A dynamic programming model for effect of worker’s type on wage arrears

Abstract

We analyze how wage arrears are affected by the worker’s type (in this paper, the worker’s type means the worker’s attitude to wage arrears). Wage arrears cause workers’ negative emotion which may lead to serious social problem and the government may intervene. In this paper, we model the process of wage arrears as a Markov decision process in which the firm is the decision maker. We develop an optimal solution approach under the assumption that the worker’s negative emotion threshold (The worker’s negative emotion increases monotonically with increasing back pay. Once the cumulative back pay exceeds a particular value, the worker will take legal actions and the government will intervene. We define the particular value as the worker’s negative emotion threshold.) is normally distributed and provide insights into how wage arrears vary with the worker’s type and the government intervention. We propose the optimal government intervention which stops wage arrears and does not disturb the normal order of the market economy. We show that the intervention depends on the worker’s type and the results imply that the government intervention should be adjusted dynamically according to different regions, industrial sectors and time periods.

Appendix

Proof of Theorem 1

$$\begin{aligned} G^{'}(x)= & {} 1-(P+x)f(x)-F(x),\end{aligned}$$

(17)

$$\begin{aligned} G^{''}(x)= & {} \left[ \frac{(x+P)(x-\mu )}{\sigma ^2}-2\right] f(x). \end{aligned}$$

(18)

Note that $\frac{(x+P)(x-\mu )}{\sigma ^2}-2=0$ has two distinct solutions

$$\begin{aligned} x_1= & {} \frac{\mu -P-\sqrt{(P+\mu )^2+8\sigma ^2}}{2},\end{aligned}$$

(19)

$$\begin{aligned} x_2= & {} \frac{\mu -P+\sqrt{(P+\mu )^2+8\sigma ^2}}{2}, \end{aligned}$$

(20)

then for $x<x_1$ or $x>x_2$, we have $G^{''}(x)>0$, which means $G^{'}(x)$ is strictly monotone increasing; for $x_1<x<x_2$, we have $G^{''}(x)<0$, which means $G^{'}(x)$ is strictly monotone decreasing. And by $G^{'}(-\infty )=1,G^{'}(\infty )=0$, we can obtain that for $x\in (-\infty ,x_1]$, $G^{'}(x)>1$; and for $x\in [x_2,\infty )$, $G^{'}(x)<0$. Hence, there exists a unique $x^*\in (x_1,x_2)$, such that for $x<x^*, G^{'}(x)>0$, which means G(x) is strictly monotone increasing and for $x>x^*, G^{'}(x)<0$, which means G(x) is strictly monotone decreasing. Obviously, G(x) attains its global maximum at $x^*$.$\square$

Proof of Corollary 1

By (3), we have

$$\begin{aligned} G^{'}(x^*)=1-(P+x^*)f(x^*)-F(x^*)=0 \end{aligned}$$

(21)

Differentiating (21) at $x^*$ with respect to $\mu$, we obtain

$$\begin{aligned} \frac{d x^*}{d\mu }=\frac{\displaystyle \frac{(x^*+P)(x^*-\mu )}{\sigma ^2}-1}{\displaystyle \frac{(x^*+P)(x^*-\mu )}{\sigma ^2}-2} \end{aligned}$$

(22)

From the proof of Theorem 1, we have

$$\begin{aligned} \frac{(x^*+P)(x^*-\mu )}{\sigma ^2}-2<0. \end{aligned}$$

(23)

Let us consider the sign of

$$\begin{aligned} \frac{(x^*+P)(x^*-\mu )}{\sigma ^2}-1. \end{aligned}$$

$\displaystyle h(x)\triangleq \frac{(x+P)(x-\mu )}{\sigma ^2}-1$ has two distinct solutions:

$$\begin{aligned} z_1= & {} \frac{\mu -P-\sqrt{(P+\mu )^2+4\sigma ^2}}{2},\end{aligned}$$

(24)

$$\begin{aligned} z_2= & {} \frac{\mu -P+\sqrt{(P+\mu )^2+4\sigma ^2}}{2}. \end{aligned}$$

(25)

For $x<z_1$ or $x>z_2$, we have $h(x)>0$; for $z_1<x<z_2$, we have $h(x)<0$. Denote

$$\begin{aligned} \beta =\frac{P+\mu }{\sigma }, \end{aligned}$$

(26)

Now we consider the sign of $G^{'}(z_1),G^{'}(z_2)$ to decide the relative position of $x^*,z_1,z_2$.

$$\begin{aligned} G^{'}(z_1)= & {} 1-\Phi \left( \frac{-\beta -\sqrt{\beta ^2+4}}{2}\right) - \frac{\beta -\sqrt{\beta ^2+4}}{2}\phi \left( \frac{-\beta -\sqrt{\beta ^2+4}}{2}\right) ,\\ G^{'}(z_2)= & {} 1-\Phi \left( \frac{-\beta +\sqrt{\beta ^2+4}}{2}\right) - \frac{\beta +\sqrt{\beta ^2+4}}{2}\phi \left( \frac{-\beta +\sqrt{\beta ^2+4}}{2}\right) . \end{aligned}$$

Differentiating $G^{'}(z_2)$ with respect to $\beta$, we have $G^{'}(z_2)$ is strictly monotone decreasing in $\beta$. And $G^{'}(z_2)\rightarrow 1-\Phi (1)-\phi (1)<0$ as $\beta \rightarrow 0$, hence $G^{'}(z_2)<0$. Note that $G^{'}(z_1)>0$.

By the property of $G^{'}(x)$, we obtain $x^*\in (z_1,z_2)$. Then

$$\begin{aligned} \frac{(x^*+P)(x^*-\mu )}{\sigma ^2}-1<0. \end{aligned}$$

Hence, $\displaystyle \frac{d x^*}{d\mu }>0$.$\square$

Proof of Corollary 2

1. Differentiating

$$\begin{aligned} G^{'}(x^*)=1-(P+x^*)f(x^*)-F(x^*)=0 \end{aligned}$$

(27)

at $x^*$ with respect to $\sigma ^2$, we obtain

$$\begin{aligned} \frac{d x^*}{d\sigma ^2}= \frac{1}{2\sigma ^2}\frac{\displaystyle \frac{(x^*-\mu )^2(x^*+P)}{\sigma ^2}-(2x^*+P-\mu )}{\displaystyle \frac{(x^*+P)(x^*-\mu )}{\sigma ^2}-2}. \end{aligned}$$

(28)

As

$$\begin{aligned} \frac{(x^*-\mu )^2(x^*+P)}{\sigma ^2}-(2x^*+P-\mu )=(x^*-\mu )\left[ \frac{(x^*+P)(x^*-\mu )}{\sigma ^2}-2\right] -(P+\mu ), \end{aligned}$$

(29)

then

$$\begin{aligned} 2\sigma ^2\frac{d x^*}{d\sigma ^2}=x^*-\mu -\frac{P+\mu }{\displaystyle \frac{(x^*+P)(x^*-\mu )}{\sigma ^2}-2}. \end{aligned}$$

(30)

Moreover,

$$\begin{aligned} G^{'}(\mu )=1-(P+\mu )f(\mu )-F(\mu )=\frac{1}{2}-\frac{\beta }{\sqrt{2\pi }}\ge 0 \Longleftrightarrow \beta \le \sqrt{\frac{\pi }{2}} \end{aligned}$$

(31)

By the property of $G^{'}(x)$, we know that $G^{'}(\mu )\ge 0$ is equivalent to $x^*\ge \mu$. Hence, by (30) and (23), we have $\displaystyle \frac{d x^*}{d\sigma ^2}>0\,$ for $\displaystyle \beta \le \sqrt{\frac{\pi }{2}}$.

2.

$$\begin{aligned} G^{'}\left( \frac{3\mu -P}{4}\right)= & {} 1-F\left( \frac{3\mu -P}{4}\right) -\frac{3(\mu +P)}{4}f\left( \frac{3\mu -P}{4}\right) \nonumber \\= & {} \Phi \left( \frac{\beta }{4}\right) -\frac{3}{4}\beta \phi \left( \frac{\beta }{4}\right) , \end{aligned}$$

(32)

Differentiating $G^{'}\left( \frac{3\mu -P}{4}\right)$ with respect to $\beta$, we can see that it attains its minimum at $\beta =\frac{4\sqrt{6}}{3}$ with value 0.0927, and thus $G^{'}\left( \frac{3\mu -P}{4}\right) >0$.

By (31), we have $\displaystyle G^{'}(\mu )<0\,$ for $\beta >\sqrt{\frac{\pi }{2}}$. By the property of $G^{'}(x)$, we obtain $x^*\in (\frac{3\mu -P}{4},\mu )\,$. $\square$

Proof of Corollary 3

The proof is similar to that of Corollary 1.$\square$

Proof of Corollary 4

From the proof of Theorem 1, we know that

$$\begin{aligned} G^{'}(0)=1-P f(0)-F(0)\le 0 \quad \text{ if } \text{ and } \text{ only } \text{ if }\quad x^*\le 0. \end{aligned}$$

(33)

Then for $P\ge \frac{1-F(0)}{f(0)}$, we have $x^*\le 0$.$\square$

Proof of Proposition 1

For the given $\sigma ^2$, differentiating (5) with respect to $\mu$, we obtain

$$\begin{aligned} \frac{d P^*}{d\mu }=\frac{\phi \left( \frac{\mu }{\sigma }\right) + \frac{\mu }{\sigma }\Phi \left( \frac{\mu }{\sigma }\right) }{\phi \left( \frac{\mu }{\sigma }\right) }>0, \end{aligned}$$

(34)

$\square$

Proof of Proposition 2

For the given $\mu$, differentiating (5) with respect to $\sigma$, we obtain

$$\begin{aligned} \frac{d P^*}{d \sigma }=\frac{\Phi \left( \frac{\mu }{\sigma }\right) \left( 1-\frac{\mu ^2}{\sigma ^2}\right) - \frac{\mu }{\sigma }\phi \left( \frac{\mu }{\sigma }\right) }{\phi \left( \frac{\mu }{\sigma }\right) } \end{aligned}$$

(35)

It is clear that for $\sigma \rightarrow 0,\frac{d P^*}{d \sigma }\rightarrow -\infty$ and for $\sigma \rightarrow \infty , \frac{d P^*}{d \sigma }\rightarrow \sqrt{\frac{\pi }{2}}>0$. $\square$

Proof of Theorem 2

First we prove (36) by mathematical induction. For $t=1,\ldots ,N$,

$$\begin{aligned} M_t((1,b))=\delta [(k-1)b+c_t], \end{aligned}$$

(36)

where $c_t$ only depends on $r,w,k,\delta ,\mu _t,\sigma _t^2,P_t,\ldots ,\mu _N,\sigma _N^2,P_N$.

By (12), we obtain

$$\begin{aligned} M_N((1,b))=\delta [(k-1)b+\max _a G_N(b+a)] \end{aligned}$$

(37)

Denote $v=b+a$ and let us consider $G_N(v)=r-w+v-(P_N+v)F_N(v)$. Differentiating $G_N(v)$ with respect to v, we have

$$\begin{aligned} G_N^{'}(v)=1-(P_N+v)f_N(v)-F_N(v). \end{aligned}$$

(38)

Through the proof of Theorem 1, we know that there exists a unique $v_N^*\in (v_{1,N},v_{2,N})$ such that $G_N^{'}(v_N^*)=0$ and $G_N(v)$ has a maximum at $v_N^*$. Here $~v_{1,N},v_{2,N}$ are the only two solutions of $G^{''}(v)=0$, satisfying

$$\begin{aligned} v_{1,N}= & {} \frac{\mu _N-P_N-\sqrt{(P_N+\mu _N)^2+8\sigma _N^2}}{2}\end{aligned}$$

(39)

$$\begin{aligned} v_{2,N}= & {} \frac{\mu _N-P_N+\sqrt{(P_N+\mu _N)^2+8\sigma _N^2}}{2}. \end{aligned}$$

(40)

Denote

$$\begin{aligned} c_N=G_N(v_N^*)=r-w+v_N^*-(P_N+v_N^*)F_N(v_N^*), \end{aligned}$$

(41)

then $a_N^*((1,b))=v_N^*-b$ is the firm’s optimal decision in state (1, b) at time N. It is uniquely determined by $G_N^{'}(b+a)=0$ and

$$\begin{aligned} M_N((1,b))=\delta [(k-1)b+c_N], \end{aligned}$$

(42)

By (41), it is clear that $c_N$ only depends on $r,w,\mu _N,\sigma _N^2,P_N$. Then (36) holds for $t=N$.

Suppose (36) holds for some $t+1(t<N)$, we have

$$\begin{aligned} G_t(v)=r-w+v-(P_t+v)F_t(v)+\delta [1-F_t(v)][(k-1)v+c_{t+1}]. \end{aligned}$$

(43)

Then

$$\begin{aligned} G_t^{'}(v)= & {} [1+\delta (k-1)][1-F_t(v)]-f_t(v)\left( [1+\delta (k-1)]v+\delta c_{t+1}+P_t\right) {,}\\ G_t^{''}(v)= & {} \left[ \frac{(v-\mu _t)([1+\delta (k-1)]v+\delta c_{t+1}+P_t)}{\sigma _t^2}-2[1+\delta (k-1)]\right] f_t(v). \end{aligned}$$

$\frac{(v-\mu _t)([1+\delta (k-1)]v+\delta c_{t+1}+P_t)}{\sigma _t^2}-2[1+\delta (k-1)]=0$ has two distinct solutions:

$$\begin{aligned} v_{1,t}= & {} \frac{\mu _t[1+\delta (k-1)]-(\delta c_{t+1}+P_t)-\sqrt{\Delta _t}}{2[1+\delta (k-1)]}<0,\end{aligned}$$

(44)

$$\begin{aligned} v_{2,t}= & {} \frac{\mu _t[1+\delta (k-1)]-(\delta c_{t+1}+P_t)+\sqrt{\Delta _t}}{2[1+\delta (k-1)]}>\mu _t, \end{aligned}$$

(45)

where

$$\begin{aligned} \Delta _t=\left( \delta c_{t+1}+P_t+\mu _t[1+\delta (k-1)]\right) ^2+8\sigma _t^2[1+\delta (k-1)]^2. \end{aligned}$$

(46)

Similar to the proof of Theorem 1, we can obtain there exists a unique $v_t^*\in (v_{1,t},v_{2,t})$ such that $G_t^{'}(v_t^*)=0$ and $G_t(v)$ has a maximum at $v_t^*$.

Denote

$$\begin{aligned} c_t= & {} G_t(v_t^*)=r-w+v_t^*-(P_t+v_t^*)F_t(v_t^*)\nonumber \\&+\delta [1-F_t(v_t^*)][(k-1)v_t^*+c_{t+1}], \end{aligned}$$

(47)

then $a_t^*((1,b))=v_t^*-b$ is the firm’s optimal decision in state (1, b) at time t. It is uniquely given by $G_t^{'}(b+a)=0$. By (12), we have $M_t((1,b))=\delta \left[ (k-1)b+c_t\right]$. Since $v_t^*$ only depends on $r,w,\delta ,k,\mu _t,\sigma _t^2,P_t,c_{t+1}$ and $c_{t+1}$ only depends on $r,w,k,\delta ,\mu _{t+1},\sigma _{t+1}^2,P_{t+1},\ldots ,\mu _N,\sigma _N^2,P_N$ as to the assumption, by (47) $c_t$ only depends on $r,w,k,\delta ,\mu _t,\sigma _t^2,P_t,\ldots ,\mu _N,\sigma _N^2,P_N$. Thus (36) is proved.

From the above, we obtain that for $t=1,\ldots ,N-1$, (43) holds. Then the optimal wage arrear policy for the firm is as follows.

$$\begin{aligned} a_t^*((s,b))= & {} \left\{ \begin{array}{ll} v_t^*-b,&{}s=1;\\ -b,&{}s=0;\\ \diamond ,&{}s=\varDelta . \end{array}\right. t=1,\ldots ,N.\end{aligned}$$

(48)

$$\begin{aligned} a_{N+1}^*((s,b))= & {} \left\{ \begin{array}{ll} \diamond ,&{}s=1;\\ -b,&{}s=0. \end{array}\right. \end{aligned}$$

(49)

Here, $v_t^*$ is uniquely determined by

$$\begin{aligned} G_t^{'}(v)= & {} [1+\delta (k-1)][1-F_t(v)]-f_t(v)\left( [1+\delta (k-1)]v+\delta c_{t+1}+P_t\right) =0,\nonumber \\ t= & {} 1,\ldots ,N-1.\end{aligned}$$

(50)

$$\begin{aligned} G_N^{'}(v)= & {} 1-(P_N+v)f_N(v)-F_N(v)=0. \end{aligned}$$

(51)

Note that (50) contains $c_{t+1}$, $c_t$ is obtained from (47) recursively and (47) contains $v_t^*$, thus $v_t^*,c_t(t=1,\ldots ,N)$ can be obtained recursively by the following.

$$\begin{aligned} \left\{ \begin{array}{rcl} 0&{}=&{}1-(P_N+v_N^*)f_N(v_N^*)-F_N(v_N^*),\\ c_N&{}=&{}r-w+v_N^*-(P_N+v_N^*)F_N(v_N^*),\\ 0&{}=&{}[1+\delta (k-1)][1-F_t(v_t^*)]-f_t(v_t^*)\left( [1+\delta (k-1)]v_t^*+\delta c_{t+1}+P_t\right) ,\\ c_t&{}=&{}r-w+v_t^*-(P_t+v_t^*)F_t(v_t^*)\\ &{} &{}{}+\delta [1-F_t(v_t^*)][(k-1)v_t^*+c_{t+1}],t=1,\ldots ,N-1. \end{array}\right. \end{aligned}$$

(52)

$\square$

Proof of Corollary 5

From theorem 2, given the state (1, b) at time N, we have

$$\begin{aligned} G_N^{'}(v_N^*)=1-(P_N+v_N^*)f_N(v_N^*)-F_N(v_N^*)=0 \end{aligned}$$

(53)

Differentiating it with respect to $P_N$, we obtain

$$\begin{aligned} \frac{d a_N^*((1,b))}{d P_N}=\frac{d v_N^*}{d P_N}=\frac{1}{\frac{(v_N^*+P_N)(v_N^*-\mu _N)}{\sigma _N^2}-2}. \end{aligned}$$

(54)

Through the proof of Theorem 2, we have $\frac{d a_N^*((1,b))}{d P_N}<0$.

For $t=1,\ldots ,N-1$,

$$\begin{aligned} G_t^{'}(v_t^*)=[1+\delta (k-1)][1-F_t(v_t^*)]-f_t(v_t^*)\left( [1+\delta (k-1)]v_t^*+\delta c_{t+1}+P_t\right) =0 \end{aligned}$$

(55)

Differentiating (55) with respect to $P_t$, we obtain

$$\begin{aligned} \frac{d a_t^*((1,b))}{d P_t}=\frac{d v_t^*}{d P_t}=\frac{1}{\frac{(v_t^*-\mu _t)([1+\delta (k-1)]v_t^*+\delta c_{t+1}+P_t)}{\sigma _t^2}-2[1+\delta (k-1)]} \end{aligned}$$

(56)

Also by Theorem 2, we have $\frac{d a_t^*((1,b))}{d P_t}<0$.$\square$

Proof of Corollary 6

Differentiating (36) with respect to $P_t$, we obtain

$$\begin{aligned} \frac{\partial M_t((1,b))}{\partial P_t}=\delta \frac{\partial c_t}{\partial P_t},t=1,\ldots ,N. \end{aligned}$$

(57)

By (41), we have

$$\begin{aligned} \frac{\partial c_N}{\partial P_N}= & {} \frac{d v_N^*}{d P_N}-\left( 1+\frac{d v_N^*}{d P_N}\right) F_N(v_N^*)-(P_N+v_N^*)f_N(v_N^*)\frac{d v_N^*}{d P_N}\nonumber \\= & {} \frac{d v_N^*}{d P_N}\left[ 1-(P_N+v_N^*)f_N(v_N^*)-F_N(v_N^*)\right] -F_N(v_N^*)\nonumber \\= & {} -F_N(v_N^*)<0 \end{aligned}$$

(58)

For $t=1,\ldots ,N-1$, by (47), we have

$$\begin{aligned} \frac{\partial c_t}{\partial P_t}= & {} \frac{d v_t^*}{d P_t}-\left( 1+\frac{d v_t^*}{d P_t}\right) F_t(v_t^*)-(P_t+v_t^*)f_t(v_t^*)\frac{d v_t^*}{d P_t}\nonumber \\&{}-\delta f_t(v_t^*)\frac{d v_t^*}{d P_t}[(k-1)v_t^*+c_{t+1}]+\delta (k-1)\left[ 1-F_t(v_t^*)\right] \frac{d v_t^*}{d P_t}\nonumber \\= & {} \frac{d v_t^*}{d P_t}\left\{ [1+\delta (k-1)]\left[ 1-F_t(v_t^*)\right] \right. \nonumber \\&\left. {}-f_t(v_t^*)\left( [1+\delta (k-1)]v_t^*+\delta c_{t+1}+P_t\right) \right\} -F_t(v_t^*)\nonumber \\= & {} -F_t(v_t^*)<0. \end{aligned}$$

(59)

Obviously, $M_t((1,b))$ is strictly monotone decreasing in $P_t$.

Proof of Corollary 7

From Theorem 2, we obtain that $c_t$ does not depend on b. Then by (36), the result is straightforward.$\square$

Proof of Theorem 3

The property of $G_t^{'}(\cdot )$ in Theorem 2 is the same as $G^{'}(\cdot )$ in Theorem 1 and the optimal government intervention in period t is the penalty level satisfying $G_t^{'}(0)=0$.

Proof of Corollary 8

By (2), if the state at time t is (1, b), $v_t^*=b+a_t^*$ is uniquely determined by $G_t^{'}(v)=0$.

For $t=N$, similar to (54), we can obtain $\displaystyle \frac{d a_N^*((1,b))}{d P}<0$.

Next, we prove: for $t=1,\ldots ,N$,

$$\begin{aligned} -1<\frac{\partial c_t}{\partial P}<0. \end{aligned}$$

(60)

By (58), $\displaystyle \frac{\partial c_N}{\partial P}=-F_N(v_N^*)$, then (60) holds for $t=N$.

For $t=1,\ldots ,N-1$, differentiating (47) with respect to P, we have

$$\begin{aligned} \frac{\partial c_t}{\partial P}= & {} \frac{d v_t^*}{d P}-\left( 1+\frac{d v_t^*}{d P}\right) F_t(v_t^*)-(P+v_t^*)f_t(v_t^*)\frac{d v_t^*}{d P}\nonumber \\&{}-\delta f_t(v_t^*)\frac{d v_t^*}{d P}[(k-1)v_t^*+c_{t+1}]+\delta \left[ 1-F_t(v_t^*)\right] \left[ (k-1)\frac{d v_t^*}{d P}+\frac{\partial c_{t+1}}{\partial P}\right] \nonumber \\= & {} \delta [1-F_t(v_t^*)]\frac{\partial c_{t+1}}{\partial P}-F_t(v_t^*)+\frac{d v_t^*}{d P}\left\{ [1+\delta (k-1)]\left[ 1-F_t(v_t^*)\right] \right. \nonumber \\&\left. {}-f_t(v_t^*)\left( [1+\delta (k-1)]v_t^*+\delta c_{t+1}+P\right) \right\} \nonumber \\= & {} \delta [1-F_t(v_t^*)]\frac{\partial c_{t+1}}{\partial P}-F_t(v_t^*). \end{aligned}$$

(61)

Then

$$\begin{aligned} \frac{\partial c_t}{\partial P}+1=[1-F_t(v_t^*)]\left[ 1+\delta \frac{\partial c_{t+1}}{\partial P}\right] \end{aligned}$$

(62)

By the mathematical induction algorithm, we can easily obtain (60) holds for $t=1,\ldots ,N$.

For $t<N$, differentiating

$$\begin{aligned} G_t^{'}(v_t^*)=[1+\delta (k-1)][1-F_t(v_t^*)]-f_t(v_t^*)\left( [1+\delta (k-1)]v_t^*+\delta c_{t+1}+P\right) =0 \end{aligned}$$

(63)

with respect to P, we have

$$\begin{aligned} \frac{\partial a_t^*}{\partial P}=\frac{\partial v_t^*}{\partial P}=\frac{\displaystyle 1+\delta \frac{\partial c_{t+1}}{\partial P}}{\displaystyle \frac{(v_t^*-\mu _t)([1+\delta (k-1)]v_t^*+\delta c_{t+1}+P)}{\sigma _t^2}-2[1+\delta (k-1)]}<0. \end{aligned}$$

(64)

Thus, $a_t^*((1,b))$ is strictly monotone decreasing in P.$\square$

Proof of Corollary 9

It is straightforward by (36) and Corollary 8.$\square$