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I am reading Comparing Open and Sealed Bid Auctions: Evidence From Timber Auctions (working paper version) and struggling with getting a log likelihood function about a Gamma-Weibull distribution. In particular, the simple version of the distribution used in paper (p.20 Eq (7)) is:

$$G(b|X,u,N,n)=1-\exp\left(-u\left(\frac{b}{\lambda(X,N,n)}\right)^{\rho(n)}\right)$$

where $\lambda(X,N,n) = \exp(X\beta)$, $\rho(n)=\exp(\gamma n)$ and $u$ has a Gamma distribution with unit mean and variance $\theta$. While I can understand the log-likelihood without $u$, the log-likelihood for auction $t$ would be the following with $u$ (simplified version of p.32):

$$\ln L_t = n\ln\theta + \ln\Gamma\left(\frac{1}{\theta}+n\right)-\ln\Gamma\left(\frac{1}{\theta}\right) + \sum_{i=1}^n\ln\left(\rho_{it}\lambda_{it}\left(\frac{b_{it}}{\lambda_{it}}\right)^{\rho_{it}-1}\right) + \left(\frac{1}{\theta} + n\right)\ln\left(1+\theta\sum^n_{i=1}\left(\frac{b_{it}}{\lambda_{it}}\right)^{\rho_{it}}\right)$$

How do I derive this equation?

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1 Answer 1

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I was not able to find the result. Here is my attempt.

The random term $u$ is not observed, and so it is not possible to rely on the conditional density $$g(b|X,u,N,n)$$ to derive the likelihood function whose parameters have to be estimated using observed explanatory variables only. The following unconditional likelihood function is of interest: $$L(b|X,N,n)= \int_0^\infty g(b|X,u,N,n)h(u|X,N,n) du,$$ with $h$ being the density of $u$. We obtain the Weibull density for $b$ from the cdf by $g=G'$ which yields $$ g(b|X,u,N,n)= \frac{u\rho(n)}{\lambda(X,N,n)}\left(\frac{b}{\lambda(X,N,n)}\right)^{\rho(n)-1} \exp\left(-u\left(\frac{b}{\lambda(X,N,n)}\right)^{\rho(n)}\right). $$ For the random term $u$, the authors consider a "Gamma distribution with unit mean and variance $\theta$, and which is independent of $X$, $N$, and $n$", that is $$ h(u|X,N,n)=h(u)= \left(\frac{1}{\theta}\right)^{1/\theta} u^{1/\theta-1} \frac{\exp(- u/\theta)}{\Gamma(1/\theta)}. $$ Once inserted into the likelihood function we have (omitting the conditioning variables): \begin{align*} L(b|X,N,n) = & \frac{\rho}{\lambda}\left(\frac{b}{\lambda}\right)^{\rho-1} \left(\frac{1}{\theta}\right)^{1/\theta} \frac{1}{\Gamma(1/\theta)} \cdot \int_0^\infty \exp\left[-u\left(\left(\frac{b}{\lambda}\right)^{\rho}+\frac{1}{\theta}\right)\right] u^{1/\theta-1} du \\ = & \frac{\rho}{\lambda}\left(\frac{b}{\lambda}\right)^{\rho-1} \left(\frac{1}{\theta}\right)^{1/\theta} \frac{1}{\Gamma(1/\theta)} \cdot \frac{\Gamma(1/\theta)}{ \left(\left(\frac{b}{\lambda}\right)^{\rho}+\frac{1}{\theta}\right)^{1/\theta}} \\ = & \frac{\rho}{\lambda}\left(\frac{b}{\lambda}\right)^{\rho-1} \left(\frac{1}{\theta}\right)^{1/\theta} \cdot \left(\left(\frac{b}{\lambda}\right)^{\rho}+\frac{1}{\theta}\right)^{-1/\theta}. \end{align*} The second equality is achieved after use of the following result: \begin{align*} & \frac{\Gamma(\beta+1)}{\alpha^{\beta+1}} = \int_0^\infty \exp(-\alpha u)u^\beta du. \\ \end{align*} For one observation, the log-likelihood is: \begin{align*} \ln L(b|X,N,n) &= \ln \left( \rho/\lambda (b/ \lambda)^{\rho-1} \right) - 1/\theta\ln \theta - 1/\theta\ln( (b/\lambda)^{\rho} + 1/\theta) \\ &= \ln \left( \rho/\lambda (b/ \lambda)^{\rho-1} \right) - 1/\theta\ln \theta - 1/\theta \left( \ln( 1+ \theta(b/\lambda)^{\rho} ) - \ln\theta \right) \\ &= \ln \left( \rho/\lambda (b/ \lambda)^{\rho-1} \right) - 1/\theta \ln( 1+ \theta(b/\lambda)^{\rho} ). \end{align*}

Summation over $n$ yields the sample log-likelihood. There are obvious differences between this expression and the one expected to be found... Where is the error ?

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