Why Use Different Utility Representations?
A utility function assigns a number to each consumption bundle. For an ordinal representation, the number describes a bundle’s rank relative to other bundles; its numerical size is not, by itself, a measure of how much satisfaction the bundle provides. This distinction allows the same preferences to be written using different utility functions. In this tutorial, the comparison is between a logarithmic representation and positive power transformations of Cobb–Douglas utility.
The comparison is useful because the formulas can look quite different. Utility values and marginal utilities may change substantially across representations. Yet, when the transformation is strictly increasing, the ranking of bundles is unchanged. For differentiable representations, the marginal rate of substitution (MRS) is also unchanged, because the transformation scales both marginal utilities by the same factor.
Assume two goods have quantities \(x>0\) and \(y>0\), and consider the positive Cobb–Douglas function \(U(x,y)=x^\alpha y^\beta\), where \(\alpha>0\) and \(\beta>0\). These conditions ensure that \(U\) is positive and differentiable for the bundles considered. The exponents need not sum to one; choosing exponents that sum to one is a common normalization, not a requirement for the results below.
Rankings Under Logarithmic and Power Transformations
The logarithmic form can make comparisons easier because a product becomes a sum. For example, comparing two values of \(x^\alpha y^\beta\) can instead be done by comparing \(\alpha\ln x+\beta\ln y\). The power form changes the scale and exponents, but not the ranking when \(r>0\). The following example checks the rankings directly under all three functions.
Worked Example: Checking Three Bundle Rankings
Let \(U(x,y)=x^{1/2}y^{1/2}\), with bundles \(A=(1,25)\), \(B=(4,4)\), and \(C=(9,1)\). Compare them using \(U\), the logarithmic representation \(L=\ln U\), and the power representation \(V=U^2\). State the method and check that the rankings agree.
Method. Evaluate each representation at each bundle, then order the resulting values. All quantities are positive, so both transformations are defined.
For \(A=(1,25)\), \(U(A)=1^{1/2}\times25^{1/2}=1\times5=5\). Thus \(L(A)=\ln5\approx1.6094\), and \(V(A)=5^2=25\).
For \(B=(4,4)\), \(U(B)=4^{1/2}\times4^{1/2}=2\times2=4\). Thus \(L(B)=\ln4\approx1.3863\), and \(V(B)=4^2=16\).
For \(C=(9,1)\), \(U(C)=9^{1/2}\times1^{1/2}=3\times1=3\). Thus \(L(C)=\ln3\approx1.0986\), and \(V(C)=3^2=9\).
Under \(U\), the order is \(A>B>C\), since \(5>4>3\). The logarithmic values also satisfy \(1.6094>1.3863>1.0986\), and the power values satisfy \(25>16>9\). Therefore, all three representations rank the bundles \(A>B>C\). The method avoids the mistake of treating utility values from different representations as directly comparable cardinal measures.
Check. The transformed values are \(\ln U\) and \(U^2\) for the same positive \(U\) values, so their order matches the original order.
A positive power changes the exponents in the Cobb–Douglas expression: \(U^r=x^{r\alpha}y^{r\beta}\). It does not change the ratio of the exponents, because \((r\alpha)/(r\beta)=\alpha/\beta\). As the next section shows, that ratio is important for comparing MRS values. A zero power would produce the constant function \(U^0=1\), which does not preserve distinct rankings. A negative power is decreasing in \(U\) and reverses strict rankings. The condition \(r>0\) is essential.
Marginal Utilities and MRS
For a differentiable utility function, the marginal utility of a good is the partial derivative with respect to that good. For the Cobb–Douglas representation \(U=x^\alpha y^\beta\), these derivatives are \(U_x=\alpha x^{\alpha-1}y^\beta\) and \(U_y=\beta x^\alpha y^{\beta-1}\). The MRS of \(x\) for \(y\), using the positive ratio convention, is \(U_x/U_y\). It measures the local trade-off along an indifference curve, rather than the absolute numerical level of utility.
The cancellation in this derivative ratio is why the MRS has a particularly simple form. It depends on the relative exponents and the quantities of the two goods; the overall level of utility cancels out.
For the logarithmic representation, \(\phi(U)=\ln U\), so \(\phi'(U)=1/U>0\). For the power representation, \(\phi(U)=U^r\), so \(\phi'(U)=rU^{r-1}>0\) when \(U>0\) and \(r>0\). In either case, the same positive factor multiplies both marginal utilities.
Why the MRS Is Preserved
The chain rule gives a direct justification. If a differentiable transformation depends on \(x\) and \(y\) only through \(U(x,y)\), differentiating it with respect to either good introduces the same outer derivative \(\phi'(U)\). Dividing the resulting marginal utilities cancels that common factor. Thus a positive increasing transformation changes the numerical scale of marginal utility, but not the local trade-off represented by their ratio.
This result applies where the functions are differentiable and the denominator marginal utility is nonzero. In the positive Cobb–Douglas model with \(\alpha,\beta>0\) and \(x,y>0\), both marginal utilities are positive, so the ratio is well-defined. A transformation that is not strictly increasing may fail to preserve rankings even if a ratio of its derivatives happens to cancel.
Worked Example: Comparing Marginal Utilities and MRS
Let \(U(x,y)=x^{1/2}y^{1/2}\), and evaluate the original, logarithmic, and squared-power representations at \((x,y)=(4,9)\). Find each pair of marginal utilities and its MRS. The method is to differentiate each representation and then take the ratio; the common mistake to avoid is assuming that marginal utilities themselves must match.
First, \(U(4,9)=4^{1/2}9^{1/2}=2\times3=6\). For the original representation,
Its MRS is \(U_x/U_y=(3/4)/(1/3)=9/4=2.25\).
For the logarithmic representation, \(L=\frac12\ln x+\frac12\ln y\). Therefore \(L_x=(1/2)/x=1/8=0.125\), and \(L_y=(1/2)/y=1/18\approx0.0556\). Its MRS is \((1/8)/(1/18)=18/8=9/4=2.25\).
For the squared-power representation, \(V=U^2=xy\). Thus \(V_x=y=9\) and \(V_y=x=4\). Its MRS is \(9/4=2.25\).
The marginal utilities are different across the three representations: for example, the \(x\)-marginal utility is \(0.75\), \(0.125\), and \(9\), respectively. However, each MRS is \(2.25\). This is the expected check: each representation describes the same local trade-off at the bundle.
Power Exponents and the MRS
A positive power transformation can make the exponents look very different, so it is useful to verify the MRS using the transformed exponents. If \(V=x^{r\alpha}y^{r\beta}\), its marginal-utility ratio is \((r\alpha)/(r\beta)\times y/x\). The factor \(r\) cancels, giving exactly \((\alpha/\beta)(y/x)\). This calculation is equivalent to the chain-rule argument, but it makes the exponent cancellation explicit.
Worked Example: Checking the Exponents After a Power Transformation
Let \(U(x,y)=x^{0.3}y^{0.7}\), choose \(V=U^3\), and evaluate both marginal-utility pairs at \((4,1)\). Confirm that the MRS agrees. This method differentiates the two displayed Cobb–Douglas forms separately; it avoids omitting the quantity factor \(y/x\) when calculating MRS.
First, \(U(4,1)=4^{0.3}\times1^{0.7}=4^{0.3}\approx1.5157\). The power representation is \(V=U^3=x^{0.9}y^{2.1}\), and \(V(4,1)=4^{0.9}\times1^{2.1}\approx3.4822\).
For \(U\), the marginal utilities are \(U_x=0.3U/x=0.3(1.5157)/4\approx0.1137\) and \(U_y=0.7U/y=0.7(1.5157)/1\approx1.0610\). Their ratio is approximately \(0.1137/1.0610\approx0.1071\). Using the exact exponents and quantities gives \((0.3/0.7)(1/4)=3/28\approx0.1071\).
For \(V=x^{0.9}y^{2.1}\), \(V_x=0.9V/x=0.9(3.4822)/4\approx0.7835\), and \(V_y=2.1V/y=2.1(3.4822)/1\approx7.3126\). Their ratio is approximately \(0.7835/7.3126\approx0.1071\). Exactly, it is \((0.9/2.1)(1/4)=3/28\approx0.1071\). The MRS agrees under both representations.
A common incorrect calculation would stop at \(0.9/2.1=3/7\), omitting \(y/x\). At this bundle \(y/x=1/4\), so the correct MRS is \((3/7)(1/4)=3/28\), not \(3/7\). The final check is that both differentiated representations give the same ratio.
Common Mistakes and Practical Checks
When comparing utility representations, keep the preference ranking separate from the numerical scale. The following checks help catch errors:
- Check the domain. The logarithm requires \(U>0\). For the stated Cobb–Douglas model, \(x>0\), \(y>0\), and positive exponents ensure this. Do not apply the logarithmic representation at a bundle where its inputs make \(U\) zero or undefined.
- Check that the transformation is increasing. The power representation \(U^r\) preserves rankings only for \(r>0\). If \(r<0\), a larger positive \(U\) produces a smaller \(U^r\); if \(r=0\), all bundles receive the same value.
- Do not expect marginal utilities to be equal. They depend on the chosen representation and its numerical scale. Compare the MRS ratio, not the marginal utility values one at a time.
- Keep the quantity ratio in the MRS. For Cobb–Douglas utility, the MRS is \((\alpha/\beta)(y/x)\), not just the ratio of exponents.
- Use one representation consistently in a calculation. Utility values from one representation and marginal utilities from another should not be combined as if they were generated by a single function.
Check Your Understanding
For each problem, check both the representation and the bundle conditions before comparing values or derivatives.
- Let \(U=x^{1/4}y^{3/4}\). Compare bundles \(A=(1,16)\) and \(B=(4,4)\) under \(U\), \(\ln U\), and \(U^2\). Give the ranking and the three utility values under each representation. (Answer: \(U(A)=8>U(B)=4\); \(\ln U(A)=\ln8\approx2.0794>\ln4\approx1.3863\); \(U(A)^2=64>16=U(B)^2\), so \(A>B\) under all three.)
- For \(U=x^{0.4}y^{0.6}\) at \((2,8)\), calculate the marginal utilities of \(L=\ln U\) and its MRS. (Answer: \(L_x=0.4/2=0.2000\), \(L_y=0.6/8=0.0750\), and \(MRS=0.2000/0.0750=8/3\approx2.6667\).)
- Let \(U=x^{0.2}y^{0.8}\), \(V=U^2\), and \((x,y)=(4,1)\). Write \(V\) using its transformed exponents and calculate its MRS. (Answer: \(V=x^{0.4}y^{1.6}\); \(MRS=(0.4/1.6)(1/4)=1/16=0.0625\).)
- For \(U=x^{1/2}y^{1/2}\), compare \(A=(1,16)\) and \(B=(4,4)\) under \(U\), \(\ln U\), and \(U^3\). Are they strictly ranked or tied? (Answer: \(U(A)=U(B)=4\); \(\ln U(A)=\ln U(B)=\ln4\); and \(U(A)^3=U(B)^3=64\). They are tied under all three.)
- Suppose \(U(A)=2\) and \(U(B)=5\), with both positive, and a proposed representation is \(W=U^{-1}\). Does it preserve their ranking? Explain using the transformed values. (Answer: No. \(W(A)=1/2\) and \(W(B)=1/5\), so \(W(A)>W(B)\) even though \(U(A)<U(B)\); the negative power reverses the ranking.)