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Some Examples of Utility Functions
Each of these functions represents preferences which are continuous, strictly monotonic, and convex.

Example 1 : Fixed Coefficients

2 goods


\begin{displaymath}u(x_1,x_2) = \min{(ax_1,bx_2)}\end{displaymath}

where $a$ and $b$ are positive constants ( and where ``$\min$'' means ``the minimum of the two items''. So if $ax_1<bx_2$, then $u(x_1,x_2)=ax_1$, and if $ax_1>bx_2$, then $u(x_1,x_2)=bx_2$. The indifference curves for these preferences are $L$-shaped, with a kink along the diagonal line through the origin with equation $ax_1=bx_2$. ( This line has a slope $a/b$. )

$n$ goods


\begin{displaymath}u(x_1,x_2,\ldots,x_n)=\min{(a_1x_1,a_2x_2,\ldots,a_nx_n)}\end{displaymath}

where each $a_i$ is a positive constant.

This utility function is continuous, although it is not continuously differentiable at the kinks. It is increasing, in the sense that $u(x)> u(x')$ if $x >>x'$ -- although it is not necessarily true that $u(x)> u(x')$ if $x>x'$. The utility function is quasi-concave, since the preferences it represents are convex ( although not strictly convex ). ( This convexity can be seen by drawing a line between any two points on or above some $L$-shaped indifference curve ; the line connecting these points must also be above the indifference curve. )

Example 2 : Perfect Substitutes


\begin{displaymath}u(x_1,x_2,\ldots,x_n)=a_1x_1 + a_2x_2 + \cdots + a_nx_n\end{displaymath}

where the $a_i$'s are all positive constants. Since the gradient of this function,

\begin{displaymath}\nabla u(x) = (a_1,a_2,\ldots,a_n)\end{displaymath}

the preferences the function represents are strictly monotonic if each of the $a_i$'s is positive.The Hessian matrix of second deriatives of the function is just the zero matrix, since all of the $a_i$'s are constants. That means that the function $u(\cdot)$ is not just quasi-concave, it's actually concave. It's actually convex, as well, since it's linear. ( But it's not strictly concave, and the preferences it represents are not strictly convex. )

In 2 dimensions, the indifference curves for these preferences are straight lines, with slope $-a_1/a_2$. In higher dimensions, the indifference surfaces are planes, or hyperplanes.

Example 3 : Quasi-Linear Preferences


\begin{displaymath}u(x)= x_1 + f(x_2,x_3,\ldots,x_n)\end{displaymath}

where $f(x_2,x_3,\ldots,x_n)$ is any increasing, concave ( not just quasi-concave ) function mapping $n-1$-dimensional vectors into real numbers. The partial derivatives of this function are $u_1=1$, and $u_i=f_i$ for any $i>1$. So if $f(\cdot)$ is non-decreasing in all its arguments, then the preferences that this utility function represents are strictly monotonic. The matrix $H(x)$ of second derivatives of this utility function is

\begin{displaymath}H(x)=\pmatrix{0 & 0 & . & 0 \cr
0 & f_{22} & . & f_{2n} \cr
0 & . & . & . \cr
0 & f_{n2} & . & f_{nn} \cr}\end{displaymath}

so that

\begin{displaymath}v'H(x)v = \sum_{i=2}^n \sum_{j=2}^n v_iv_jf_{ij}\end{displaymath}

and $v'Hv \leq 0$ for any direction vector $v$, if the function $f(\cdot)$ is concave.

In two dimensions, the slope of an indifference curve through any bundle $(x_1,x_2)$ is $-1/f_2(x_2)$. Concavity of the function $f(\cdot)$ means that $f_22(x_2) \leq 0$, so that the indifference urve gets steeper as we move up it ( and to the left ). In this case, notice that the slope of an indifference curve is independent of the level of consumption of good 1, since it depends only on $x_2$. That means that if we move right horizontally, the slopes of the indifference curves stay constant.

Example 4 : Cobb-Douglas Preferences

There are several ways of representing the same preferences here. One way is

\begin{displaymath}u(x)=x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}\end{displaymath}

where the $a_i$'s are all positive constants. I could take the natural logarithm of $u(x)$ above, which is a monotonically increasing transformation, to get

\begin{displaymath}U(x)=a_1 \ln{x_1} + a_2\ln{x_2} + \cdots a_n \ln{x_n}\end{displaymath}

For given $a_1, a_2,\ldots,a_n$, the functions $u(x)$ and $U(x)$ represent exactly the same preferences. A third transformation is to take $u(x)$ to the power $1/A$, where

\begin{displaymath}A \equiv a_1 + a_2 + \cdots + a_n\end{displaymath}

to get

\begin{displaymath}\tilde{u}(x) = x_1^{b_1}x_2^{b_2}\cdots x_n^{b_n}\end{displaymath}

where

\begin{displaymath}b_i \equiv {a_i \over A} \equiv {a_i \over a_1 + a_2 + \cdots + a_n}\end{displaymath}

Notice that the new exponents $b_i$ have been constructed so that

\begin{displaymath}b_1 + b_2 + \cdots + b_n = 1\end{displaymath}

which turns out to be a fairly convenient property of the representation $\tilde{u}(\cdot)$.

Why is this property of $\tilde{u}(\cdot)$ so convenient? Suppose that I multiply each element in the consumption bundle $(x_1,x_2,\ldots,x_n)$ by the same constant $k$. Then

\begin{displaymath}\tilde{u}(kx)=\tilde{u}(kx_1,kx_2,\ldots,kx_n)=(kx_1)^{b_1}(k...
...dots k^{b_n}][x_1^{b_1}x_2^{b_2}\cdots
x_n^{b_n}]=k\tilde{u}(x)\end{displaymath}

where, in the above string of equations, I used the fact that $z^az^b=z^{a+b}$, and the fact that the $b_i$'s sum to 1. A function $f(x)$ such that $f(kx)=kf(x)$ for all $x$ is called homogeneous of degree 1. So the particular form $\tilde{u}(x)$ is often used, because this representation is homogeneous of degree 1.

Note that the other functions $u(\cdot)$ and $U(\cdot)$ representing Cobb-Douglas preferences are not homogeneous of degree 1. A function, such as $u(\cdot)$ or $U(\cdot)$, which is a monotonically increasing transformation of a homogeneous-of-degree-1 function is called homothetic. So ``homothetic'' is a property of the underlying Cobb-Douglas preferences. ``Homogeneous of degree 1'' is a property only of one of the functions (  $\tilde{u}(\cdot)$ ) representing those preferences.

Left to the reader : check whether the preferences in examples 1,2 and 3 are homothetic.

To check that Cobb-Douglas preferences are strictly monotonic and convex, it is easiest to use the logarithmic representation $U(\cdot)$. Taking the first partial derivatives,

\begin{displaymath}\nabla U(x) = ({a_1 \over x_1},{a_2 \over x_2},\ldots,{a_n \over x_n})\end{displaymath}

so that all the partial derivatives are positive ( at least, when the consumption levels of each good are positive ) and the preferences are strictly monotonic. Taking derivatives yet again, the Hessian matrix of second derivatives of the function $U(\cdot)$ is

\begin{displaymath}\pmatrix{-{1 \over x_1^2} & 0 & . & 0 \cr
0 & -{1 \over x_2^2} & . & 0 \cr
0 & 0 & . & 0 \cr
0 & 0 & . & -{1 \over x_n^2}\cr}\end{displaymath}

so that $U(\cdot)$ is a concave function, which means that all other representations of these preferences are quasi-concave, and that the preferences themselves are convex.

In two dimensions, the slope of an indifference curve is $-u_1/u_2$, or

\begin{displaymath}MRS = {a_1 \over a_2}{x_2 \over x_1}\end{displaymath}

since

\begin{displaymath}u_1= a_1x_1^{a_1-1}x_2^{a_2}\end{displaymath}


\begin{displaymath}u_2= a_2x_1^{a_1}x_2^{a_2-1}\end{displaymath}

Left to the reader : check that $U_1/U_2=\tilde{u}_1
/\tilde{u}_2 = (a_1/a_2)(x_2/x_1)$.

As we move up and to the left along an indifference curve, $x_1$ falls and $x_2$ rises, so that the curve gest steeper.

Also, the $MRS$, the slope of the indifference curve, depends only on the ratio of consumption of the two goods, $x_1/x_2$. This propperty, that the slope of an indifference curve does not vary as all elements of the consumption bundle are increased by the same prroportion, will hold for any homothetic preferences.

In two dimensions, what this means is that the slope of an indifference curve is unchanged as we move along any diagonal through the origin, since the ratio $x_1/x_2$ is constant along any such diagonal.

Example 5 : Constant Elasticity of Substitution


\begin{displaymath}u(x) = [a_1x_1^{\rho} + a_2x_2^{\rho} + \cdots +
a_nx_n^{\rho]^{1/\rho}}\end{displaymath}

where the $a_i$'s are positive constants, and $\rho$ is a constant, which is less than or equal to 1.

So it's o.k. for $\rho$ to be negative -- but it cannot exceed 1.

Now if $\rho > 0$, I can take a monotonically increasing transform of $u(x)$ by taking it to the power $\rho$, to get

\begin{displaymath}U(x) = a_1x_1^{\rho} + a_2x_2^{\rho} + \cdots + a_nx_n^{\rho}\end{displaymath}

but this only works if $0 < \rho < 1$. If $\rho \leq 0$, then taking $u(x)$ to the power $\rho$ is not a monotonically increasing transformation, so that $U(x)$ will not represent the same preferences.

Using the chain rule to take the partial derivatives of $u(\cdot)$,

\begin{displaymath}u_i = {1 \over \rho}[a_1x_1^{\rho} + a_2x_2^{\rho} + \cdots +
a_nx_n^{\rho}]^{1/\rho-1}(\rho)a_ix_i^{\rho-1}\end{displaymath}

which simplifies ( a little ) to

\begin{displaymath}u_i=[a_1x_1^{\rho} + a_2x_2^{\rho} + \cdots +
a_nx_n^{\rho}]^{1/\rho-1}a_ix_i^{\rho-1}\end{displaymath}

which must be non-negative, so that the preferences represented by $u(\cdot)$ are strictly monotonic.

Left to the reader : $u(kx)=ku(x)$ so that $u(\cdot)$ is homogeneous of degree 1, and the preferences represented by $u(x)$ are homothetic.

Using the other representation $U(x)$,

\begin{displaymath}\nabla U(x) =
\rho(a_1x_1^{\rho-1},a_2x_2^{\rho-2},\ldots,a_nx_n^{\rho-1})\end{displaymath}

( again demonstrating that CES preferences are strictly monotonic ). This means that the matrix of second derivatives of $U$ is

\begin{displaymath}\rho(\rho-1)\pmatrix{a_1x_1^{\rho-2} & 0 & . & 0 \cr
0 & a_2x...
...} & . & 0 \cr
0 & 0 & . & 0 \cr
0 & 0 & . & a_nx_n^{\rho-2}\cr}\end{displaymath}

which makes it a straightforward exercise ( left to the reader ) to show that $U(\cdot)$ is a concave function if $0 < \rho \leq
1$. This expression for the Hessian also shows why the requirement was imposed that $\rho \leq 1$ : if $\rho$ exceeded 1, then preferences would not be convex. In fact, when $\rho=1$, CES preferences become example 2, perfect substitutes, which are just on the borderline of representing convex preferences.

When $\rho<0$, taking $u(\cdot)$ to the power $\rho$ is no longer a monotonically increasing transformation. But letting $\tilde{U}(x)=-[u(x)]^{\rho}$ is a monotonically increasing transformation. [ Why? If $\rho<0$, then the derivative with respect to $a$ of $-a^{\rho}$ is $-\rho a^{\rho-1} > 0$. ] The matrix of second derivatives of $\tilde{U}$ is

\begin{displaymath}-\rho(\rho-1)\pmatrix{a_1x_1^{\rho-2} & 0 & . & 0 \cr
0 & a_2...
...} & . & 0 \cr
0 & 0 & . & 0 \cr
0 & 0 & . & a_nx_n^{\rho-2}\cr}\end{displaymath}

which must be negative definite when $\rho<0$. [ Again, this demonstration is left to the reader. ]

So Constant Elasticity of Substitution preferences are strictly monotonic and convex if $\rho \leq 1$.

A problem : the only restriction imposed on $\rho$ is that it be less than or equal to 1. But when $\rho=0$, the expression for $u(\cdot)$ does not make much sense.

A solution : Look at the slope of the indifference curves, in two dimensions, the $MRS$, When $1 \geq \rho > 0$, then

\begin{displaymath}MRS = {U_1 \over U_2} = {a_1 \over a_2}{x_1^{\rho-1} \over x_2^{\rho
-1}}\end{displaymath}

Left to the reader : This is also the correct expression for the $MRS$ if $\rho<0$ and $\tilde{U}$ is used as a representation of the preferences, or if $\rho \not= 0$ and $u(\cdot)$ is used as the representation of the preferences.

So what happens to this $MRS$ as $\rho$ gets close to zero? This expression approaches

\begin{displaymath}{a_1 \over a_2}{x_2 \over x_1}\end{displaymath}

which is an expression that's been used before here, in the section immediately above. That's the $MRS$ for Cobb-Douglas preferences.

It turns out that Cobb-Douglas preferences are a special case of CES preferences, the limiting case of CES preferences as $\rho$ approaches 0.

This first graph illustrates indifference curves when the elasticity of substitution is greater than 1.

This second graph illustrates indifference curves when the elasticity of substitution is less than 1.

6. Stone-Geary Preferences

These preferences are not defined on the whole of $R_+^n$. They are only defined on consumption bundles for which $x_i \geq s_i$ for each element $i$, where the $s_i$'s are constants, usually described as subsistence levels of consumption. Stone-Geary preferences can be represented by the utility function

\begin{displaymath}u(x) = (x_1-s_1)^{a_1}(x_2-s_2)^{a_2}\cdots(x_n-s_n)^{a_n}\end{displaymath}

where the $a_i$'s are again posiitve constants. [ Warning : These preferences are also discussed in exercise $1.56$ of the text. I am using somewhat different notation than the text : their $a_i$'s are my $s_i$'s, and their $b_i$'s are my $a_i$'s. ]

Notice that Cobb-Douglas preferences are a special case of Stone-Geary preferences : simply set all the subsistence levels $s_i$ equal to 0.I can also use the same monotonically increasing transformations of $u(\cdot)$ as I used for Cobb-Douglas, namely, $U(x) =
\ln{u(x)}$ so that

\begin{displaymath}U(x) = a_1\ln{(x_1-s_1)} + a_2\ln{(a_2-s_2)} + \cdots +
a_n\ln{(a_n-s_n)}\end{displaymath}

and $\tilde{u}(x)=[u(x)]^{1/A}$, with $A \equiv a_1 + a_2 + \cdots + a_n$, so that

\begin{displaymath}\tilde{u}(x) = (x_1-s_1)^{b_1}(x_2-s_2)^{b_2}\cdots(x_n-s_n)^{b_n}\end{displaymath}

where $b_1 + b_2 + \cdots + b_n = 1$.

From the representation $U(x)$, it should be clear [ but is left to the reader ] that Stone-Geary preferences are strictly monotonic on the consumption set $\{x \vert x_i > s_i\}$.

Since the Hessian of $U(x)$ is a diagonal matrix, with elements of the form

\begin{displaymath}-{a_i \over (x_i-s_i)^2}\end{displaymath}

along the diagonal, the function $U(\cdot)$ is concave, so that Stone-Geary preferences are quasi-concave.

But, unlike Cobb-Douglas ( or CES ) preferences, Stone-Geary preferences are not necessarily homothetic.[Left to the reader : what restrictions on the $s_i$'s would make these preferences homothetic? ]

The slope of an indifference curve for these preferences is

\begin{displaymath}MRS = {a_1 \over a_2}{x_2-s_2 \over x_1-s_1}\end{displaymath}

so that, in general, the slopes of the indifference curves are not constant along a diagonal line through the origin. But these slopes are constant along any diagonal line through the ``subsistence point'' $(s_1,s_2)$.

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Sam Bucovetsky 2002-01-14