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Consider an exchange economy with two goods, e.g. home furniture (x) and electrical equipment (y). The interesting thing about these goods is that, when a family owns a bundle, all members of the family enjoy the same bundle (it is like a "club good" but only for the family).

There are two families. In each family, there are different members with different preferences over bundles. Assume that all preferences are monotonically-increasing and strictly convex.

An allocation is a pair of bundles, $(x_1,y_1)$ for family 1 and $(x_2,y_2)$ for family 2.

An allocation is called envy-free if:

  • All members of family 1 believe that $(x_1,y_1)$ is at least as good as $(x_2,y_2)$;
  • All members of family 2 believe that $(x_2,y_2)$ is at least as good as $(x_1,y_1)$.

An allocation is called Pareto-efficient if there is no other allocation of bundles to families such that all members of all families weakly prefer and at least one member of one family strictly prefers.

Under what conditions does a Pareto-efficient envy-free allocation exist?

If each family has a single member, then a Pareto-efficient envy-free allocation exists; this is a famous theorem of Varian. Has this theorem been generalized from individuals to families?

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  • $\begingroup$ Very strong definition of envy-freeness. One would guess you would somehow aggregate the preferences first and then claim that there is no envy according to the aggregated preferences. $\endgroup$ – Giskard Dec 23 '15 at 16:36
  • $\begingroup$ @denesp indeed, I thought about aggregating preferences, e.g. using a social welfare function. But, every selection of such a function would be arbitrary and not sufficiently motivated. $\endgroup$ – Erel Segal-Halevi Dec 23 '15 at 18:16
  • $\begingroup$ @ErelSegal-Halevi Do you want us to also assume that the utility of each member of each family is weakly increasing in the quantity of $x$ and $y$ their family receives? If so, I have a very unsatisfying condition for you under which a Pareto-efficient, envy-free allocation exists: Suppose that, for each family, each member of that family has the same preferences... :P $\endgroup$ – Shane Dec 27 '15 at 19:23
  • $\begingroup$ @Shane weak monotonicity seems like a reasonable assumption. If, in each family, all members have the same preferences, then each family is actually like a single agent, so we are back at the standard setting... $\endgroup$ – Erel Segal-Halevi Dec 28 '15 at 19:16
  • $\begingroup$ What about the case where $x_1 = x_2$ and $y_1 = y_2$? Assuming weak monotonicity, then this must be Pareto and envy-free. From there, we could maybe make some small epsilon changes? $\endgroup$ – Kitsune Cavalry♦ Dec 28 '15 at 20:06
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Right now I'm not sure about the equivalence of the relabeling, and therefore the usefulness of this anwer -- see comments below.

This is the beginning of an answer and an attempt to demonstrate how strong the necessary assumptions would have to be to guarantee existence.

Let's transform the problem into one that's equivalent but a bit easier to work with. Instead of indexing over families, let's instead index over the agents (members of families). The key to this relabeling is understanding that families can be written as constraints: If agents $i$ and $j$ belong to the same family, then $x_i=x_j$ and $y_i = y_j$.

Now we're back in the standard environment with individual agents (not families) but with these familial constraints. Recall the proof of Varian's theorem, which you link in the question. It uses the existence of a competitive equilibrium from equal incomes. In this context, we would need the existence of a competitive equilibrium from equal incomes in which the familial constraints were also met. This is going to be very difficult to do. For instance, consider $i$ and $j$ are in a family, and $$ u_i=x_i + \varepsilon y_i \:\: \text{ and } \:\: u_j = \varepsilon x_j + y_j $$ where $\varepsilon>0$ is tiny. These preferences are monotonic and convex. Basically, one family member cares about $x$ and the other cares about $y$. If each of the two agents is purchasing $x$ and $y$ to maximize his or her utility, you would not expect $x_i^* = x_j^*$ or $y_i^* = y_j^*$ in the competitive equilibrium (see addendum at end).

This is why you certainly need some assumption on preference similarities within families (at least to use a version of Varian's proof). My sense is that if you give me any arbitrarily small difference in preferences between family members, I can construct an example around it where there exists no CEEI in which they choose the same allocation. And then, at the very least, you can't use Varian's proof.

Two questions:

  1. Do you agree that my reformulation of the problem is formally equivalent to yous?
  2. Can you think of any assumption weaker than assuming preference homogeneity within the family that I can try to invalidate with a counter-example?

Addendum: Remember that in a competitive equilibrium, each agent's marginal rate of substitution (MRS) equals the price ratio. Here, my agents have constant and different MRS's, so there can exist no competitive equilibrium with a price ratio that equals both of their MRS's. If each agent has an MRS that varies, then perhaps they could happen to be equal at the equilibrium price ratio. So maybe you could get away with some notion of local homogeneity of familial preferences. But you need to have them be locally homogenous at the competitive equilibrium, which is exactly what you're trying to prove exists, so it would be a bit circular.

Important note: As mentioned previously, I'm assuming that the only way to prove existence is how Varian did it, via CEEI. There may be other proof techniques that skirt these issues, but I suspect not.

Beyond CEEI: As the OP points out in the comments, proving existence of PEEFs through CEEI as Varian does is somewhat restrictive. I do not have a lot to say about proving existence of PEEFs directly, but the following is readily apparent: For any allocation satisfying your Pareto efficiency condition (ignore envy-freeness for the moment), for any $i,j$ such that $x_i, x_j, y_i, y_j > 0$, $$MRS_i = MRS_j$$ If this weren't true, there would be a Pareto improvement. Competitive equilibrium essentially equates the MRS's through the price ratio, but you still need to equate these MRS's just to find a Pareto efficient allocation. I think the familial constraints will make this very difficult -- it's not hard to come up with an environment and familial constraints such that there exists no Pareto efficient equilibrium satisfying those constraints. In any case, this could be another partial step towards an answer: Forget about envy-freeness. First try to come up with an assumption on preferences (and maybe on familial constraints) that guarantees the existence of a Pareto efficient allocation that satisfies familial constraints. Then worry about envy.

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    $\begingroup$ I share your intuition that a CEEI with familial constraints usually does not exist. But, there are many more PEEF allocations than CEEI allocations. In many cases, a CEEI is essentially unique while there are many different PEEFs. As an example (without familial constraints), take $u_1 = 2 x_1 + y_1$ and $u_2 = x_2 + 2 y_2$ and the total endowments are (4,4). The only CEEI allocation is [(4,0) ; (0,4)]. However, the range of PEEF allocations goes from [(3,0); (1,4)] to [(4,1); (0,3)]. CEEI is only a single point in that range. $\endgroup$ – Erel Segal-Halevi Dec 29 '15 at 10:16
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    $\begingroup$ I found in Varian's original paper: sciencedirect.com/science/article/pii/0022053174900751 proofs of existence of PEEF allocations, which do not rely on CEEI and so are valid even in situations in which a CEEI does not exist (the preferences are not strictly convex). So far, I haven't managed to understand these proofs, but they may be relevant. $\endgroup$ – Erel Segal-Halevi Dec 29 '15 at 10:19
  • $\begingroup$ @ErelSegal-Halevi In your example, any allocation in which both agents get strictly positive quantities of both goods is Pareto inefficient, no? I'm struggling to understand your ranges. More generally, though, I agree with you. I've added an additional section on proving PEEFs directly (without CEEI). I don't think you'll find it particularly satisfying, but it's about all that's obvious to me right now. $\endgroup$ – Shane Dec 29 '15 at 16:23
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    $\begingroup$ you are right. The ranges of PEEF allocations in my example are: $[(x_1,0),(4-x_1,4)]$ where $x_1\in [3,4]$, and $[(4,4-y_2),(0,y_2)]$ where $y_2\in [3,4]$ $\endgroup$ – Erel Segal-Halevi Dec 30 '15 at 7:02
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    $\begingroup$ Hmm... All of my indexing is over individuals, even with $x_i,x_j,y_i,y_j$. So I was just saying: "take any two individuals, same family or not, ... " But now I'm wondering whether this relabeling actually does work, because I'm not directly accounting for the fact that if $i$ and $j$ are in the same family, and $x_i = x_j = 1$, then that only uses 1 unit of good $x$, not 2. Now I'm questioning the equivalence of the relabeling. Families are not just a constraint (in that people must share the same goods), they're also a benefit, in that goods are public/shared within the family. $\endgroup$ – Shane Dec 31 '15 at 17:49
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Suppose there are two families: Family U has $n_u$ members, and family V has $n_v$ members. Utility function of member $i$ of family U is: \begin{eqnarray*} u_i(x_u, y_u) = a_ix_u + y_u \end{eqnarray*} where all $a_i$s are positive for all $i\in\{1,2,\ldots, n_u\}$,

and utility function of member $j$ of family V is: \begin{eqnarray*} v_j(x_v, y_v) = b_jx_v + y_v \end{eqnarray*} where all $b_j$s are positive for all $j\in\{1,2,\ldots, n_v\}$.

Also, suppose $\min_i a_i \geq \max_j b_j$.

Suppose the total endowment vector of $X$ and $Y$ is $(\omega_X, \omega_Y)$.

For any $\theta \in [\max_j b_j, \min_i a_i]$, define $m := \displaystyle\frac{\theta\omega_X}{2} + \frac{\omega_Y}{2} $.

Check that if $\displaystyle\frac{m}{\theta} \leq \omega_X$, then $\displaystyle (x_u, y_u) = \left(\frac{m}{\theta}, 0\right)$ and $\displaystyle (x_v, y_v) = \left(\omega_X - \frac{m}{\theta}, \omega_Y\right)$ is Pareto efficient envy free allocation, and on the other hand if $\displaystyle\frac{m}{\theta} > \omega_X$, then $\displaystyle (x_u, y_u) = \left(\omega_X, m-\theta\omega_X\right)$ and $\displaystyle (x_v, y_v) = \left(0, m\right)$ is Pareto efficient envy free allocation.

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  • $\begingroup$ What is the meaning of the requirement $\min_i a_i \geq \max_j b_j$.? $\endgroup$ – Erel Segal-Halevi Feb 16 '17 at 13:41
  • $\begingroup$ All the members of family U have higher MRS then all the members of family V. $\endgroup$ – Amit Feb 16 '17 at 14:18
  • $\begingroup$ I think for 2 families and linear preferences, this requirement can be removed. I have to work on the details yet. $\endgroup$ – Erel Segal-Halevi Feb 16 '17 at 14:54
  • $\begingroup$ I think it will be difficult to remove this requirement because we want allocation to be envy free. The conditions may not appear neat even if somehow it is relaxed. But this result holds for a larger class of utility functions. It will be a good idea to extend the result to include preferences of other type. For example: A version of it can also be proved for Cobb Douglas preferences. $\endgroup$ – Amit Feb 16 '17 at 15:09
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Suppose the preferences of all agents in all families are monotone and convex (the standard assumptions of consumer theory).

Then, a Pareto-efficient envy-free allocation always exists when there are two families. However, it might not exist when there ar three or more families.

Proofs and examples can be found in this working paper.

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The problem statement seems to imply that X and Y cannot be substitutes (an electrical device cannot be used as home furniture).

A Pareto-efficient envy-free allocation exists when:

For at least one agent, at least some goods have negative utility or are complements, and agents can chose to not consume.

Example:

  1. Agents A and B are in family F1.
  2. The utility function of agent A is:

Ua = -X1-X2-Y1-Y2

  1. The utility function of agent B is:

Ub = X1-X2+Y1-Y2

  1. Agents C and D are in family 2.
  2. Agent C has a utility function:

Uc = -X1-X2-Y1-Y2

  1. Agent D has utility function:

Ud = -X1+X2-Y1+Y2

Solution:

F1 prefers (X1, Y1) and agent A will chose to not consume any good.

F2 prefers (X2, Y2) and agent C chosen to not consume any good.

These are really semantic arguments and there is no meaningful equilibrium without assuming shared preferences.

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  • $\begingroup$ Could you perhaps make your statements more precise? For instance, what are "negative complements"? And please offer at least a heuristic argument supporting the claims, if not a full proof, so that we can understand your reasoning. $\endgroup$ – Shane Jan 5 '16 at 16:14
  • $\begingroup$ It looks from your utility functions like Agents A and B care about the other family's consumption? And I don't follow the idea of "choosing not to consume." Are you supposing that a member of family 1 can choose to consume anywhere in $[0,x_1]$? $\endgroup$ – Shane Jan 5 '16 at 18:32
  • $\begingroup$ Edited the answer. Youre correct on the second point. If agents are required to consume then the argument doesn't apply. $\endgroup$ – D J Sims Jan 6 '16 at 4:04

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