Matching models with transferable utility

Alfred Galichon (NYU+ScPo)

'math+econ+code' masterclass on equilibrium transport and matching models in economics

With python code examples

© 2018-2022 by Alfred Galichon. Past and present support from NSF grant DMS-1716489, ERC grant CoG-866274 are acknowledged, as well as inputs from contributors listed here.

If you reuse material from this masterclass, please cite as:
Alfred Galichon, 'math+econ+code' masterclass on equilibrium transport and matching models in economics, June 2021. https://github.com/math-econ-code/mec_equil

References

Textbooks

Papers

Loading the libraries

First, let's load the libraries:

Generating worker and firm data

We will consider an academic job market, where candidate professors match with colleges--or rather, academic positions, as each position can only be filled by one applicant.

A professor $x$ has characteristics $\xi(x) \in \mathbb{R}^d$ where $\xi$ includes measures of skills, such as general skills

College offer positions $y$ with characteristics $\zeta(y) \in \mathbb{R}^d$, where $\zeta_k$=how much one unit of characteristics $k$ is valued (in dollar terms) for this position, for $k\in\{0,...,7\}$ and $\zeta_6,\zeta_7$ = college location's geographic coordinates.

Preferences

We assume that the amenity that worker $x$ associates with working for position $y$ is equal to minus the Euclidian distance between her location and the position's location, that is $$\alpha_{xy} = - \sqrt{(\xi_{x6} - \zeta_{y6})^2+(\xi_{x7} - \zeta_{y7})^2}.$$

We assume that the output $\gamma_{xy}$ that worker $x$ produces if she takes position $y$ is a CES function where specific/general skills are more or less substituable $$\gamma_{xy}=\left( \left(\sum_{k\in\{0,...,2\} }(\xi_{xk} \zeta_{yk})^{r_g} \right)^{r/{r_g}}+ \left(\sum_{k\in\{3,...,5\} }(\xi_{xk} \zeta_{yk})^{r_s} \right)^{r/{r_s}} \right)^{1/r}.$$

The total output is defined as $\Phi _{xy}=\alpha _{xy}+\gamma _{xy}$. It measures the complemenarity between workers and firms.

Generating demand and supply

We shall take 50 applicants, 30 positions, $r_g = .8$, $r_s = .6$ and $r = .7$.

The class TU_model

We encode the problem into:

We create a mkt object with the data generated above:

Wage determination

Let $w_{xy}$ be the wage (endogenous, determined at equilibrium) that college $y$ would be willing to pay to employee $x$, to be determined. Start by assuming that there are no taxes, so if $x$ and $y$ match, then their payoff are respectively: $U_{xy} = \alpha_{xy} + w_{xy}\\ V_{xy}= \gamma_{xy} - w_{xy} $ while we assume that these payoffs are zero if $x$ and $y$ do not match.

Matching patterns

Assume that there are $n_{x}$ employees of type $x$, and $m_{y}$ colleges of type $y$. The number of matched $xy$ pairs, denoted $\mu_{xy}$, is determined at equilibrium. It should satisfy the populations constraints:

$\left\{ \begin{array}{l} \sum_{y}\mu _{xy}\leq n_{x} \\ \sum_{x}\mu _{xy}\leq m_{y}.% \end{array} \right. $

In the sequel it will be convenient to denote $\mathcal{X}_0 = \mathcal{X} \cup \{ 0 \}$ and $\mathcal{Y}_0 = \mathcal{Y} \cup \{ 0 \}$, and introduce

$\left\{ \begin{array}{l} \mu_{x0} = n_x - \sum_{y}\mu _{xy}\leq n_{x} \\ \mu_{0y} = m_y - \sum_{x}\mu _{xy}\leq m_{y}. \end{array} \right. $

the number of unassigned individuals of each type.

Next, we shall be looking for equilibrium conditions on $\mu_{xy}$ and $w_{xy}$.

The Becker model

Introduce $u_x$ and $v_y$ the indirect utilities of $x$ and $y$, respectively:

$u_x = \max_{y}\left\{ \alpha _{xy}+w_{xy}, 0 \right\}$

$v_y = \max_{x}\left\{ \gamma _{xy}-w_{xy},0\right\}$

We have

$u_x + v_y \geq \Phi_{xy}$, with equality if $\mu_{xy}>0$, and

$u_x \geq 0$ with equality if $\mu_{x0} >0$, and

$v_y \geq 0$ with equality if $\mu_{0y} >0$.

Equilibrium in the Becker model

To recap, $\left( \mu ,u,v\right) $ is an equilibrium matching iff

(1) quantities $\mu $ satisfy the populations constraints:

$\left\{ \begin{array}{l} \sum_{y}\mu _{xy} + \mu_{x0} = n_{x} \\ \sum_{x}\mu _{xy} + \mu_{0y} = m_{y} \end{array} \right. $

(2) utilities $(u,v)$ satisfy pairwise stability:

$\left\{ \begin{array}{l} u_{x}+v_{y}\geq \Phi _{xy}, \forall x,y \\ u_{x}\geq 0, \forall x \\ v_{y}\geq 0, \forall y \end{array} \right. $

(3) the complementarity conditions hold:

$\left\{ \begin{array}{l} \mu _{xy}>0\implies u_{x}+v_{y}=\Phi _{xy} \\ \mu _{x0}>0\implies u_{x}=0 \\ \mu _{0y}>0\implies v_{y}=0. \end{array} \right. $

Linear programming formulation

Noting that the above conditions are complementary slackness conditions in linear programming, one has:

Theorem (Becker-Shapley-Shubik). In the equilibrium problem above, $\mu$ and $(u,v)$ are the resepctive optimal solutions associated with the following primal and dual linear programming problems:
$ \begin{array}{l} \max_{\mu \geq 0} &&\sum_{xy}\mu _{xy}\Phi _{xy} \\ &&\sum_{y}\mu _{xy}\leq n_{x}~\left[ u_{x}\geq 0\right] \\ &&\sum_{x}\mu _{xy}\leq m_{y}~\left[ v_{y}\geq 0\right] \end{array} $

and
$\begin{array}{l} \min_{u\geq 0,v\geq 0} &&\sum n_{x}u_{x}+\sum m_{y}v_{y} \\ s.t.~ &&u_{x}+v_{y}\geq \Phi _{xy}~\left[ \mu _{xy}\geq 0\right] \end{array} $

Interpretation as a welfare theorem: coincidence of the central planner's solution and the decentralized solution.

Computation

Setting $z=vec\left( \mu\right)$, the Linear Programming problem then becomes

$ \begin{array}{l} & \max_{z\geq0}vec\left( \Phi\right) ^{\top}z\\ s.t.~ & \left( I_{\mathcal{X}} \otimes 1_{\mathcal{Y} }^{\top}\right) z\leq n \nonumber\\ & \left( 1_{\mathcal{X}}^{\top} \otimes I_{\mathcal{Y}}\right) z \leq m \nonumber \end{array} $

which is ready to be passed on to a linear programming solver. See chapter 3 of [OTME].

We compute the model and access to $(u,v)$ in the following manner:

The Choo-Siow model

Introduce logit random utilities in the problems of workers and firms. We have:

$u_x = \mathbb{E} \left[ \max_{y}\left\{ \alpha _{xy}+w_{xy} + T \varepsilon_y, T \varepsilon_0 \right\} \right] $

$v_y = \mathbb{E} \left[ \max_{x}\left\{ \gamma _{xy}-w_{xy} + T \eta_y,T \eta_0\right\} \right]$

where $\epsilon$ and $\eta$ are random vectors of i.i.d. Gumbel distributions. We have

$u_{x}=T\log \left( 1+\sum_{y}\exp \left( \frac{\alpha _{xy}+w_{xy}}{T}% \right) \right) $

$v_{y}=T\log \left( 1+\sum_{x}\exp \left( \frac{\gamma _{xy}-w_{xy}}{T}% \right) \right). $

Choice probabilities

We have

$\frac {\mu_{xy}} {n_x} = \Pr \left( y \text{ chosen by } x \right) =\frac{\exp \left( \frac{\alpha _{xy}+w_{xy}}{T}\right) }{1+\sum_{y}\exp \left( \frac{\alpha _{xy}+w_{xy}}{T} \right) } = \exp \left( \frac{\alpha _{xy}+w_{xy}-u_{x}}{T}\right)$

$ \frac {\mu_{x0}} {n_x} = \Pr \left( 0 \text{ chosen by } x \right) = \frac{1}{1+\sum_{y}\exp \left( \frac{\alpha _{xy}+w_{xy}}{T}\right) } =\exp \left( -\frac{u_{x}} {T} \right) $

and

$\frac {\mu_{xy}} {m_y} = \Pr \left( x \text{ chosen by } y \right) =\frac{\exp \left( \frac{\gamma _{xy} - w_{xy}}{T}\right) }{1+\sum_{x}\exp \left( \frac{\gamma _{xy} - w_{xy}}{T} \right) } = \exp \left( \frac{\gamma _{xy} - w_{xy}-v_{y}}{T}\right)$

$ \frac {\mu_{0y}} {m_y} = \Pr \left( 0 \text{ chosen by } y \right) = \frac{1}{1+\sum_{x}\exp \left( \frac{\gamma _{xy} - w_{xy}}{T}\right) } =\exp \left( -\frac{v_{y}} {T} \right) $

Matching functions

Write

$ \frac {\mu_{xy}^2 } {n_x m_y} = \Pr \left( y \text{ chosen by } x \right) . \Pr \left( x \text{ chosen by } y \right) \\ =\exp \left( \frac{\alpha _{xy} + w_{xy}-u_{x}}{T}\right) . \exp \left( \frac{\gamma _{xy} - w_{xy}-v_{y}}{T}\right) = \exp \left( \frac{\Phi _{xy} - u_{x}-v_{y}}{T}\right) $

and as a result, $\mu_{xy} = M_{xy}(\mu_{x0}\mu_{0y})$, where the functions
$M_{xy}(\mu_{x0}\mu_{0y}):= \sqrt{\mu_{x0} \mu_{0y}} \exp \left( \frac{\Phi _{xy}}{2T}\right) $

relate the number of matched pairs of type $xy$ to number of unmatched agents of respective types $x$ and $y$. They are called matching functions.

Introduce
$ a_{x} = u_{x}-T\ln n_{x}\text{ and }b_{y} = v_{y}-T\ln m_{y}$
so that we have

$\mu_{xy} = \exp(\frac {\Phi_{xy} - a_x - b_y} {2T} )$
$\mu_{x0} = \exp(\frac {- a_x} {T}),~\mu_{0y} = \exp(\frac {- b_y} {T})$

Theorem (Choo-Siow). At equilibrium, the quantities $a_x$ and $b_y$ defined above solve the following system of equations
$\left\{ \begin{array} [c]{l}% n_x = \exp(\frac {- a_x} {T}) + \sum_{y\in \mathcal{Y}}\exp(\frac {\Phi_{xy} - a_x - b_y} {2T} )\\ m_y = \exp(\frac {- b_y} {T}) + \sum_{x\in \mathcal{X}}\exp(\frac {\Phi_{xy} - a_x - b_y} {2T} ) \end{array} \right.$

We create a method M_x_y which we append to the TU_model class:

The Choo-Siow model as an optimization problem

One can reformulate the previous equations are the first order conditions of the following optimization problem:

Theorem (Galichon-Salanié). Choo and Siow's equations are the first order conditions associated to the following convex optimization problem:
$W(\Phi) = \min_{a,b} f(a,b) $
where
$f(a,b):= \left\{ \sum_x a_x n_x + \sum_y b_y m_y + 2T \sum_{xy} e^{ \frac {\Phi_{xy} - a_x - b_y} {2T} }+ T \sum_x e^{\frac {-a_x} {T}} + T \sum_y e^{\frac {-b_y} {T}} \right\}$
whose dual program is:
$\begin{array}{l}W(\Phi) = \max_\mu &&\sum_{x,y} \mu_{xy} \Phi_{xy}- 2T \sum_{x,y} \mu_{xy} \log \mu_{xy} - T \sum_x \mu_{x0} \log \mu_{x0} - T \sum_y \mu_{0y} \log \mu_{0y}\\ s.t.~ &&\sum_{y} \mu_{xy} + \mu_{x0} = n_x \\ &&\sum_{x} \mu_{xy} + \mu_{0y} = m_y. \\ \end{array} $

Remark

Note that, setting $A_x =\exp(-a_x / (2T)) $ and $B_y =\exp(-b_y / (2T)) $, the problem rewrites equivalently as
$\min_{A,B} F(A,B)$
where
$F(A,B):= \left\{ - \sum_x n_x \log A_x - \sum_y m_y \log B_y + 2 \sum_{xy} K_{xy} A_x B_y + \sum_x A^2_x + \sum_y B^2_y \right\}$
where $K_{xy} = T \exp( \Phi_{xy} / (2T) ).$
One has:
$\frac{ \partial F} {\partial A_x} = 2 A_x - \frac {n_x} {A_x} + \sum_y K_{xy} B_y$
$\frac{ \partial F} {\partial B_y} = 2 B_y - \frac {m_y} {B_y} + \sum_x K_{xy} A_x,$

We now build a method to solve for equilibrium via the optimization problem:

Run it on our example:

The Choo-Siow model as an equilibrium problem with GS

Can we reformulate as a competitive equilibrium problem with Gross Substitutes? Let's explore the model we are trying to solve.

We see that the matrix is of the form

$\begin{pmatrix} H_{11} & H_{12} \\ H_{21} & H_{22} \end{pmatrix}$

where $H_{11}$ and $H_{22}$ are diagonal matrices with positive diagonal terms, and $H_{12}$ and $H_{21}$ are matrices with nonnegative entries.

Further, note that $H_{12}$ and $H_{21}$ are transpose to each other.

The change-of-sign trick

Define $p\in \mathbb{R}^{\mathcal{X}\cup\mathcal{Y}}$ by
$p_x = a_x, x\in\mathcal{X}$
$p_y = -b_y, y\in\mathcal{Y}$
and define $q \in \mathbb{R}^{\mathcal{X}\cup\mathcal{Y}}$ by
$q_x = -n_x, x\in\mathcal{X}$
$q_y = b_y, y\in\mathcal{Y}.$

Define
$\left\{ \begin{array} [c]{l}% Q_x(p) = - \exp(\frac {- p_x} {T}) - \sum_{y\in \mathcal{Y}}\exp(\frac {\Phi_{xy} - p_x + p_y} {2T} )\\ Q_y(p) = \exp(\frac {p_y} {T}) + \sum_{x\in \mathcal{X}}\exp(\frac {\Phi_{xy} - p_x +p_y} {2T} ) \end{array} \right. $

so that the equilibrium problem reformulates as
$Q(p)=q.$

In terms of Jacobian of the system, this amounts to

$\begin{pmatrix} H{11} & - H{12} \

Coordinate update algorithm

Let's initialize $p_{x}^{0}=-T\ln n_{x}$ and $p_{y}^{0}=-\infty $.

Then update $p_{y}^{1}$ so that $Q_{y}\left( \left( p_{x}^{0}\right) _{x},p_{y}^{1}\right) =q_{y}$($=m_{y})$.

we solve $\sum \exp \left( \Phi _{xy}-p_{x}^{0}+p_{y}^{1}\right) +\exp \left( p_{y}^{1}\right) =m_{y}$

Because the value of $p_{y}^{1}$ is finite, we have $p_{y}^{0}=-\infty \leq p_{y}^{1}$

Then update $p_{x}^{1}$ so that $Q_{x}\left( p_{x}^{1},\left( p_{y}^{1}\right) _{y}\right) =q_{x}$($=-n_{x}$). We have

$\sum_{y}\exp \left( \Phi _{xy}-p_{x}^{1}+p_{y}^{1}\right) +\exp \left( -% \frac{p_{x}^{1}}{T}\right) =n_{x}\geq \exp \left( -\frac{p_{x}^{1}}{T}% \right) $

thus $\exp \left( -\frac{p_{x}^{0}}{T}\right) =n_{x}\geq \exp \left( -\frac{% p_{x}^{1}}{T}\right) $

thus $p_{x}^{0}\leq p_{x}^{1}$

We have $m_{y}\geq \exp \left( p_{y}/T\right) $, thus $p_{y}\leq T\ln m_{y}$.

Set $p_{y}=-b_{y}$ for $y\in Y$

$\left\{ \begin{array}{l} Q_{x}\left( p\right) = -\sum_{y}\exp \left( \frac{\Phi _{xy}-p_{x}+p_{y}}{2T}\right) -\exp \left( -\frac{p_{x}}{T}\right) \\ Q_{y}\left( p\right) = \sum_{x}\exp \left( \frac{\Phi _{xy}-p_{x}+p_{y}}{2T}% \right) +\exp \left( \frac{p_{y}}{T}\right) \end{array} \right.$

We have

$DQ= \begin{pmatrix} \frac{\partial e{x}}{\partial p{x}} & \frac{\partial e{x}}{\partial p{y}} \ \left( \frac{\partial e{x}}{\partial p{y}}\right) ^{\top } & \frac{% \partial e{y}}{\partial p{y}}%

\end{pmatrix}

\begin{pmatrix} diag\left( \frac{1}{2T}\exp \left( \frac{\Phi _{xy}-p_{x}+p_{y}}{2T}\right) + \frac{1}{T}\exp \left( -\frac{p_{x}}{T}\right) \right) & \\ -\frac{1}{2T}\exp \left( \frac{\Phi _{xy}-p_{x}+p_{y}}{2T}\right) & diag\left( \frac{1}{2T}\exp \left( \frac{\Phi _{xy}-p_{x}+p_{y}}{2T}\right) + \frac{1}{T}\exp \left( \frac{p_{y}}{T}\right) \right) \end{pmatrix}

$

Gauss-Seidel algorithm

The algorithm runs as follows:

These equations can be solved explicitly. Introduce $K_{xy}=\exp \left( \frac{\Phi _{xy}}{2T}\right)$ and new unknowns
$A_{x}=\exp \left( -\frac{a_{x}}{2T}\right),~B_{y}=\exp \left( -\frac{b_{y}}{2T}\right)$

we have

$\sum_{y}K_{xy}A_{x}B_{y}+A_{x}^{2} = n_{x} \\ \sum_{x}K_{xy}A_{x}B_{y}+B_{y}^{2} = m_{y} $

thus

$ A_{x} = \sqrt{n_{x}+\left( \frac{1}{2}\sum_{y}K_{xy}B_{y}\right) ^{2}}- \frac{1}{2}\sum_{y}K_{xy}B_{y} \\ B_{y} = \sqrt{m_{y}+\left( \frac{1}{2}\sum_{x}K_{xy}A_{y}\right) ^{2}}- \frac{1}{2}\sum_{x}K_{xy}A_{y} $

We create a method implementing the Gauss-Seidel algorithm:

Run it using:

Matching with linear taxes

We now consider a version of the Becker model with a "flat tax": assume that the gross wage $w_{xy}$ is taxed at a constant rate $\tau \in [0,1)$: $U_{xy} = \alpha_{xy} + (1-\tau) w_{xy}\\ V_{xy} = \gamma_{xy} - w_{xy}.$

Dupuy, Galichon, Jaffe and Kominers (2020) have shown the following result:

Theorem (DGJK). The equilibrium matching in the problem above with a flat tax $\tau$ above is the optimal matching in a TU matching problem with surplus function
$\Phi^\tau_{xy} = \alpha_{xy} + (1-\tau ) \gamma_{xy}.$

Proof

At equilibrium there is a wage $w_{xy}$ such that:

$x$ gets indidirect utility $u_{x}=\max_{y}\left\{ \alpha _{xy}+\left( 1-\tau \right) w_{xy},0\right\} $, and

$y$ gets indirect utility $v_{y}=\max_{x}\left\{ \gamma _{xy}-w_{xy},0\right\} $

Equilibrium consists thus of looking for matching patterns $\mu _{xy}$, wages $w_{xy}$, and indirect utilities $(u_x,v_y)$ such that:

(i) populations constraints are met:
$\left\{ \begin{array}{l} \sum_{y}\mu _{xy}+\mu _{x0}=n_{x} \\ \sum_{x}\mu _{xy}+\mu _{0y}=m_{y} \end{array} \right. $

(ii) stability conditions hold:
$u_{x}\geq \alpha _{xy}+\left( 1-\tau \right) w_{xy}\text{ and } v_{y}\geq \gamma _{xy} - w_{xy} \\ u_{x}\geq 0,v_{y}\geq 0$

(iii) complementary conditions hold:
$\mu_{xy}>0 \implies u_{x} = \alpha _{xy}+\left( 1-\tau \right) w_{xy}\text{ and } v_{y} = \gamma _{xy} - w_{xy} \\ \mu_{x0}>0 \implies u_{x} =0 \\ \mu_{0y}>0 \implies v_{y} = 0$

Proof (ctd)

Denote $\tilde{v}_{y}=\left( 1-\tau \right) v_{y}$ and $\tilde{\gamma}% _{xy}=\left( 1-\tau \right) \gamma _{xy}$ and $\tilde{w}_{xy}=\left( 1-\tau \right) w_{xy}$ the indirect utility of the firm and the output measured in post-tax dollars, then we have

$\sum_{y}\mu _{xy}+\mu _{x0}=n_{x}$

$\sum_{x}\mu _{xy}+\mu _{0y}=m_{y}$

$u_{x}\geq 0,\tilde{v}_{y}\geq 0$

$u_{x}\geq \alpha _{xy}+\tilde{w}_{xy}$ with equality if $\mu _{xy}>0$

$\tilde{v}_{y}\geq \tilde{\gamma}_{xy}-\tilde{w}_{xy}$ with equality if $\mu _{xy}>0$.

Therefore, $\mu$ is the optimal matching associated with surplus function
$\tilde{\Phi}_{xy} = \alpha_{xy}+\left( 1-\tau \right) \gamma _{xy}$.

Embedding in a Choo-Siow model

Consider the logit random utility version of the previous problem:

$u_x = \mathbb{E} \left[ \max_{y}\left\{ \alpha _{xy}+(1 - \tau) w_{xy} + T \varepsilon_y, T \varepsilon_0 \right\} \right] $

$v_y = \mathbb{E} \left[ \max_{x}\left\{ \gamma _{xy}-w_{xy} + T \eta_y,T \eta_0\right\} \right]$

where $\epsilon$ and $\eta$ are random vectors of i.i.d. Gumbel distributions.

Defining as before $\tilde{v}_{y}=\left( 1-\tau \right) v_{y}$ and $\tilde{\gamma}_{xy}=\left( 1-\tau \right) \gamma _{xy}$ and $\tilde{w}_{xy}=\left( 1-\tau \right) w_{xy}$\, we have
$u_{x}=T\log \left( 1+\sum_{y}\exp \left( \frac{\alpha _{xy}+\tilde{w}_{xy}}{T}\right) \right) \\ \tilde{v}_{y}=(1-\tau) T\log \left( 1+\sum_{x}\exp \left( \frac{\tilde{\gamma} _{xy}-\tilde{w}_{xy}}{(1-\tau)T} \right) \right). $

As a result, the model is a Choo-Siow model with surplus $\alpha_{xy} + (1-\tau) \gamma_{xy}$ and scaling parameter $1-\tau / 2$

Define:
$\begin{array}{l}W\left( \theta ,\lambda \right) = \max_\mu &&\sum_{x,y} \mu_{xy} \left( \alpha \theta +\gamma \lambda \right)- (\theta+\lambda)T \sum_{x,y} \mu_{xy} \log \mu_{xy} - \theta T \sum_x \mu_{x0} \log \mu_{x0} - \lambda T \sum_y \mu_{0y} \log \mu_{0y}\\ s.t.~ &&\sum_{y} \mu_{xy} + \mu_{x0} = n_x \\ &&\sum_{x} \mu_{xy} + \mu_{0y} = m_y \\ \end{array}$
so that the equilibrium matching model with taxes solves for an equilibrium with $\theta = 1$ and $\lambda = 1-\tau$.

Comparative statics

Proposition (DGJK). $W$ is a convex and positive homogenous function, and we have:
$\left\{ \begin{array}{l} \frac{\partial W}{\partial \theta } (\theta,\lambda)=\sum_{xy} \mu_{xy} \alpha_{xy} -T \sum_x \mu_{x0} \log \mu_{x0} =:A\\ \frac{\partial W}{\partial \lambda } (\theta,\lambda)=\sum_{xy} \mu_{xy} \gamma_{xy} -T \sum_y \mu_{0y} \log \mu_{0y} =:\Gamma \end{array} \right. $

therefore $\frac{\partial \Gamma }{\partial \lambda }\geq 0$

Similarly, $W=\theta A+\lambda \Gamma $ and $\Gamma =\theta \frac{\partial A% }{\partial \lambda }+\Gamma +\lambda \frac{\partial \Gamma }{\partial \lambda }$

hence $\frac{\partial \Gamma }{\partial \lambda }=-\frac{\theta }{\lambda }% \frac{\partial A}{\partial \lambda }$

so with $\theta =1$, one has

$\frac{\partial A}{\partial \lambda }=-\lambda \frac{\partial \Gamma }{% \partial \lambda }\leq 0$

and

$\frac{\partial \left( A+\Gamma \right) }{\partial \lambda }=\left( 1-\lambda \right) \frac{\partial \Gamma }{\partial \lambda }$

As a result, the sum of amenities $A$ increases with the tax wage, but the total welfare $A+\Gamma$ decreases, so the increase in $A$ is not enough to compensate the decrease in $\Gamma$.