# 1 Preliminaries: Data importation

library("readxl")
my_data <- read_excel("mydatax.xlsx")
m = my_dataMarkup hist(m, probability = TRUE, col = "lightgrey") mm = seq(from = min(m), to = max(m), length.out = 10^4) lines(mm, col = 2, lwd = 2) Minimum markup in the data $$m_{min}$$: (mmin = min(m)) ## [1] 1.001387 Maximum markup in the data $$m_{max}$$: (mn = max(m)) ## [1] 9.977231 Number of observations $$n$$: (nn = length(m)) ## [1] 2457 # 2 Markup estimation ## 2.1 Underlying productivity distribution: Pareto We assume that firm productivities follow a Pareto distribution $$\phi\sim \mathcal{P}(\underline{\phi}, k)$$ with CDF: $$$G_{P}(\phi) = 1 - \underline{\phi}^k \phi^{-k} \label{pareto}$$$ Parameters of Pareto distribution: lower bound $$\underline\phi>0$$ and shape $$k>0$$. ### 2.1.1 CREMR demand The CDF of the markup distribution implied by the assumptions of Pareto productivity and CREMR demand $$p(x)=\frac{\beta }{x}\left( x-\gamma \right)^{\frac{\sigma-1}{\sigma}}$$ is: \begin{align} B(m)={}&1 - \underline{\phi}^k \left(\frac{1}{\beta}\frac{\sigma}{\sigma-1}\left(\frac{\frac{\sigma-1}{\sigma}m}{1-\frac{\sigma-1}{\sigma}m}\gamma\right)^{\frac{1}{\sigma}}\right)^{-k}\;\;\; \text{with} \;\;\; m \in \left[\underline{m}, \frac{\sigma}{\sigma - 1}\right]\\ ={}&1-\omega^{k}\left(\frac{(\sigma-1)m}{m+\sigma-\sigma m}\right)^{-\frac{k}{\sigma}} \end{align} where $$\omega=\frac{\underline{\phi}\beta}{\gamma^{\frac{1}{\sigma}}}\frac{\sigma-1}{\sigma}=\left(\frac{(\sigma-1)\underline{m}}{\underline{m}+\sigma-\sigma \underline{m}}\right)^{\frac{1}{\sigma}}$$ Let $$b(m)$$ denote the the pdf of $$B(m)$$: \begin{align} b(m) &= \frac{k\omega^k}{m(m+\sigma-m\sigma)}\left(\frac{(\sigma-1)m}{m+\sigma-\sigma m}\right)^{-\frac{k}{\sigma}}\\ &=\left(\frac{\underline{m}}{\underline{m}+\sigma-\sigma \underline{m}}\right)^{\frac{k}{\sigma}}\frac{k}{m(m+\sigma-m\sigma)}\left(\frac{m}{m+\sigma-\sigma m}\right)^{-\frac{k}{\sigma}} \end{align} The log-likelihood function is: \begin{align} L(\theta) = \sum_{i = 1}^n \log(b(m_i))={}&\sum_{i = 1}^n\log\left(\left(\frac{\underline{m}}{\underline{m}+\sigma-\sigma \underline{m}}\right)^{\frac{k}{\sigma}}\frac{k}{m(m+\sigma-m\sigma)}\left(\frac{m}{m+\sigma-\sigma m}\right)^{-\frac{k}{\sigma}}\right)\\ ={}&n\left(\log k+\frac{k}{\sigma}\log\underline{m}-\frac{k}{\sigma}\log(\underline{m}+\sigma-\sigma\underline{m})\right)+\sum_{i = 1}^n\left(-\frac{\sigma+k}{\sigma}\log m_i+\frac{k-\sigma}{\sigma}\log(m_i+\sigma-m_i\sigma)\right) \end{align} The log-likelihood function is monotonically increasing in $$\underline{m}$$, hence $$\hat{\underline{m}}=m_{min}$$. Differentiating the log-likelihood with respect to $$k$$ yields: \begin{align} \frac{dL}{dk} &=n(\frac{1}{k}+\log\omega-\frac{1}{\sigma}\log(\sigma-1))+\sum_{i = 1}^n\left(-\frac{1}{\sigma}\log m_i+\frac{1}{\sigma}\log(m_i+\sigma-m_i\sigma)\right)\\ &\Rightarrow \hat k=\frac{\sigma}{\frac{1}{n}\sum_{i = 1}^n(\log m_i-\log(m_i+\sigma-m_i\sigma))+\log(\sigma-1)-\sigma\log\omega} \end{align} And differentiating the log-likelihood with respect to $$\sigma$$ yields: \begin{align} \frac{dL}{d\sigma} &=\frac{nk}{\sigma^2}\left(\log(\sigma-1)-\frac{\sigma}{\sigma-1}\right)+\sum_{i = 1}^n\left(\frac{k}{\sigma^2}\left(\log m_i+\log(m_i+\sigma-m_i\sigma)\right)-\frac{(k-\sigma)\sigma(m_i-1)}{m_i+\sigma-m_i\sigma}\right) \end{align} It is not obvious to get a closed-form solution for $$\hat{k}$$ and $$\hat{\sigma}$$, but we can optimize numerically. We define the objective function (i.e.Â minus log-likelihood) for $$b(m)$$ (to be minimized): log_likelihood_b_CREMR = function(theta, m){ k = theta[1] sigma = theta[2] -sum(log((k*(mmin/(mmin+sigma-sigma*mmin))^(k/sigma)/(m*(m+sigma-m*sigma)))*(m/(m+sigma-m*sigma))^(-k/sigma))) } For the first optimization, we use the following starting values. Let $$m_{max}$$ denote the largest markup value (in the data), we have that $m_{max} < \frac{\sigma}{\sigma - 1},$ and $\sigma < \frac{m_{max}}{m_{max} - 1}.$ Therefore, we propose using $k_{\text{start}} = 2 \;\;\; \text{and} \;\;\; \sigma_{\text{start}} = \frac{1}{2} \left( 1 + \frac{m_{max}}{m_{max}-1} \right).$ The estimation of $$\sigma$$, $$k$$ is simply: # Estimation (CREMR+Pareto) (theta_startCREMR = c(2, 0.5*(1 + mn/(mn-1)))) ## [1] 2.000000 1.055696 (estimCREMR = optim(par = theta_startCREMR, log_likelihood_b_CREMR, m = m)) ##par
## [1] 1.232891 1.111174
##
## $value ## [1] 3021.716 ## ##$counts
## function gradient
##       87       NA
##
## $convergence ## [1] 0 ## ##$message
## NULL

The estimated parameters are therefore:

(kCREMR=estimCREMR$par[1]) ## [1] 1.232891 (sigmaCREMR=estimCREMR$par[2])
## [1] 1.111174
(omegaCREMR=((sigmaCREMR-1)*mmin/(mmin+sigmaCREMR-sigmaCREMR*mmin))^(1/sigmaCREMR))
## [1] 0.1386922

To illustrate this estimation, we plot the empirical distribution (using a histogram) and the estimated pdf:

hist(m, probability = TRUE, col = "lightgrey")
mm = seq(from = min(m), to = max(m), length.out = 10^4)
yy = (kCREMR*((sigmaCREMR-1)*mmin/(mmin+sigmaCREMR-sigmaCREMR*mmin))^(kCREMR/sigmaCREMR)/(mm*(mm+sigmaCREMR-mm*sigmaCREMR)))*((sigmaCREMR-1)*mm/(mm+sigmaCREMR-mm*sigmaCREMR))^(-kCREMR/sigmaCREMR)
lines(mm, yy, col = 2, lwd = 2)

We calculate the Akaike Information Criterion (AIC): $$AIC=2p-2\log(\hat L)$$ where $$p$$ is the number of estimated parameters and $$\hat L$$ is the maximum value of the likelihood function.

#Three estimated parameters
pCREMR = 3
(AIC_CREMR = 2*pCREMR - 2*(-estimCREMR\$value))
## [1] 6049.433

### 2.1.2 Linear demand

Assume the demand is linear $$p(x)=\alpha-\beta x$$. Maximum output is $$\bar x=\frac{\alpha}{2\beta}$$. The markup is $$m(x)=\frac{\alpha-\beta x}{\alpha-2\beta x}$$. Maximum markup is infinite: $$m(x)\to\infty$$ as $$x\to\bar x$$. Minimum markup $$\underline{m}$$.

The CDF of the markup distribution implied by the assumptions of Pareto productivity and linear demand is:

\begin{align} B(m) &=1-\left(\frac{2m-1}{\alpha\underline{\phi}}\right)^{-k} \;\;\; \text{with} \;\;\; m \in \left(\underline{m}, \infty\right), \end{align}

where $$k > 0$$, $$\underline{\phi}>0$$, $$\alpha > 0$$, and $$\underline{\phi}=\phi(\underline{m})=\frac{2\underline{m}-1}{\alpha}$$.

Let $$b(m)$$ denote the the pdf of $$B(m)$$:

\begin{align} b(m) &= 2k(2\underline{m}-1)^k\left(2m-1\right)^{-k-1} \end{align}

The log-likelihood function is:

\begin{align} L(\theta) = \sum_{i = 1}^n \log(b(m_i))=\sum_{i = 1}^n\log\left(2k(2\underline{m}-1)^k\left(2m_i-1\right)^{-k-1}\right)&=\sum_{i = 1}^n\left(\log(2k)+k\log(2\underline{m}-1)-(k+1)\log(2m_i-1)\right)\\ &=n\left(\log(2k)+k\log(2\underline{m}-1)\right)-(k+1)\sum_{i = 1}^n\log(2m_i-1) \end{align}

The log-likelihood function is monotonically increasing in $$\underline{m}$$, hence $$\hat{\underline{m}}=m_{min}$$.

Differentiating $$L$$ with respect to $$k$$, we obtain a closed-form expression for the MLE estimator of $$k$$:

\begin{align} \frac{dL}{dk} &=\frac{n}{k}+n\log(2\underline{m}-1)-\sum_{i = 1}^n\log(2m_i-1)=0\Rightarrow \hat k=\frac{1}{\frac{1}{n}\sum_{i = 1}^n\log(2m_i-1)-\log(2\hat{\underline{m}}-1)} \end{align}

MLE estimation:

(omegaLIN=2*mmin-1)
## [1] 1.002774
(kkLIN = nn/(sum(log(2*m-1))-nn*log(omegaLIN)))
## [1] 1.001127
hist(m, probability = TRUE, col = "lightgrey")
mm = seq(from = min(m), to = max(m), length.out = 10^4)
xx = 2*kkLIN*(omegaLIN^kkLIN)*(2*mm-1)^(-kkLIN-1)
lines(mm, xx, col = 3, lwd = 2)

We calculate the Akaike Information Criterion (AIC):

#Two estimated parameters
pLIN = 2
(AIC_LIN = 2*pLIN - 2*(sum(log(2*kkLIN*(omegaLIN^kkLIN)*(2*m-1)^(-kkLIN-1)))))
## [1] 6428.425

### 2.1.3 LES demand

Assume the demand is LES $$p(x)=\frac{\delta}{x+\gamma}$$. The markup is $$m(x)=\frac{x+\gamma}{\gamma}$$. Maximum markup is again infinite: $$m(x)\to\infty$$ as $$x\to\infty$$.

The CDF of the markup distribution implied by the assumptions of Pareto productivity and LES demand is:

\begin{align} B(m) &=1-\left(\frac{\gamma}{\delta\underline{\phi}}m^2\right)^{-k} \;\;\; \text{with} \;\;\; m \in \left(\underline{m}, \infty\right), \end{align}

where $$k > 0$$, $$\underline{\phi}>0$$, $$\gamma>0$$, $$\delta > 0$$, and $$\underline{\phi}=\frac{\gamma}{\delta}\underline{m}^2$$.

Let $$b(m)$$ denote the the pdf of $$B(m)$$:

\begin{align} b(m)&=2k\left(\underline{m}\right)^{2k}m^{-2k-1} \end{align}

The log-likelihood function is:

\begin{align} L(\theta) = \sum_{i = 1}^n \log(b(m_i))=\sum_{i = 1}^n\log\left(2k\left(\underline{m}\right)^{2k}m^{-2k-1}\right)&=\sum_{i = 1}^n\left(\log(2k)+2k\log\underline{m}-(2k+1)\log(m_i)\right)\\ &=n(\log(2k)+2k\log\underline{m})-(2k+1)\sum_{i = 1}^n\log(m_i) \end{align}

The estimating pocedure is the same as for linear above. The log-likelihood function is monotonically increasing in $$\underline{m}$$, hence $$\hat{\underline{m}}=m_{min}$$.

Furthermore, we can get a closed-form solution for the MLE estimator of $$k$$. Differentiate $$L$$ with respect to $$k$$:

\begin{align} \frac{dL}{dk} &=\frac{n}{k}+2n\log\underline{m}-2\sum_{i = 1}^n\log(m_i)=0\Rightarrow \hat k=\frac{1}{\frac{2}{n}\sum_{i = 1}^n\log(m_i)-2\log\hat{\underline{m}}} \end{align}

MLE estimation:

(omegaLES=mmin^2)
## [1] 1.002776
(kkLES = 1/((2/nn)*sum(log(m))-log(omegaLES)))
## [1] 0.7466883
hist(m, probability = TRUE, col = "lightgrey")
mm = seq(from = min(m), to = max(m), length.out = 10^4)
yy = 2*kkLES*omegaLES^(kkLES)*(mm)^(-2*kkLES-1)
lines(mm, yy, col = 4, lwd = 2)

We calculate the Akaike Information Criterion (AIC):

#Two estimated parameters
pLES = 2
(AIC_LES = 2*pLES - 2*(sum(log((2*kkLES*(omegaLES^(kkLES))*(m)^(-2*kkLES-1))))))
## [1] 6244.631

### 2.1.4 Translog demand

Assume the demand is translog $$x(p)=\frac{1}{p}(\gamma-\eta\log p)$$. The markup is $$m(x)=1+W(\frac{e^{\frac{\gamma}{\eta}}}{\eta}x)$$. Maximum markup is again infinite: $$m(x)\to\infty$$ as $$x\to\infty$$.

The CDF of the markup distribution implied by the assumptions of Pareto productivity and translog demand is:

\begin{align} B(m) &=1-\left(\frac{e^{-1-\frac{\gamma}{\eta}}}{\underline{\phi}}me^{m}\right)^{-k} \;\;\; \text{with} \;\;\; m \in \left(?, \infty\right), \end{align}

where $$k > 0$$, $$\underline{\phi}>0$$, $$\gamma>0$$, $$\eta > 0$$, and $$\underline{\phi}=\phi(\underline{m})=\underline{m}e^{\underline{m}}e^{-1-\frac{\gamma}{\eta}}$$

Let $$b(m)$$ denote the the pdf of $$B(m)$$:

\begin{align} b(m) &= k\left(\underline{m}e^{\underline{m}}\right)^{k}(m+1)m^{-k-1}e^{-km} \end{align}

The log-likelihood function is:

\begin{align} L(\theta) = \sum_{i = 1}^n \log(b(m_i))=\sum_{i = 1}^n\log\left(k\left(\underline{m}e^{\underline{m}}\right)^{k}(m_{i}+1)m_{i}^{-k-1}e^{-km_{i}}\right)&=\sum_{i = 1}^n\left(\log k+k\log\left(\underline{m}e^{\underline{m}}\right)+\log(m_{i}+1)-(k+1)\log m_{i}-km_{i}\right)\\ &=n\log k +nk\log\left(\underline{m}e^{\underline{m}}\right) +\sum_{i = 1}^n\log(m_{i}+1)-(k+1)\sum_{i = 1}^n\log m_{i}-k\sum_{i = 1}^n m_{i} \end{align}

The estimating pocedure is the same as for linear and LES above. The log-likelihood function is monotonically increasing in $$\underline{m}$$, hence $$\hat{\underline{m}}=m_{min}$$.

Then we can again solve for the MLE estimator of $$k$$ in closed form. Differentiate $$L$$ with respect to $$k$$:

\begin{align} \frac{dL}{dk} &=\frac{n}{k}+n\log\left(\underline{m}e^{\underline{m}}\right)-\sum_{i = 1}^n\log m_{i}-\sum_{i = 1}^n m_{i}=0\Rightarrow \hat k=\frac{1}{\frac{1}{n}\sum_{i = 1}^n(m_{i}+\log m_{i})-\log\left(\hat{\underline{m}}e^{\hat{\underline{m}}}\right)} \end{align}

MLE estimation:

(omegaTLOG=mmin*exp(mmin))
## [1] 2.72583
(kkTLOG = 1/((1/nn)*sum(log(m)+m)-log(omegaTLOG)))
## [1] 0.4868533
hist(m, probability = TRUE, col = "lightgrey")
mm = seq(from = min(m), to = max(m), length.out = 10^4)
yy = kkTLOG*(omegaTLOG^(kkTLOG))*(1+mm)*mm^(-kkTLOG-1)*exp(-kkTLOG*mm)
lines(mm, yy, col = 5, lwd = 2)