# How does this imply that a Pareto optimum maximizes a weighted average of utility functions?

I'm reading a passage from Asset Pricing and Portfolio Choice Theory by Kerry Back, and I don't understand some of it. I would appreciate any help anyone could provide me.

In the passage, Back is proving that a Pareto optimum maximizes a weighted average of utility functions, and we are assuming that the utility functions are concave (representing risk averse investors).

From the text: "Consider an allocation $$(\tilde{w_1}', ..., \tilde{w_H}')$$. Define $$\bar{u_h} = E[u_h(\tilde{w_h}')]$$. If the allocation is Pareto optimal, then it must solve $$\max_{\tilde{w_2}, ..., \tilde{w_H}} E[u_1(\tilde{w_m} - \sum_{h=2}^H \tilde{w_h})]$$ subject to $$E[u_h(\tilde{w_h})] \geq \bar{u_h}$$ for $$h = 2, ..., H$$. (4.1)" Here, $$\tilde{w_m}$$ is the market (total) wealth, i.e. the sum of the wealths of the $$H$$ individual investors, and it's a random variable. I don't see why this statement is true. Couldn't it be that there's some allocation that is Pareto optimal that isn't the one that maximizes the first investor's expected utility? My understanding was that Pareto optimal just meant that there was no way to make one person better off without making another person worse off.

More from the text: "The Lagrangean for this problem is $$\max_{\tilde{w_2}, ..., \tilde{w_H}} E[u_1(\tilde{w_m} - \sum_{h=2}^H \tilde{w_h})] + \sum_{h=2}^H \lambda_h E[u_h(\tilde{w_h})] - \sum_{h=2}^H \lambda_h \bar{u_h}$$. Because of concavity, the optimum for (4.1) maximizes the Lagrangean." I don't see what concavity has to do with this or why it is that the optimum for (4.1) also maximizes the Lagrangean.

More from the text: "Taking $$\lambda_1 = 1$$ shows that a Pareto-optimal allocation must solve $$\max \sum_{h=1}^H \lambda_h E[u_h(\tilde{w_h})]$$ subject to $$\sum_{h=1}^H \tilde{w_h} = \tilde{w_m}$$ (4.2) for some $$(\lambda_1, ..., \lambda_H)$$." Of course, I see that this is the conclusion we want, but I don't see how it follows from the previous step. There wasn't explicitly a $$\lambda_1$$ in the previous step, so where are we taking $$\lambda_1 = 1$$? In the previous step, the indexing of the constraint terms with the lagrange multipliers started at 2, not 1. And how are we showing that the Pareto optimum solves this maximization problem? I don't see how this follows from the previous step.

Thank you for any help! I'd really appreciate it!

## 1 Answer

Couldn't it be that there's some allocation that is Pareto optimal that isn't the one that maximizes the first investor's expected utility? My understanding was that Pareto optimal just meant that there was no way to make one person better off without making another person worse off.

Let me consider the case of a simple resource economy such that we don't have to bother with random variables or expected utilities. However the arguments carry directly over to that case also.

Let $$W$$ be the total amount of endowment available. The following gives the definition of a feasible allocation.

Definition: we say that $$(w_1, \ldots, w_H) \in \mathbb{R}^{nH}_+$$ is a feasible allocation if $$\sum_{i = 1}^H w_i = W$$, and we denote by $$F$$ the set of all feasible allocations.

Now, assume that each individual has a strictly increasing and continuous utility function $$u_i: \mathbb{R}^n_+ \to \mathbb{R}$$.

Definition: A feasible allocation $$(w_1, \ldots, w_H) \in F$$ is Pareto optimal if there does not exist another feasible allocation $$(w_1',\ldots, w_H') \in F$$ such that for all $$i = 1,\ldots, H$$, $$u_i(w_i) \ge u_i(w_i'),$$ with at least one inequality strict.

Consider the set of all utility values that can be reached by the individuals $$2$$ up to $$H$$ for some feasible allocation. $$\overline{U} = \{(u_2(w_2), \ldots, u_H(w_H)) \in \mathbb{R}^{H-1}|\exists (w_1, w_2,\ldots, w_H) \in F)\}.$$ Fix values $$(\overline{u}_2,\ldots, \overline{u}_{H}) \in \overline{U}$$ and consider the problem of maximising the utility of individual 1 subject to the constraint that the utilities of all other agents should be at least equal to their utility in the vector $$\overline{U}$$: \begin{align*} \max_{w_1, \ldots, w_H} &u_1(w_1), \tag{1}\\ \text{ subject to } &\sum_{i = 1}^H w_i = W,\\ & u_h(w_h) \ge \overline{u}_h, \forall h = 2,\ldots, H \end{align*} In some sense, this problem fixed the utility values of all but one individual at $$\overline{u}_h$$. Then it looks for the allocation that maximizes the utility of the first individual. Any such solution should be Pareto optimal, as otherwise individual 1 should be able to get a higher utility. This is shown in the following theorem.

Theorem Assume the utility functions $$u_1(.), \ldots, u_H(.)$$ are strictly monotone and continuous. Then $$(w_1^\ast, \ldots, w_H^\ast) \in F$$ is Pareto optimal if and only if there exist values $$(\overline{u}_2,\ldots, \overline{u}_H) \in \overline{U}$$ such that $$(w_1^\ast, \ldots, w_H^\ast)$$ solves $$(1)$$.

Proof: Let $$(w_1^\ast, \ldots, w_H^\ast) \in F$$ be Pareto optimal. Define $$\overline{u}_h = u_h(w_h^\ast)$$ for all $$h = 2,\ldots, H$$. We have that $$(\overline{u}_2,\ldots, \overline{u}_H) \in \overline{U}$$. Then if $$(w_1^\ast, \ldots, w_H^\ast)$$ does not solve $$(1)$$, there must be an allocation $$(w_1, \ldots, w_H) \in F$$ such that for $$h \in \{2,\ldots, H\}$$: $$u_h(w_h) \ge \overline{u}_h = u_h(w_h^\ast)$$ and: $$u_1(w_1) > u_1(w_1^\ast).$$ However, this implies that $$(w_1, \ldots, w_H)$$ Pareto dominates the allocation $$(w_1^\ast, \ldots, w_H^\ast)$$, a contradiction.

For the reverse, assume that $$(w_1^\ast,\ldots, w_H^\ast) \in F$$ solves $$(1)$$ and assume, towards a contradiction that there is an allocation $$(w_1, \ldots, w_H) \in F$$ that Pareto dominates $$(w_1^\ast, \ldots, w_H^\ast)$$. This gives that for all $$h \in \{2,\ldots, H\}$$: $$u_h(w_h) \ge u_h(w^\ast_h) \ge \overline{u}_h,$$ and $$u_1(w_1) \ge u_1(w_1^\ast),$$ with at least one inequality strict. If the strict inequality is for agent $$1$$, then we see that $$(w_1, \ldots, w_H)$$ improves upon the solution $$(w_1^\ast, \ldots w_H^\ast)$$ for problem $$(1)$$ which is a contradiction. As such, assume that $$u_k(w_k) > u_h(w_k^\ast) \ge \overline{u}_k$$ for some $$k \ne 1$$.

Given this, by strict monotonicity, there must be at least one good in $$w_k$$ whose amount is strictly above the amount in $$w_k^\ast$$. Let $$j$$ be this good and let $$1_j$$ be the vector of all zeros except at position $$j$$.

Now take a $$\varepsilon > 0$$ small enough such that: $$u_k(w_k - \varepsilon 1_j) \ge \overline{u}_k,$$ Such $$\varepsilon> 0$$ can be found as $$u_k$$ is continuous. Then define $$(\hat w_1, \ldots, \hat w_H) \in F$$ such that $$\hat w_1 = w_1 + \varepsilon 1_j$$, $$\hat w_k = w_k - \varepsilon 1_j$$ and $$\hat w_h = w_h$$ for all $$h \in \{2,\ldots, H\} \setminus \{k\}$$. Then we have that for all $$h \in \{2,\ldots H\}$$: $$u_h(\hat w_h) \ge \overline{u}_h,$$ and, $$u_1(\hat w_1) > u_1(w_1) \ge u_1(w_1^\ast)$$ this means that $$(\hat w_1,\ldots, \hat w_H)$$ provides a better solution to $$(1)$$ compared to $$(w_1^\ast, \ldots, w_H^\ast)$$ which contradicts optimality of the latter. $$\square$$

Because of concavity, the optimum for (4.1) maximizes the Lagrangean." I don't see what concavity has to do with this or why it is that the optimum for (4.1) also maximizes the Lagrangean.

The Langrangian first order conditions are sufficient if the objective function is concave and the constraints produce a convex set. This is true if the utility functions are concave. If you don't know how to get from an optimization problem to the Lagrangian (and vice versa), there are many resources online.

In the previous step, the indexing of the constraint terms with the lagrange multipliers started at 2, not 1. And how are we showing that the Pareto optimum solves this maximization problem? I don't see how this follows from the previous step.

Consider the Lagrangian corresponding to $$(1)$$: $$(w_1^\ast, \ldots, w_H^\ast) \in \arg\max_{w_1, \ldots, w_H} u_1(w_1) + \sum_{i = 2}^H \lambda^i [u_i(w_i) - \overline{u}_h] + \mu\left[ \sum_{i = 1}^H w_i - W\right]$$ Here I explicitly added the constraint set to the problem. In your case, it is already substituted for into the utility function of the first individual.

Now, let us define $$\lambda_1 = 1$$. Then we can rewrite the Lagrangian optimisation problem as: $$(w_1^\ast, \ldots, w_H^\ast) \in \arg\max_{w_1, \ldots, w_H} \sum_{i = 1}^H \lambda_i u_i(w_i) + \mu \left[\sum_{i = 1}^H w_i - W\right] - \sum_{i = 2}^H \lambda_i \overline{u}_h.$$ The last sum is a constant term. So deleting it from this problem does not change the optimal solution. This gives: $$(w_1^\ast, \ldots, w_H^\ast) \in \arg\max_{w_1, \ldots w_H} \sum_{i = 1}^H \lambda_i u_i(w_i) + \mu\left[\sum_{i = 1}^H w_i - W\right].$$ Now, fixing the values $$(\lambda_1,\ldots, \lambda_H)$$, this is the Lagrangian corresponding to the following optimization problem: \begin{align*} \max_{w_1,\ldots, w_H} &\sum_{i = 1}^H \lambda_i u_i(w_i),\\ \text{ subject to: } & \sum_{i = 1}^H w_i = W. \end{align*} This problem maximises a weighted sum of utilities subject to the resource constraint. This is in fact a more general result. If all utility functions are continuous strictly monotonic and concave then every Pareto efficient allocation results from the maximisation of some weighted sum of the utilities of all the individuals in the economy. The proof of this relies on a supporting hyperplane argument.