我即将完成我的论文，但我不明白为什么我所包含的 4 个附录中只有 3 个被呈现在目录中。我正在使用 latextemplatex 的模板“硕士/博士论文”，由于大小问题，我无法将其上传到这里。这是我的 main.tex 文件的代码：
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% Masters/Doctoral Thesis
% LaTeX Template
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11pt, % The default document font size, options: 10pt, 11pt, 12pt
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%toctotoc, % Uncomment to add the main table of contents to the table of contents
%parskip, % Uncomment to add space between paragraphs
%nohyperref, % Uncomment to not load the hyperref package
headsepline, % Uncomment to get a line under the header
%chapterinoneline, % Uncomment to place the chapter title next to the number on one line
%consistentlayout, % Uncomment to change the layout of the declaration, abstract and acknowledgements pages to match the default layout
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\usepackage{mathpazo} % Use the Palatino font by default
\begin{document}
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% DEDICATION
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% ACKNOWLEDGEMENTS
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\begin{acknowledgements}
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% LIST OF CONTENTS/FIGURES/TABLES PAGES
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\tableofcontents % Prints the main table of contents
\listoffigures % Prints the list of figures
\listoftables % Prints the list of tables
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% THESIS CONTENT - CHAPTERS
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\mainmatter % Begin numeric (1,2,3...) page numbering
\pagestyle{thesis} % Return the page headers back to the "thesis" style
% Include the chapters of the thesis as separate files from the Chapters folder
% Uncomment the lines as you write the chapters
\include{Chapters/Chapter0}
\include{Chapters/Chapter1}
\include{Chapters/Chapter2}
\include{Chapters/Chapter3}
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% BIBLIOGRAPHY
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\printbibliography[heading=bibintoc]
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% THESIS CONTENT - APPENDICES
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\appendix % Cue to tell LaTeX that the following "chapters" are Appendices
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\include{Appendices/AppendixA}
\include{Appendices/AppendixB}
\include{Appendices/AppendixC}
\include{Appendices/AppendixD}
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\end{document}
这是附录文件 C 和 D 的代码（A 和 B 具有与 C 相同的结构）。
% Appendix C
\chapter{Steady State Distributions}
\section{Without Minimum Wages}
\label{app:A1}
In this appendix detailed derivations for $v(y)$, $u(x)$ and $h(x,y)$ are worked out. We start with the balance conditions, which are no more that accounting identities, that are met in every point in time
\begin{align*}
\int h_t(x,y') dy' + u_t(x) = l_t(x)\\
\int h_t(x',y) dx' + v_t(y) = n_t(y)
\end{align*}
and flow equations in discrete time.
\begin{footnotesize}
\begin{align*}
h_{t+1}(x,y) & = h_{t}(x,y) & + k v_t(y) u_t(x)\Delta & + sk \int_{\underline{y}}^y h_t(x,y')dy' v_t(y)\Delta & - \delta h_t(x,y)\Delta & - sk \int_y^{\overline{y}}v_t(y')dy' h_t(x,y)\Delta\\
u_{t+1}(x) & = u_{t}(x) & - k V_t u_t(x)\Delta && + \delta h_{x,t}(x)\Delta \\
v_{t+1}(y) & = v_{t}(y) & - k v(y) U_t\Delta &- sk \int_{\underline{y}}^y h_{y,t}(y')dy' v_t(y)\Delta &+ \delta h_{y,t}(y)\Delta & + sk \int_y^{\overline{y}}v_t(y')dy' h_{y,t}(y)\Delta
\end{align*}
\end{footnotesize}
Where $h_{x,t}(x) = \int_{\underline{y}}^{\overline{y}}h_{t}(x,y')dy'$ and $h_{y,t}(y) = \int_{\underline{x}}^{\overline{x}}h_{t}(x',y)dx'$. The expressions are easier to work with in continuous time so I rearrange stocks to the LHS and flows to the RHS, Divide by $\Delta$ and take the limit as $\Delta \rightarrow 0$ to have
\begin{footnotesize}
\begin{align*}
\dot{h}_{t}(x,y) & = k v_t(y) u_t(x) & + sk \int_{\underline{y}}^y h_t(x,y')dy' v_t(y) & - \delta h_t(x,y) & - sk \int_y^{\overline{y}} v_t(y')dy' h_t(x,y)\\
\dot{u}_{t}(x) & = - k V_t u_t(x) && + \delta h_{x,t}(x) \\
\dot{v}_{t}(y) & = - k v_t(y) U_t &- sk \int_{\underline{y}}^y h_{y,t}(y')dy' v_t(y) &+ \delta h_{y,t}(y) & + sk \int_y^{\overline{y}}v_t(y')dy' h_{y,t}(y)
\end{align*}
\end{footnotesize}
Before working out specific expressions for every type of firm and worker, it will be useful to calculate aggregate balance conditions, just aggregate over the set of firm productivities and worker abilities to have
\begin{align*}
H_t + U_t = L_t\\
H_t + V_t = N_t
\end{align*}
Where $H_t = \int \int h_t(x',y') dx'dy'$. $L_t$ is exogenous and given $N_t$, $V_t$ is pinned down, at this point both conditions are knotted by $H_t$, so we can write
\begin{align*}
L_t - U_t = N_t - V_t
\end{align*}
And deriving with respect to time we have that $\dot{U}_t=\dot{V}_t$. Also, we need to have the expressions for the aggregate flows. Integrate $v(y)$, $u(x)$ and $h(x,y)$ over the variables that they depend on. Furthermore, in the aggregate all the workers that quit are the same as those who are poached, i.e. $\int_{\underline{y}}^y h_t(x,y')dy' v_t(y) = \int_y^{\overline{y}}v_t(y')dy'h_t(x,y)$, hence
\begin{align*}
\left.\begin{matrix}
\dot{H_t} & = k V_tU_t - \delta H_t\\
\dot{U_t} & = -k V_tU_t + \delta H_t\\
\dot{V_t} & = -k V_tU_t + \delta H_t
\end{matrix}\right\} \overset{SS}{\Longrightarrow}
\delta H_t = k V_tU_t
\end{align*}
Since we are in the S.S., time dependence can be dropped from the notation. Now, Plug the aggregate balance conditions to have $H$ as a function of $N, \; k$ (endogenous objects) and $\delta,\; L$ (parameters), $\delta H = k (N - H)(L - H)$, this is a quadratic equation on $H$
\begin{align*}
k H^2 - \left(\delta + k L + k N \right)H + k LN = 0
\end{align*}
Which solves as
\begin{align*}
H = \frac{1}{2} \left\{\left(\frac{\delta}{k} + L + N\right) - \sqrt{\left(\frac{\delta}{k} + L + N\right)^2 - 4LN} \right\}
\end{align*}
Below, it is the proof of why only the negative part is taken. First consider the positive part, i.e.
\begin{footnotesize}
\begin{align*}
H(N) = \frac{1}{2} \left\{\left(\frac{\delta}{k} + L + N\right) + \sqrt{\left(\frac{\delta}{k} + L + N\right)^2 - 4LN} \right\} & > \frac{1}{2} \left\{\left(\frac{\delta}{k} + L + N\right) + \sqrt{\left(\frac{\delta}{k} + L + N\right)^2} \right\} \Leftrightarrow \\
\frac{1}{2} \left\{\left(\frac{\delta}{k} + L + N\right) + \sqrt{\left(\frac{\delta}{k} + L + N\right)^2 - 4LN} \right\} & > \left(\frac{\delta}{k} + L + N\right) \\
\end{align*}
\end{footnotesize}
Which means that the number of employed is higher than the number of people in the economy, an absurdity. Turning to the negative part, I would like to work out the maximum and minimum values as a function of $N$. The minimum value can be easily worked out as
\begin{align*}
H\left(0\right) = \frac{1}{2} \left\{\left(\frac{\delta}{k} + L \right) - \sqrt{\left(\frac{\delta}{k} + L\right)^2 } \right\} = 0
\end{align*}
And the maximum
\begin{align*}
\lim_{N\rightarrow\infty} H(N) & = \lim_{N\rightarrow\infty} \frac{1}{2} \left\{\left(\frac{\delta}{k} + L + N\right) - \sqrt{\left(\frac{\delta}{k} + L + N\right)^2 - 4LN} \right\}
\end{align*}
Which is undetermined, dividing and multiplying by the complement and later on dividing by $N$ in the numerator and denominator we have
\begin{align*}
\lim_{N\rightarrow\infty} H(N) & = \lim_{N\rightarrow\infty} \frac{1}{2} \left\{ \frac{4LN}{\left(\frac{\delta}{k} + L + N\right) + \sqrt{\left(\frac{\delta}{k} + L + N\right)^2 - 4LN}} \right\}\\
& = \lim_{N\rightarrow\infty} \frac{1}{2} \left\{ \frac{4L}{\left(\frac{\delta}{Nk} + \frac{L}{N} + 1\right) + \sqrt{\left(\frac{\delta}{Nk} +\frac{L}{N} + 1\right)^2 + \frac{4L}{N}}} \right\}\\
& = \frac{1}{2} \left\{ \frac{4L}{ 1 + \sqrt{\ 1}} \right\} = L
\end{align*}
Which means that as the number of firms tends to infinity the number of employed workers tends to the number of people in the economy. Also, it would be convenient to check if the function is increasing in the whole domain
\begin{align*}
\frac{\partial H}{\partial N} = \frac{1}{2}\left\{ 1 - \frac{2\left(\frac{\delta}{k} + L + N\right)-4L}{\sqrt{\left(\frac{\delta}{k} + L + N\right)^2 - 4LN}} \right\}>0
\end{align*}
Using the balance conditions $U$ and $V$ can easily be derived. With this expressions at hand we can work out their desegregated counterparts. First, consider the S.S. and drop the time dependence,so
\begin{align*}
0 & = k v(y) u(x) & + sk \int_{\underline{y}}^y h(x,y')dy' v(y) & - \delta h(x,y) & - sk \int_y^{\overline{y}}v(y')dy'h(x,y)\\
0 & = - k V u(x) && + \delta h_x(x) \\
0 & = - k v(y) U &- sk \int_{\underline{y}}^y h_y(y')dy' v(y) &+ \delta h_y(y) & + sk \int_y^{\overline{y}}v(y')dy' h_y(y)
\end{align*}
Now, consider the last expression and rearrange such that U-to-H and H-to-U flows are on the LHS and H-to-H flows are in the RHS. Also, for notational convenience let's write $F_z(z)=\int_{\underline{z}}^z f_z(z')dz'$ for any function over the $z$-characteristic and denote its complement counterpart as $\overline{F}_z(z) = F_z(\overline{z})-F_z(z)= \int_{z}^{\overline{z}}f_z(z')dz'$
\begin{align*}
k v(y) U - \delta h_y(y) = - sk H_y(y) v(y) + sk \overline{V}(y) h_y(y)
\end{align*}
Integrate both sites from $\underline{y}$ to $y$ to have
\begin{align} \label{eq: H-V}
k V(y) U - \delta H_y(y) = sk H_y(y)\overline{V}(y)
\end{align}
And Integrate the balance condition from $\underline{y}$ to $y$ to have
\begin{align*}
H_y(y) + V(y) = N(y)
\end{align*}
Then solve last expression for V(y) and plug it into the aggregate flow equation for vacancies to arrive at
\begin{align*}
k \left( N(y) - H_y(y) \right) U -\delta H_y(y) & = sk H_y(y)\left(V - N(y) + H_y(y)\right)\\
k U N(y) - k U H_y(y) -\delta H_y(y) & = sk V H_y(y) - sk N(y) H_y(y) + sk H^2(y)\\
sk H^2(y) + \big( \delta + k U + sk V \big. & \left. - sk N(y) \right) H_y(y) - k U N(y) = 0
\end{align*}
Again, this is a quadratic equation in $H_y(y)$, which depends solely on $k$ and $N(y)$ and the rest of variables have been previously worked out, as we can see below
\begin{align*}
\underset{A}{\underbrace{sk}} H_y^2(y) + \underset{B\left(N(y)\right)}{\underbrace{\left(\delta + k U + sk V - sk N(y) \right)}}H_y(y) - \underset{C\left(N(y)\right)}{\underbrace{k U N(y)}} = 0
\end{align*}
Then
\begin{align*}
H_y(y) = \frac{-B\left(N(y)\right) + \sqrt{B^2\left(N(y)\right)+4AC\left(N(y)\right)}}{2A}
\end{align*}
The negative part can safely be discarded as
\begin{footnotesize}
\begin{align*}
H_y(y) = \frac{-B\left(N(y)\right) - \sqrt{B^2\left(N(y)\right)+4AC\left(N(y)\right)}}{2A} < \frac{-B\left(N(y)\right) - \sqrt{B^2\left(N(y)\right)}}{2A} = -\frac{2B\left(N(y)\right)}{2A}<0
\end{align*}
\end{footnotesize}
Whereas the positive part is always greater than zero
\begin{footnotesize}
\begin{align*}
H_y(y) = \frac{-B\left(N(y)\right) + \sqrt{B^2\left(N(y)\right)+4AC\left(N(y)\right)}}{2A} > \frac{-B\left(N(y)\right) + \sqrt{B^2\left(N(y)\right)}}{2A} = 0
\end{align*}
\end{footnotesize}
Now, derive the quadratic equation implicitly with respect to to $y$ to find $h_y(y)$
\begin{align*}
2sk H_y(y)h_y(y) + \left(\delta + k U + s k V - sk N(y)\right)h_y(y) - s k n(y)H_y(y) -k U n(y) =0
\end{align*}
Solving for $h_y(y)$
\begin{align*}
h_y(y) = \frac{k U + s k H_y(y)}{\left(\delta + sk V + s k H_y(y) - sk N(y)\right) + \left( k U + s k H_y(y) \right)}\cdot n(y)
\end{align*}
Where
\begin{itemize}
\item $\left(\delta + sk V + s k H_y(y) - sk N(y)\right) = \left(\delta + sk \overline{V}(y)\right)$: are flows out of $H_y(y)$ and
\item $k U + s k H_y(y)$: are flows into $H_y(y)$
\end{itemize}
It's worth noting that $H_y(y)$ depends on $N(y)$ and so does $h_y(y)$.
At this point we are ready to come with an expression for $v(y)$. Consider again the expression coming from the integrated flow of vacancies in the S.S.
\begin{align*}
k V(y) U - \delta H_y(y) = sk H_y(y)\overline{V}(y)
\end{align*}
It is just left to solve for $V(y)$ and derive to reach the desire result
\begin{align*}
V(y) = \frac{\delta + sk V}{k U + sk H_y(y)}H_y(y)
\end{align*}
And deriving
\begin{align*}
v(y) & = \frac{\delta + sk V}{\left(k U + skH_y(y)\right)^2}k U h_y(y)
\end{align*}
Which is not very intuitive. In order to have an expression in terms of flows, change $h_y(y)$ by its last derived expression; plug the definition of $V(y)$ and use the integrated flow of vacancies in the S.S. to arrive to the desired result
\begin{align*}
v(y) = \frac{\delta + sk \overline{V}(y)}{\left(\delta + sk \overline{V}(y)\right) + \left( k U + s k H_y(y) \right)}\cdot n(y)
\end{align*}
Once we have worked out a close form expression for the number of vacancies and the number of workers in $y$-type firms, we can deal with the number of unemployed and the number of workers with $x$-characteristic. From the differential equation for unemployed, substitute the balance condition for $h_x(x)$, such that
\begin{align*}
k V u(x) = \delta h_x(x) \Leftrightarrow k V u(x) & = \delta \left(l(x)-u(x)\right) \Leftrightarrow \left(\delta +k V\right) u(x) = \delta l(x)\\
u(x) & = \frac{\delta}{\left(\delta +k V\right)}l(x)
\end{align*}
With this expression and basic algebra we work out $h_x(x)$
\begin{align*}
h_x(x) = \frac{k V}{\left(\delta +k V\right)}l(x)
\end{align*}
Finally, we are ready to calculate $h(x,y)$. The steps to arrive at the solution are basically the same as those to compute $h_y(y)$. THen, consider the last expression and rearrange such that U-to-H and H-to-U flows are on the LHS and H-to-H flows are in the RHS.
\begin{align*}
\delta h(x,y) - k v(y)u(x) & = sk \int_{\underline{y}}^y h(x,y')dy' v(y) - sk \int_y^{\overline{y}}v(y')dy'h(x,y)
\end{align*}
Integrate both sites from $\underline{y}$ to $y$ to have
\begin{align*}
\delta \int_{\underline{y}}^y h(x,y')dy' - k u(x) V(y) & = -sk \int_{\underline{y}}^y h(x,y')dy'\overline{V}(y)
\end{align*}
rearranging
\begin{align*}
\left( \delta +sk \overline{V}(y) \right) \int_{\underline{y}}^y h(x,y')dy' & = k u(x) V(y) \\
\int_{\underline{y}}^y h(x,y')dy' & = \frac{k u(x) V(y)}{\left( \delta +sk \overline{V}(y) \right) }
\end{align*}
And deriving with respect to to $y$ we arrive at the final form
\begin{align*}
h(x,y) = \frac{\delta + s k V}{\left(\delta + sk \overline{V}(y)\right)^2}\cdot k u(x) v(y)
\end{align*}
Which is difficult to interpret. However, we can prove that the following interesting result holds, $h(x,y) = \frac{1}{H}h_x(x)h_y(y)$. Start by plugging in $h(x,y)$ the expressions for $k u(x) = \frac{\delta}{V}h_x(x)$, $v(y)$, and solve for $\left(\delta + sk \overline{V}(y)\right)$ in equation \ref{eq: H-V}, so that we get
\begin{align*}
h(x,y) = \frac{\left(\delta + s k V\right)}{\left(\frac{k V(y)U}{H_y(y)}\right)^2}\cdot \frac{\left(\delta + s k V\right)}{\left(k U + sk H_y(y)\right)^2}\cdot k U h_y(y) \frac{\delta}{V}h_x(x)
\end{align*}
Solve for $\left(k U + sk H_y(y)\right)$ in \ref{eq: H-V} and use the fact that $k U = \frac{\delta}{V}H$
\begin{align*}
h(x,y) = \frac{\left(\delta + s k V\right)}{\left(\frac{k V(y)U}{H_y(y)}\right)^2}\cdot \frac{\left(\delta + s k V\right)}{\left(\frac{\left(\delta + sk V\right)H_y(y)}{V(y)}\right)^2}\cdot \left(\frac{\delta}{V}\right)^2 H h_x(x)h_y(y)
\end{align*}
Cancelling out terms and substituting
\begin{align*}
h(x,y) = \left(\frac{\delta}{k UV}\right)^2 H h_x(x)h_y(y) = \frac{1}{H^2}H h_x(x) h_y(y) = \frac{1}{H}h_x(x)h_y(y)
\end{align*}
Now, we are ready to calculate the distribution of wages from the flow equation
\begin{align*}
\frac{dG_t(w|x,y)\cdot h_t(x,y)}{dt} & = k v_t(y) u_t(x) + s k \int_{\underline{y}}^q h(x,y')dy'\cdot v(y) \\
& - \delta G_t(w|x,y)\cdot h_t(x,y) - s k \int_q^{\overline{y}}v(y')dy'G_t(w|x,y)\cdot h_t(x,y)=0
\end{align*}
rearranging
\begin{align*}
\left(\delta + s k \int_q^{\overline{y}}v(y')dy'\right)G(w|x,y)\cdot h(x,y) = k v(y) u(x) + s k \int_{\underline{y}}^q h(x,y')dy' v(y)
\end{align*}
solving for $G_t(w|x,y)$ we have
\begin{align*}
G(w|x,y) = \frac{\left(k v(y) u(x) + s k \int_{\underline{y}}^q h(x,y')dy' v(y)\right)}{\left(\delta + s k \int_q^{\overline{y}}v(y')dy'\right)} \cdot \frac{1}{h(x,y)}
\end{align*}
Substitute $h(x,y)$ by the product of the marginals $\frac{1}{H}h_x(x)h_y(y)$. Also use the flow equation for the unemployed $k Vu(x)=h_x(x)$ and the aggregate flow equation $k VU=\delta H$ to arrive at $u(x)=\frac{U}{H}h_x(x)$, then after cancelling terms
\begin{align*}
G(w|y) = \frac{\left(k U + s k \int_{\underline{y}}^q h_y(y')dy'\right) }{\left(\delta + s k \int_q^{\overline{y}}v(y')dy'\right)} \cdot \frac{v(y)}{h_y(y)}
\end{align*}
Which shows what intuition could have told us in advance, namely that the distribution of wages does not depend on $x$.
% ---------------------------------DISTRIBUTIONS WITH MINIMUM WAGES-----------------------------------------
\section{With Minimum Wages}
To find out the distributions of matches, vacancies and unemployed under the minimum wage, $\tilde{h}(x,y)$, $\tilde{v}(y)$ and $\tilde{u}(x)$ respectively, it will just suffice to rewrite them in terms of the old ones. Under the minimum wage some meetings that could have ended in a match are not going to be possible, as the flow revenue of the match it is not enough to pay the minimum wage. Then the balance conditions can be rewritten as
\begin{align*}
l(x) & = \underset{\tilde{u}(x)}{\underbrace{u(x) + \int_{\underline{y}}^{\hat{y}(x)} h(x,y') dy'}} + \underset{\tilde{h}_x(x)}{\underbrace{\int_{\hat{y}(x)}^{\overline{y}} h(x,y') dy'}} \\
n(y) & = \underset{\tilde{v}(y)}{\underbrace{v(y) + \int_{\underline{x}}^{\hat{x}(y)} h(x',y) dx'}} + \underset{\tilde{h}_y(y)}{\underbrace{\int_{\hat{x}(y)}^{\overline{x}} h(x',y)dx'}}.
\end{align*}
Because the previous analysis without minimum wages, we know that under random search there is no sorting under the model assumptions. The implication being to express $h(x,y)$ as the product of two functions that describe abilities and productivities independently, namely $h(x,y) = \frac{1}{H}h_y(y)h_x(x)$, then the balance conditions can be rewritten as
\begin{align*}
l(x) & = \underset{\tilde{u}(x)}{\underbrace{u(x) + h_x(x)\int_{\underline{y}}^{\hat{y}(x)} \frac{h_y(y')}{H} dy'}} + \underset{\tilde{h}_x(x)}{\underbrace{h_x(x)\int_{\hat{y}(x)}^{\overline{y}}\frac{h_y(y')}{H} dy'}} \\
n(y) & = \underset{\tilde{v}(y)}{\underbrace{v(y) + h_y(y)\int_{\underline{x}}^{\hat{x}(y)} \frac{h_x(x')}{H} dx'}} + \underset{\tilde{h}_y(y)}{\underbrace{h_y(y)\int_{\hat{x}(y)}^{\overline{x}} \frac{h_x(x')}{H} dx'}}
\end{align*}.
And then as
\begin{align*}
l(x) & = \underset{\tilde{u}(x)}{\underbrace{u(x) + h_x(x)F_y\left(\hat{y}(x)\right)}} + \underset{\tilde{h}_x(x)}{\underbrace{h_x(x)\overline{F}_y\left(\hat{y}(x)\right)}} \\
n(y) & = \underset{\tilde{v}(y)}{\underbrace{v(y) + h_y(y)F_x\left(\hat{x}(y)\right)}} + \underset{\tilde{h}_y(y)}{\underbrace{h_y(y)\overline{F}_x\left(\hat{x}(y)\right)}}.
\end{align*}
Integrating over abilities in the first condition and over productivities in the second we work out the aggregate balance conditions
\begin{align*}
L & = \underset{\tilde{U}}{\underbrace{U + \int_{\underline{x}}^{\overline{x}}h_x(x')F_y\left(\hat{y}(x')\right) dx'}} +
\underset{\tilde{H}}{\underbrace{\int_{\underline{x}}^{\overline{x}}h_x(x')\overline{F}_y\left(\hat{y}(x')\right)dx'}} \\
N & = \underset{\tilde{V}}{\underbrace{V +
\int_{\underline{y}}^{\overline{y}} h_y(y')F_x\left(\hat{x}(y')\right) dy'}} +
\underset{\tilde{H}}{ \underbrace{\int_{\underline{y}}^{\overline{y}}h_y(y')\overline{F}_x\left(\hat{x}(y')\right)dy'}}.
\end{align*}
For the join distribution of jobs under the minimum wage it will suffice to solve the follow equation for jobs:
\begin{align*}
0 & = k \tilde{v}(y) \tilde{u}(x) & + sk \int_{\underline{y}}^q \tilde{h}(x,y')dy' \tilde{v}(y) \\
& - \delta \tilde{G}_t(w|x,y)\cdot \tilde{h}(x,y) & - sk \int_q^{\overline{y}}\tilde{v}(y')dy'\tilde{G}_t(w|x,y)\cdot \tilde{h}(x,y).
\end{align*}
Now the number of vacancies and unemployed people will be substituted by their minimum wage counterparts, i.e. $\tilde{v}(y)$ and $\tilde{u}(y)$, since those vacancies lost by the unemployed or workers with low ability will be at the disposal of the rest, and seemingly the same argument applies for those unemployed that will not be able to cover vacancies in low productive firms. Hence, following the same procedure as before we will arrive at
\begin{align*}
\tilde{h}(x,y) & = \frac{\delta + s k \tilde{V}}{\left(\delta + sk \overline{\tilde{V}}(y)\right)^2}\cdot k \tilde{u}(x) \tilde{v}(y)\\
\tilde{G}(w|x,y) & = \frac{\left(k \tilde{v}(y) \tilde{u}(x) + s k \int_{\underline{y}}^q \tilde{h}(x,y')dy' \tilde{v}(y)\right)}{\left(\delta + s k \int_q^{\overline{y}}\tilde{v}(y')dy'\right)} \cdot \frac{1}{\tilde{h}(x,y)}
\end{align*}
Lack of assortative matching will not be the case upon introducing minimum wages, now the minimum viable productivity of a firm (or worker) to form a match will be a function of the worker ability (firm productivity). In this respect minimum wages will introduce negative sorting in our analysis.
% Appendix D
\chapter{Proofs of Lemmas}
\section{Lemma 1}
\label{app:B1}
Free-entry implies that the firm with the minimum viable productivity to hire a worker cannot make any surplus out of the match, i.e. $\Pi_1\left(x,\underline{\hat{y}}(x)\right)=0$, otherwise a firm with marginally less productivity could enter the market, hire a worker and make profits, being a contradiction; then $P\left(x,\underline{\hat{y}}(x)\right) = W_1 \left( \phi_0\left( x, \underline{\hat{y}}(x)\right), x, \underline{\hat{y}}(x)\right)$.
With respect to $W_1 \left( \phi_0\left( x, \underline{\hat{y}}(x)\right), x, \underline{\hat{y}}(x)\right)=W_0(x,m)$, the same argument along the above lines can be devised. If $W_1 \left( \phi_0\left( x, \underline{\hat{y}}(x)\right), x, \underline{\hat{y}}(x)\right)=P\left(x,\underline{\hat{y}}(x)\right)>W_0(x,m)$ then another firm with marginally less productivity could enter and make profits, once again a contradiction.
\section{Lemma 2}
\label{app:B2}
Making use of LEMMA.1 we can equate $P\left(x,\underline{\hat{y}}(x)\right) = W_1 \left( \phi_0\left( x, \underline{\hat{y}}(x)\right), x, \underline{\hat{y}}(x)\right)$, which means that
\begin{align*}
\underline{\hat{y}} &(x)x + \delta W_0(x) + sk \int_{\underline{\hat{y}}(x)}^{t(x,y)} P(x,y)v(y')dy' + sk \int_{t(x,y)}^{\overline{y}} W_{1}(m,x,y')v(y')dy'\\
& = \phi_0\left( x, \underline{\hat{y}}(x)\right) + \delta W_0(x) + sk \int_{\underline{\hat{y}}(x)}^{t(x,y)} P(x,y) v(y')dy' + sk \int_{t(x,y)}^{\overline{y}}W_1(m,x,y') v(y')dy'.
\end{align*}
And after cancelling terms we arrive at:
\begin{align*}
\underline{\hat{y}} &(x)x = \phi_0\left( x, \underline{\hat{y}}(x)\right).
\end{align*}
For convenience define the threshold $x'$ such that $\phi_0\left( x', y_{inf}\right) = m$. There are two cases of interest:
CASE.1: $x<x'$
\begin{align*}
\phi_0\left( x',\underline{\hat{y}}(x)\right) = m \Leftrightarrow \underline{\hat{y}}(x) = \frac{m}{x}.
\end{align*}
CASE.2: $x \geq x'$
\begin{align*}
\phi_0 \left( x', y_{inf}\right) > m \Leftrightarrow \underline{\hat{y}}(x) = y_{inf}.
\end{align*}
\end{document}
希望有人可以帮助解决这个问题。
韩国