare \(G=(2,8)\), \(H=(4,3)\), and \(J=(1,12)\) under \(u\) and under \(v(x,y)=\ln(u(x,y))\). Report logarithmic values to four decimal places.
Method. First calculate the utility products. Check that they are positive, then take the natural logarithm of each product and compare the results.
For \(G\), \(u(G)=2(8)=16\). For \(H\), \(u(H)=4(3)=12\). For \(J\), \(u(J)=1(12)=12\). All utility values are positive, so the logarithms are defined. The original values show \(G\) strictly preferred to both \(H\) and \(J\), with \(H\) and \(J\) tied.
The transformed values are \(v(G)=\ln(16)\approx2.7726\), \(v(H)=\ln(12)\approx2.4849\), and \(v(J)=\ln(12)\approx2.4849\). The log values retain the same ranking and tie. Values are rounded only at the final step, to four decimal places.
Check. The inputs to each logarithm are positive, and the transformed values confirm the original comparisons. The common mistake to avoid is taking the logarithm of a utility value that is zero or negative; that operation is undefined over the real numbers and cannot serve as this transformation.
Transformations That Do Not Preserve Preferences
The requirement is not simply that a transformation “changes utility consistently.” It must be strictly increasing on the relevant range. A strictly decreasing transformation reverses every strict ranking. A merely nondecreasing transformation can collapse distinct utility levels into a tie, so it may lose preference information. A transformation that is not monotonic may reverse some comparisons while leaving others intact.
| Transformation | Condition or issue | Effect on preferences |
|---|---|---|
| \(f(u)=cu\) | \(c>0\) | Preserves rankings and ties |
| \(f(u)=u+b\) | Any real constant \(b\) | Preserves rankings and ties |
| \(f(u)=\ln(u)\) | All relevant utility values must be positive | Preserves rankings and ties on its domain |
| \(f(u)=-u\) | Strictly decreasing | Reverses strict rankings |
| \(f(u)=\max(u,0)\) | Nondecreasing but not strictly increasing | Can turn distinct utility values into a tie |
For example, suppose \(u(A)=3\) and \(u(B)=1\). The incorrect calculation \(v=-u\) gives \(v(A)=-3\) and \(v(B)=-1\). Since \(-3<-1\), the transformed function ranks \(B\) above \(A\), reversing the original ranking. The correction is to recognize that \(f(u)=-u\) is decreasing, not increasing, so it does not preserve the original preferences.
A related trap is to accept a non-strictly increasing transformation without checking whether it merges distinct utility values. If \(u(A)=1\) and \(u(B)=2\), then \(f(u)=\max(u,0)\) gives \(f(u(A))=1\) and \(f(u(B))=2\); this pair remains distinct. But if \(u(A)=-2\) and \(u(B)=-1\), the same transformation gives \(0\) for both. It has turned a strict preference into indifference. Nondecreasing alone is therefore not enough to guarantee an equivalent representation.
Practical Checks
When a problem proposes a new utility function, a short sequence of checks prevents most errors:
- Express the new function as \(f(u)\), if possible, and identify the utility values on which \(f\) will be used.
- Check whether \(f\) is strictly increasing throughout that range. Do not infer this merely from a few sample calculations.
- For transformations with restricted domains, such as logarithms, verify that every relevant input is in the domain.
- Test both strict rankings and ties. Equal original values must stay equal, and distinct ordered values must not be reversed or collapsed.
- Do not interpret the numerical size or sign of transformed utility values as a preference comparison apart from their order.
These checks also clarify what a transformation does not establish. It changes the numerical representation, not the underlying preference relation. The same bundle comparisons and indifference relationships remain when the transformation is strictly increasing. This conclusion does not depend on the transformation being linear or differentiable; it follows from preserving order on the utility range.
Check Your Understanding
For each problem, compare the original utility values with the proposed transformed values. Answers are included for checking.
- A utility function gives bundles \(A\), \(B\), and \(C\) values \(7\), \(4\), and \(4\). Under \(v=3u\), calculate all three transformed values and state the ranking and tie. (Answer: \(21,12,12\); \(A\) is preferred to \(B\) and \(C\), which remain tied.)
- Two bundles have utility values \(-1\) and \(3\). Apply \(v=u+5\), then state whether the strict ranking changes. (Answer: The transformed values are \(4\) and \(8\); the bundle originally valued at \(3\) remains preferred.)
- A positive-valued utility function assigns \(9\), \(4\), and \(1\) to bundles \(P\), \(Q\), and \(R\). Apply \(v=\ln(u)\), reporting four decimal places, and state the ranking. (Answer: \(2.1972\), \(1.3863\), and \(0.0000\); \(P\) is preferred to \(Q\), which is preferred to \(R\).)
- For bundles \(M\) and \(N\), let \(u(M)=-4\) and \(u(N)=-2\). Apply \(f(u)=\max(u,0)\). Does this transformation preserve their strict ranking? (Answer: No. Both transformed values equal \(0\), so the strict ranking is collapsed into a tie.)
- A utility function assigns \(u(S)=6\) and \(u(T)=2\). One proposed transformation is \(f(u)=5-u\). Compute the transformed values and decide whether it is an equivalent representation. (Answer: \(f(u(S))=-1\) and \(f(u(T))=3\); it is not equivalent because the decreasing transformation reverses the ranking.)