Microeconometric Analysis of Residential Water Demand

Abstract

This paper develops a structural model for obtaining price elasticities and evaluating consumer’s response to changes in nonlinear tariffs when only panel data on household consumption are available. The model and the empirical strategy address problems implied by nonlinear tariffs, existence of a fixed cost, and use of limited data, giving rise to a random effects model with a nonlinear individual effect. Results show that the estimated model does well at fitting data and demand is inelastic, although elasticity varies by initial consumption block. Then, I estimate welfare consequences of implementing several demand policies.

Appendices

Appendices: Likelihood of the Approximate Model

Let $\log y_{i}=(\log y_{i1},\log y_{i2},\ldots ,\log y_{iT_{i}})$ be the vector of observations of a household $i$.¹³ The individual likelihood of the approximate model is equal to the sum of three integrals that, as shown below, have closed form:

$$\begin{aligned} f(\log y_{i1},\log y_{i2},\ldots ,\log y_{iT_{i}})&= \int \limits _{-\infty }^{\log 30+\log c-\log \nu }f(\log y_{i1},\log y_{i2},\ldots ,\log y_{iT_{i}}/\beta _{i})f(\beta _{i})d\beta _{i} \\&\quad +\int \limits _{\log 30+\log c-\log \nu }^{\log 30+\log (c+p_{2})-\log \nu } f(\log y_{i1},\log y_{i2},\ldots ,\log y_{iT_{i}}/\beta _{i})f(\beta _{i})d\beta _{i} \\&\quad +\int \limits _{\log 30+\log (c+p_{2})-\log \nu }^{+\infty } f(\log y_{i1},\log y_{i2},\ldots ,\log y_{iT_{i}}/\beta _{i})f(\beta _{i})d\beta _{i} \end{aligned}$$

1. First integral (optimal demand is $\log x_{i}=\beta _{i}-\log c+\log \nu$):

$\overline{\log y}_{i}$ is a sufficient statistic for $\beta _{i}$ (see Arellano 2003, p. 25). Thus, it holds

$$\begin{aligned} f(\log y_{i}/\beta _{i},\overline{\log y}_{i})={f(\log y_{i},\overline{\log y}_{i}/\beta _{i}) \over f(\overline{\log y}_{i}/\beta _{i})}={f(\log y_{i}/\beta _{i}) \over f(\overline{\log y}_{i}/\beta _{i})}=f(\log y_{i}/\overline{\log y}_{i}) \end{aligned}$$

and $f(\log y_{i}/\beta _{i})=f(\log y_{i}/\overline{\log y}_{i})f(\overline{\log y}_{i}/\beta _{i})$. After replacing and rearranging terms, the first integral is equal to:

$$\begin{aligned} f(\log y_{i}/\overline{\log y}_{i})f(\overline{\log y}_{i})\int \limits _{-\infty }^{\log 30+\log c-\log \nu }f(\beta _{i}/\overline{\log y}_{i})d\beta _{i}&= f(\log y_{i})p(\beta _{i}\le \log 30\\&\quad +\log c-\log \nu /\overline{\log y}_{i}) \end{aligned}$$

On the one hand, as $\log y_{it}=\log x_{i}+\varepsilon _{it}$, where $\log x_{i} \sim N\left( \mu _{\beta }+\mu ,\text { }\sigma _{\beta }^{2}\right)$, $\mu \equiv \log \nu -\log c$, and $\varepsilon _{it} \sim iid\text { }N\left( 0,\text { }\sigma _{\varepsilon }^{2}\right)$, $\log y_{i}$ follows a normal multivariate distribution:

$$\begin{aligned} \log y_{i}=\left( \begin{array}{c} \log y_{i1} \\ . \\ . \\ \log y_{iT_{i}} \end{array} \right) \sim N\left[ \underset{\overline{\mu }}{\underbrace{ \left( \begin{array}{c} \mu _{\beta }+\mu \\ . \\ . \\ \mu _{\beta }+\mu \end{array} \right) }},\underset{\Omega }{\underbrace{\left( \begin{array}{cccc} \sigma _{\beta }^{2}+\sigma _{\varepsilon }^{2} &{} \sigma _{\beta }^{2} &{}.&{} \sigma _{\beta }^{2} \\ \sigma _{\beta }^{2} &{} . &{} . &{} \sigma _{\beta }^{2} \\ . &{} . &{} . &{} . \\ \sigma _{\beta }^{2} &{} . &{} \sigma _{\beta }^{2} &{} \sigma _{\beta }^{2}+\sigma _{\varepsilon }^{2} \end{array} \right) }}\right] \end{aligned}$$

where $\overline{\mu }$ is a $T_{i}\times 1$ vector and $\Omega$ is a $T_{i}\times T_{i}$ variance-covariance matrix. Thus, $f(\log y_{i})=(2\pi )^{-(T_{i}/2)}(\det \Omega )^{-(1/2)}\exp \left[ -{1 \over 2}(\log y_{i}-\overline{\mu })^{\prime }\Omega ^{-1}(\log y_{i}-\overline{\mu })\right]$.

On the other hand, $p(\beta _{i}\le \log 30-\mu /\overline{\log y}_{i})=\Phi \left[ {\log 30-\mu -\gamma (\overline{\log y}_{i}-\mu )-(1-\gamma )\mu _{\beta } \over \sigma _{\beta }\sqrt{(1-\gamma )}}\right]$ because $\overline{\log y}_{i}$ and $\beta _{i}$ have a normal joint distribution and, so, $\beta _{i}/\overline{\log y}_{i}\sim N\left[ (1-\gamma )\mu _{\beta }\right. \left. +\gamma (\overline{\log y}_{i}-\mu ),\text { }\sigma _{\beta }^{2}(1-\gamma )\right]$, where $\gamma ={\sigma _{\beta }^{2} \over \sigma _{\beta }^{2}+{\sigma _{\varepsilon }^{2} \over T_{i}}}$.

2. Second integral (optimal demand is $\log x_{i}=\log 30$):

$$\begin{aligned}&\int \limits _{\log 30+\log c-\log \nu }^{\log 30+\log (c+p_{2})-\log \nu }f(\log y_{i}/\beta _{i})f(\beta _{i})d\beta _{i}\\&\quad \quad \quad =\left[ \Pi _{t=1}^{T_{i}}{1 \over \sigma _{\varepsilon }}\phi \left( { \log y_{it}-\log 30 \over \sigma _{\varepsilon }}\right) \right] \pi _{2}\left[ \log (c+p_{2})-\log (c)\right] \end{aligned}$$

where $\pi _{2}$ is the probability of optimizing at the kink (see Appendix).

3. Third integral (optimal demand is $\log x_{i}=\beta _{i}-\log (c+p_{2})+\log \nu$):

Like in the first integral,

$$\begin{aligned} \int \limits _{\log 30+\log (c+p_{2})-\log \nu }^{+\infty }f(\log y_{i}/\beta _{i})f(\beta _{i})d\beta _{i}=f(\log y_{i})\text { }p[\beta _{i}\ge \log 30+\log (c+p_{2})-\log \nu /\overline{\log y}_{i}] \end{aligned}$$

On the one hand, $\log y_{i}$ has the following normal multivariate distribution:

$$\begin{aligned} \log y_{i}=\left( \begin{array}{c} \log y_{i1} \\ . \\ . \\ \log y_{iT_{i}} \end{array} \right) \sim N\left[ \underset{\overline{\mu ^{\prime }}}{ \underbrace{\left( \begin{array}{c} \mu _{\beta }+\mu ^{\prime } \\ . \\ . \\ \mu _{\beta }+\mu ^{\prime } \end{array} \right) }},\underset{\Omega }{\underbrace{\left( \begin{array}{cccc} \sigma _{\beta }^{2}+\sigma _{\varepsilon }^{2} &{} \sigma _{\beta }^{2} &{} . &{} \sigma _{\beta }^{2} \\ \sigma _{\beta }^{2} &{} . &{} . &{} \sigma _{\beta }^{2} \\ . &{} . &{} . &{} . \\ \sigma _{\beta }^{2} &{} . &{} \sigma _{\beta }^{2} &{} \sigma _{\beta }^{2}+\sigma _{\varepsilon }^{2} \end{array} \right) }}\right] \end{aligned}$$

where $\overline{\mu ^{\prime }}$ is a $T_{i}\times 1$ vector and $\Omega$ is a $T_{i}\times T_{i}$ variance-covariance matrix. Thus, $f(\log y_{i})=(2\pi )^{-(T_{i}/2)}(\det \Omega )^{-(1/2)}\exp \left[ -{1 \over 2}(\log y_{i}-\overline{\mu ^{\prime }})^{\prime }\Omega ^{-1}(\log y_{i}-\overline{\mu ^{\prime }})\right]$.

On the other hand, $p(\beta _{i}\ge \log 30-\mu ^{\prime }/\overline{\log y}_{i})=1-\Phi \left[ {\log 30-\mu ^{\prime }-\gamma (\overline{\log y}_{i}-\mu ^{\prime })-(1-\gamma )\mu _{\beta } \over \sigma _{\beta }\sqrt{(1-\gamma )}}\right]$ given that $\beta _{i}/\overline{\log y}_{i}\sim N\left[ (1-\gamma )\mu _{\beta }+\gamma (\overline{\log y}_{i}-\mu ^{\prime }),\text { }\sigma _{\beta }^{2}(1-\gamma )\right]$ with $\gamma ={\sigma _{\beta }^{2} \over \sigma _{\beta }^{2}+{\sigma _{\varepsilon }^{2} \over T_{i}}}$.

Finally, the log-likelihood of the approximate model is: $L^{a}(\log y_{1},\log y_{2},\ldots ,\log y_{N})=\sum _{i=1}^{N}\log f(\log y_{i})$.

Standard Errors

Let the log-likelihood of the auxiliary model be $L^{aux}\left[ \theta ;\log y\right] ={\sum \limits _{i=1}^{N}} l_{i}(\theta )$, where $\theta$ is the vector $r \times 1$ of auxiliary parameters. Let define $\widehat{\theta }=\underset{\theta }{\arg \max }L^{aux}\left[ \theta ;\log y\right]$, where plim $\widehat{\theta }=\theta _{0}$. The robust estimate of the asymptotic variance is $\widehat{Var}(\widehat{\theta })={1 \over N}\widehat{H}^{-1}\widehat{W}\widehat{H}^{-1}$, where $\widehat{H}$ and $\widehat{W}$ are consistent estimates of:

$$\begin{aligned} H&= \underset{N\rightarrow \infty }{\lim }{1 \over N}\underset{i=1}{\overset{N}{\sum }}{\partial ^{2}l_{i}(\theta _{0}) \over \partial \theta \partial \theta ^{\prime }}\\ W&= \underset{N\rightarrow \infty }{\lim }{1 \over N}\underset{i=1}{\overset{N}{\sum }}\left( {\partial l_{i}(\theta _{0}) \over \partial \theta }{ \partial l_{i}(\theta _{0}) \over \partial \theta ^{\prime }}\right) \end{aligned}$$

The function $L^{aux}\left[ \theta ;\log y\right]$ is a pseudo-likelihood. So, in general, the information equality does not hold and $W\ne H$. Let $\alpha$ be the vector $k \times 1$ of structural parameters. The variance of $\widehat{\alpha }=\underset{\alpha }{\arg \max }L^{aux} \left[ \widetilde{\theta }(\alpha );\log y\right]$ is:

$$\begin{aligned} \widehat{Var}(\widehat{\alpha })={1 \over N}\left( \widehat{D}^{\prime }\widehat{H} \widehat{D}\right) ^{-1}\widehat{D}^{\prime }\widehat{W}\widehat{D}\left( \widehat{D}^{\prime } \widehat{H}\widehat{D}\right) ^{-1} \end{aligned}$$

where $\widehat{D}$ is the matrix $r \times k$ of numerical partial derivatives evaluated in $\widehat{\alpha }$: $\widehat{D}={\partial \widetilde{\theta }(\widehat{\alpha }) \over \partial \alpha ^{\prime }}$

If the number of parameters in $\alpha$ and $\theta$ is the same (like in this paper), then $\widehat{Var}(\widehat{\alpha })={1 \over N}\widehat{D}^{-1}\widehat{Var}(\widehat{ \theta })\widehat{D}^{-1\prime }$. Once I obtain $\widehat{Var}(\widehat{\alpha })$, I use the delta method to recover the standard errors of $c$ and $\nu$ because they are reparameterized as $c=e^{c^{\prime }}$ and $\nu ={e^{\nu ^{\prime }} \over 1+e^{\nu ^{\prime }}}$, respectively.

Probabilities

1.1 Before Including the Time-Varying Term $\varepsilon _{it}$

Under the assumption $\beta _{i}\sim N\left( \mu _{\beta },\sigma _{\beta }^{2}\right)$, the probability of the optimal demand (4) is:

• First block: $\pi _{1}= p\left[ \beta _{i}<\log \left( {30c \over \nu }+FC\right) \right] =\Phi \left[ {\log \left( {30c \over \nu } +FC\right) -\mu _{\beta } \over \sigma _{\beta }}\right]$

• Second block: $\pi _{3}\!=\!p\left[ \beta _{i}>\log \left( {30(c+p_{2}) \over \nu }+FC-30p_{2}\right) \right] \!=\!\Phi \left[ {-\log \left( {30(c+p_{2}) \over \nu }+FC-30p_{2}\right) +\mu _{\beta } \over \sigma _{\beta }}\right]$

• Kink: $\pi _{2}=1-\pi _{1}-\pi _{3}=\Phi \left[ {\log \left( {30(c+p_{2}) \over \nu }+FC-30p_{2}\right) -\mu _{\beta } \over \sigma _{\beta }}\right] -\Phi \left[ {\log \left( {30c \over \nu }+FC\right) -\mu _{\beta } \over \sigma _{\beta }}\right]$

The kink is a mass-point in the demand distribution ($\pi _{2}$ is equal to the probability of the interval of values of $\beta _{i}$ for which the kink is the optimum).

1.2 After Including the Time-Varying Term $\varepsilon _{it}$

Obtaining the unconditional probability $p(\log y_{it}\le r)$ requires, first, to derive the accumulated distribution function of $\log y_{it}$ conditional on $\varepsilon _{it}$. Let $r$ be any value from the distribution of $\log y_{it}$ (so, $R=e^{r}$) and $\overline{r}\equiv \log 30+\varepsilon _{it}$. $p(\log y_{it}\le r \mid \varepsilon _{it})$ is a function of $r$, $\overline{r}$ and $\varepsilon _{it}$ in this way:

• 1. If $r<\overline{r}$, $\log y_{it}=\log \nu -\log c+\log \Big (e^{\beta _{i}}-FC\Big )+\varepsilon _{it}$. Then:

  $$\begin{aligned} p(\log y_{it}\le r \mid \varepsilon _{it})&= p\left[ \log \nu -\log c+\log \Big (e^{\beta _{i}}-FC\Big )+\varepsilon _{it}\le r \mid \varepsilon _{it}\right] \\&= p\left[ \beta _{i}\le \log \left( {Rc \over \nu e^{\varepsilon _{it}}}+FC\right) \right] =\Phi \left[ {\log \left( {Rc \over \nu e^{\varepsilon _{it}}}+FC\right) -\mu _{\beta } \over \sigma _{\beta }}\right] \equiv F_{1}(\log y_{it} \mid \varepsilon _{it}) \end{aligned}$$

• 2. If $r=\overline{r}$, $\log y_{it}=\log 30+\varepsilon _{it}$. Then:

  $$\begin{aligned}&p(\log y_{it}\le \overline{r} \mid \varepsilon _{it})=p(\log y_{it}< \overline{r} \mid \varepsilon _{it})+p(\log y_{it}=\overline{r} \mid \varepsilon _{it}) \\&\quad =p\left[ \beta _{i}\le \log \left( {\overline{R}c \over \nu e^{\varepsilon _{it}}}+FC\right) \right] +p\left[ \log \left( {30c \over \nu }+FC\right) <\beta _{i}<\log \left( {30(c+p_{2}) \over \nu }+FC-30p_{2}\right) \right] \\&\quad =\Phi \left[ \frac{\log \left( {\overline{R}c \over \nu e^{\varepsilon _{it}}}+FC\right) -\mu _{\beta }}{\sigma _{\beta }}\right] +\Phi \left[ {\log \left( \frac{ 30(c+p_{2})}{\nu }+FC\right) -\mu _{\beta } \over \sigma _{\beta }}\right] \\&\qquad -\Phi \left[ {\log \left( {30c \over \nu }+FC\right) -\mu _{\beta } \over \sigma _{\beta }}\right] \equiv F_{2}(\log y_{it} \mid \varepsilon _{it}) \end{aligned}$$

• 3. If $r>\overline{r}$, $\log y_{it}=\log \nu -\log (c+p_{2})+\log (e^{\beta _{i}}-FC+30p_{2})+\varepsilon _{it}$. Then:

  $$\begin{aligned}&p(\log y_{it}\le r \mid \varepsilon _{it})=p(\log y_{it}<\overline{r} \mid \varepsilon _{it})+p(\log y_{it}=\overline{r} \mid \varepsilon _{it})+p(\overline{r}<\log y_{it}\le r \mid \varepsilon _{it})\\&\quad =\Phi \left[ {\log \left( {\overline{R}c \over \nu e^{\varepsilon _{it}}}+FC\right) -\mu _{\beta } \over \sigma _{\beta }}\right] +\Phi \left[ {\log \left( {30(c+p_{2}) \over \nu }+FC\right) -\mu _{\beta } \over \sigma _{\beta }}\right] \\&\qquad -\Phi \left[ {\log \left( {30c \over \nu }+FC\right) -\mu _{\beta } \over \sigma _{\beta }}\right] +\Phi \left[ {\log \left( {R(c+p_{2}) \over \nu e^{\varepsilon _{it}}}+FC-30p_{2}\right) -\mu _{\beta } \over \sigma _{\beta } }\right] \\&\qquad -\Phi \left[ {\log \left( {\overline{R}(c+p_{2}) \over \nu e^{\varepsilon _{it}}}+FC-30p_{2}\right) -\mu _{\beta } \over \sigma _{\beta }} \right] \equiv F_{3}(\log y_{it} \mid \varepsilon _{it}) \end{aligned}$$

To summarize,

$$\begin{aligned} F(\log y_{it}/\varepsilon _{it})=\left\{ \begin{array}{l} F_{1}(\log y_{it}/\varepsilon _{it}) \quad if \quad r<\overline{r} \\ F_{2}(\log y_{it}/\varepsilon _{it} \quad if \quad r=\overline{r}\\ F_{3}(\log y_{it}/\varepsilon _{it}) \quad if \quad r>\overline{r} \end{array} \right. \end{aligned}$$

Then, $p(\log y_{it}\le r)={1 \over M}{\sum \limits _{j=1}^{M}} p(\log y_{it}\le r \mid \varepsilon _{jt})$. Conditional probabilities are calculated by simulating $M=100,000$ draws from the normal distribution $N(0,\widehat{\sigma }_{\varepsilon }^{2})$, where $\widehat{\sigma }_{\varepsilon }^{2}$ is the estimate from Table 8. Using $F(\log y_{it}/\varepsilon _{jt})$ and the estimated parameters, I obtain the conditional probability for each $\varepsilon _{jt}$. In this way, I compute the probability of consuming at the first block, $p(first\text { }block)\equiv p(\log y_{it}\le \log 30)$, by setting $r=\log 30$. The probability of consuming at the second block is $p(second\text { }block)=1-p(first\text { }block)$. I repeat this process 100 times and I average $p(first\text { }block)$ and $p(second\text { }block)$ to obtain the final probabilities.

The Model for the Tariff Since 2005

The utility maximization problem for the new tariff (see Panel B of Table 1) is:

$$\begin{aligned}&\underset{\left\{ rg,\text { }x\right\} }{Max}\,\,\,\, u(rg,x)=(rg)^{1-\nu }x^{\nu }\\&\text {subject to:} \left\{ \begin{array}{l} m=rg+cx+p_{1}x\quad if\quad x\le 25 \\ m=rg+25p_{1}+cx+(x-25)p_{2}\quad if\quad 25<x\le 45 \\ m=rg+25p_{1}+20p_{2}+cx+(x-45)p_{3}\quad if\quad 45<x\le 90 \\ m=rg+25p_{1}+20p_{2}+45p_{3}+cx+(x-90)p_{4}\quad if\quad x>90 \end{array} \right. \end{aligned}$$

Optimal demand:

• Block 1: $x_{1}(m)={\nu \over c+p_{1}}m$.

• Block 2: $x_{2}(m)={\nu \over c+p_{2}}\left( m-25p_{1}+25p_{2}\right)$.

• Block 3: $x_{3}(m)={\nu \over c+p_{3}}\left( m-25p_{1}-20p_{2}+45p_{3}\right)$.

• Block 4: $x_{4}(m)={\nu \over c+p_{4}}\left( m-25p_{1}-20p_{2}-45p_{3}+90p_{4}\right)$.

After linearizing and assuming $m_{i}=e^{\beta _{i}}$, with $\beta _{i}\sim N\left( \mu _{\beta },\sigma _{\beta }^{2}\right)$,

$$\begin{aligned} \log x_{i}=\left\{ \begin{array}{l} \log { \nu -}\log { (c+p}_{1}{ )+\beta _{i}}\\ { \quad if \quad \beta }_{i}{ <}\log \left( {25(c+p_{1}) \over \nu }\right) \\ \log { 25} \\ { \quad if}\quad \log \left( {25(c+p_{1}) \over \nu }\right) { <\beta }_{i}{ <}\log \left( {25(c+p_{2}) \over \nu } { +25p_{1}-25p}_{2}\right) \\ \log { \nu -}\log { (c+p}_{2}{ )+}\log \left( { e}^{\beta _{i}}{ -25p_{1}+25p}_{2}\right) \\ { \quad if}\quad \log \left( \frac{25(c+p_{2})}{\nu }{+25p_{1}-25p}_{2}\right) { <\beta }_{i}{ <}\log \left( {45(c+p_{2}) \over \nu }{ +25p_{1}-25p}_{2}\right) {} \\ \log { 45} \\ { \quad if}\quad \log \left( {45(c+p_{2}) \over \nu }{ +25p_{1}-25p}_{2}\right) { <\beta }_{i}{ <}\log \left( {45(c+p_{3}) \over \nu }{ +25p_{1}+20p}_{2} { -45p}_{3}\right) \\ \log { \nu -}\log { (c+p}_{3}{ )+} \log \left( { e}^{\beta _{i}}{ -25p_{1}-20p}_{2} { +45p}_{3}\right) \\ { \quad if}\quad \log \left( {45(c+p_{3}) \over \nu }{ +25p_{1}+20p}_{2}{ -45p}_{3}\right) { <\beta }_{i} { <}\log \left( {90(c+p_{3}) \over \nu }{ +25p_{1}+20p}_{2}{ -45p}_{3}\right) \\ \log { 90} \\ { \quad if}\quad \log \left( {90(c+p_{3}) \over \nu }{ +25p_{1}+20p}_{2}{ -45p}_{3}\right) { <\beta }_{i} { <}\log \left( {90(c+p_{4}) \over \nu }{ +25p_{1}+20p}_{2}{ +45p}_{3}{ -90p}_{4}\right) \\ \log { \nu -}\log { (c+p}_{4}{ )+} \log \left( { e}^{\beta _{i}}{ -25p_{1}-20p}_{2} { -45p}_{3}{ +90p}_{4}\right) \\ { \quad if \quad \beta }_{i}{ >}\log \left( {90(c+p_{4}) \over \nu }{ +25p_{1}+20p}_{2}{ +45p}_{3}{ -90p}_{4}\right) \end{array} \right. \end{aligned}$$

The probability of each segment of the optimal demand is:

• $\pi _{1}\equiv p\left[ \beta _{i}<\log \left( \frac{25(c+p_{1})}{\nu } \right) \right] =\Phi \left[ \frac{\log 25+\log (c+p_{1})-\log \nu -\mu _{\beta }}{\sigma _{\beta }}\right]$

• $\pi _{2}\equiv p\left[ \log \left( \frac{25(c+p_{1})}{\nu }\right) <\beta _{i}<\log \left( \frac{25(c+p_{2})}{\nu }+25p_{1}-25p_{2}\right) \right] =$ $\Phi \left[ \frac{\log \left( \frac{25(c+p_{2})}{\nu } +25p_{1}-25p_{2}\right) -\mu _{\beta }}{\sigma _{\beta }}\right] -\Phi \left[ \frac{\log 25+\log (c+p_{1})-\log \nu -\mu _{\beta }}{\sigma _{\beta }} \right]$

• $\pi _{3}\equiv p\left[ \log \left( \frac{25(c+p_{2})}{\nu } +25p_{1}-25p_{2}\right) <\beta _{i}<\log \left( \frac{45(c+p_{2})}{\nu } +25p_{1}-25p_{2}\right) \right] =$ $\Phi \left[ \frac{\log \left( \frac{45(c+p_{2})}{\nu }+25p_{1}-25p_{2}\right) -\mu _{\beta }}{ \sigma _{\beta }}\right] - \Phi \left[ \frac{\log \left( \frac{25(c+p_{2})}{ \nu }+25p_{1}-25p_{2}\right) -\mu _{\beta }}{\sigma _{\beta }}\right]$

• $\pi _{4}\!\equiv \! p\left[ \log \left( \frac{45(c+p_{2})}{\nu } +25p_{1}-25p_{2}\right) \!<\!\beta _{i}\!<\!\log \left( \frac{45(c+p_{3})}{\nu } +25p_{1}+20p_{2}-45p_{3}\right) \right]$ $= \Phi \left[ \frac{\log \left( \frac{45(c+p_{3})}{\nu }+25p_{1}+20p_{2}-45p_{3}\right) -\mu _{\beta }}{\sigma _{\beta }}\right] -\Phi \left[ \frac{\log \left( \frac{45(c+p_{2})}{\nu }+25p_{1}-25p_{2}\right) -\mu _{\beta }}{\sigma _{\beta }}\right]$

• $\pi _{5}\equiv p\left[ \log \left( \frac{45(c+p_{3})}{\nu } +25p_{1}+20p_{2}-45p_{3}\right) <\beta _{i}<\log \left( \frac{90(c+p_{3})}{ \nu }+25p_{1}+20p_{2}-45p_{3}\right) \right] =\Phi \left[ \frac{\log \left( \frac{90(c+p_{3})}{\nu }+25p_{1}+20p_{2}-45p_{3} \right) -\mu _{\beta }}{\sigma _{\beta }}\right] -\Phi \left[ \frac{\log \left( \frac{45(c+p_{3})}{\nu }+25p_{1}+20p_{2}-45p_{3}\right) -\mu _{\beta } }{\sigma _{\beta }}\right]$

• $\pi _{6}\equiv p\left[ \log \left( \frac{90(c+p_{3})}{\nu } +25p_{1}+20p_{2} -45p _{3}\right) <\beta _{i}<\log \left( \frac{90(c+p_{4})}{\nu } +25p_{1}+20p_{2}+45p_{3} -90p_{4}\right) \right] =$ $\Phi \left[ \frac{\log \left( \frac{ 90(c+p_{4})}{\nu }+25p_{1}+20p_{2}+45p_{3}-90p_{4}\right) -\mu _{\beta }}{ \sigma _{\beta }}\right] -\Phi \left[ \frac{\log \left( \frac{90(c+p_{3})}{ \nu }+25p_{1}+20p_{2}-45p_{3}\right) -\mu _{\beta }}{\sigma _{\beta }}\right]$

• $\pi _{7}\equiv p\left[ \beta _{i}>\log \left( \frac{90(c+p_{4})}{\nu } +25p_{1}+20p_{2}+45p_{3}-90p_{4}\right) \right] =$ $1-\Phi \left[ \frac{\log \left( \frac{90(c+p_{4})}{\nu } +25p_{1}+20p_{2}+45p_{3}-90p_{4}\right) -\mu _{\beta }}{\sigma _{\beta }} \right]$

To obtain the unconditional probability $p(\log y_{it}\le r)$, I follow the procedure explained in Appendix. Let $r$ be any value from the distribution of $\log y_{it}$ (so, $R=e^{r}$), $\overline{r}_{1}\equiv \log 25+\varepsilon _{it}$, $\overline{r}_{2}\equiv \log 45+\varepsilon _{it}$ and $\overline{r}_{3} \equiv \log 90+\varepsilon _{it}$. $p(\log y_{it}\le r \mid \varepsilon _{it})$ is a function of $r$, $\overline{r}_{1}$, $\overline{r}_{2}$, $\overline{r}_{3}$ and $\varepsilon _{it}$:

• 1. If $r<\overline{r}_{1}$, $\log y_{it}=\log \nu -\log (c+p_{1})+\beta _{i}+\varepsilon _{it}$. Then:

  $$\begin{aligned} p(\log y_{it}\le r \mid \varepsilon _{it})=\Phi \left[ \frac{r-\varepsilon _{it}-\log \nu +\log (c+p_{1})-\mu _{\beta } }{\sigma _{\beta }}\right] \equiv F_{1}(\log y_{it} \mid \varepsilon _{it}) \end{aligned}$$

• 2. If $r=\overline{r}_{1}$, $\log y_{it}=\log 25+\varepsilon _{it}$. Then:

  $$\begin{aligned} p(\log y_{it}\le \overline{r}_{1} \mid \varepsilon _{it})=\Phi \left[ \frac{\overline{r}_{1}-\varepsilon _{it}-\log \nu +\log (c+p_{1})-\mu _{\beta }}{\sigma _{\beta }}\right] +\pi _{2}\equiv F_{2}(\log y_{it} \mid \varepsilon _{it}) \end{aligned}$$

• 3. If $\overline{r}_{1}<r<\overline{r}_{2}$, $\log y_{it}=\log \nu -\log (c+p_{2})+\log \Big (e^{\beta _{i}}-25p_{1}+25p_{2}\Big )+\varepsilon _{it}$. Then:

  $$\begin{aligned}&p(\log y_{it}\le r \mid \varepsilon _{it})=\Phi \left[ \frac{\log \left( \frac{R(c+p_{2})}{ \nu e^{\varepsilon _{it}}}+25p_{1}-25p_{2}\right) -\mu _{\beta }}{\sigma _{\beta }}\right] -\Phi \left[ \frac{\log \left( \frac{\overline{R}_{1} (c+p_{2})}{\nu e^{\varepsilon _{it}}}+25p_{1}-25p_{2}\right) -\mu _{\beta }}{ \sigma _{\beta }}\right] \\&\quad +F_{2}(\log y_{it}\mid \varepsilon _{it}) \equiv F_{3}(\log y_{it}/\varepsilon _{it}) \end{aligned}$$

• 4. If $r=\overline{r}_{2}$, $\log y_{it}=\log 45+\varepsilon _{it}$. Then:

  $$\begin{aligned}&p(\log y_{it}\le \overline{r}_{2} \mid \varepsilon _{it})=\pi _{4}+\Phi \left[ \frac{\log \left( \frac{\overline{R}_{2}(c+p_{2})}{\nu e^{\varepsilon _{it}}} +25p_{1}-25p_{2}\right) -\mu _{\beta }}{\sigma _{\beta }}\right] \\&\quad -\Phi \left[ \frac{\log \left( \frac{\overline{R}_{1}(c+p_{2})}{\nu e^{\varepsilon _{it}}} +25p_{1}-25p_{2}\right) -\mu _{\beta }}{\sigma _{\beta }}\right] +F_{2}(\log y_{it}\mid \varepsilon _{it})\equiv F_{4}(\log y_{it} \mid \varepsilon _{it}) \end{aligned}$$

• 5. If $\overline{r}_{2}<r<\overline{r}_{3}$, $\log y_{it}=\log \nu -\log (c+p_{3})+\log \Big (e^{\beta _{i}}-25p_{1}-20p_{2}+45p_{3}\Big )+\varepsilon _{it}$. Then:

  $$\begin{aligned}&p(\log y_{it}\le r \mid \varepsilon _{it})=F_{4}(\log y_{it} \mid \varepsilon _{it})+\Phi \left[ \frac{\log \left( \frac{R(c+p_{3})}{ \nu e^{\varepsilon _{it}}}+25p_{1}+20p_{2}-45p_{3}\right) -\mu _{\beta }}{ \sigma _{\beta }}\right] \\&\quad -\Phi \left[ \frac{\log \left( \frac{ \overline{R}_{2}(c+p_{3})}{\nu e^{\varepsilon _{it}}} +25p_{1}+20p_{2}-45p_{3}\right) -\mu _{\beta }}{\sigma _{\beta }}\right] \equiv F_{5}(\log y_{it} \mid \varepsilon _{it}) \end{aligned}$$

• 6. If $r=\overline{r}_{3}$, $\log y_{it}=\log 90+\varepsilon _{it}$. Then:

  $$\begin{aligned}&p(\log y_{it}\le \overline{r}_{3} \mid \varepsilon _{it})=\pi _{6}+\Phi \left[ \frac{\log \left( \frac{\overline{R}_{3}(c+p_{3})}{\nu e^{\varepsilon _{it}}} +25p_{1}+20p_{2}-45p_{3}\right) -\mu _{\beta }}{\sigma _{\beta }}\right] \\&\quad -\Phi \left[ \frac{\log \left( \frac{\overline{R}_{2}(c+p_{3})}{\nu e^{\varepsilon _{it}}}+25p_{1}+20p_{2}-45p_{3}\right) -\mu _{\beta }}{\sigma _{\beta }}\right] +F_{4}(\log y_{it} \mid \varepsilon _{it}) \equiv F_{6}(\log y_{it}\mid \varepsilon _{it}) \end{aligned}$$

• 7. If $r>\overline{r}_{3}$, $\log y_{it}=\log \nu -\log (c+p_{4})+\log (e^{\beta _{i}}-25p_{1}-20p_{2}-45p_{3}+90p_{4})+\varepsilon _{it}$. Then:

  $$\begin{aligned}&p(\log y_{it}\le r\mid \varepsilon _{it})=\Phi \left[ \frac{\log \left( \frac{R(c+p_{4})}{ \nu e^{\varepsilon _{it}}}+25p_{1}+20p_{2}+45p_{3}-90p_{4}\right) -\mu _{\beta }}{\sigma _{\beta }}\right] \\&\quad -\Phi \left[ \frac{\log \left( \frac{\overline{R}_{3}(c+p_{4})}{\nu e^{\varepsilon _{it}}} +25p_{1}+20p_{2}+45p_{3}-90p_{4}\right) -\mu _{\beta }}{\sigma _{\beta }} \right] +F_{6}(\log y_{it}\mid \varepsilon _{it}) \equiv F_{7}(\log y_{it}\mid \varepsilon _{it}) \end{aligned}$$

Then, $p(\log y_{it}\le r)=\frac{1}{M} {\sum \limits _{j=1}^{M}} p(\log y_{it}\le r \mid \varepsilon _{jt})$. Conditional probabilities are calculated by simulating $M=100,000$ draws from the normal distribution $N(0,\widehat{\sigma }_{\varepsilon }^{2})$, where $\widehat{\sigma }_{\varepsilon }^{2}$ is the estimate from Table 8. Using $F(\log y_{it}\mid \varepsilon _{jt})$, I calculate $p(\log y_{it}\le r \mid \varepsilon _{jt})$ for each value of $\varepsilon _{jt}$. Thus, the probabilities of being at each block are:

• $p(block$ $1)=p(\log y_{it}\le \log 25)$

• $p(block$ $2)=p(\log 25<\log y_{it}\le \log 45)=p(\log y_{it}\le \log 45)-p(\log y_{it}<\log 25)$

• $p(block$ $3)=p(\log 45<\log y_{it}\le \log 90)=p(\log y_{it}\le \log 90)-p(\log y_{it}<\log 45)$

• $p(block$ $4)=p(\log y_{it}>\log 90)=1-p(block 1)-p(block 2)-p(block 3)$

Final probabilities are obtained after repeating this process 100 times.