<p>I was not able to find the result. Here is my attempt.</p>
<p>The random term <span class="math-container">$u$</span> is not observed, and so it is not possible to rely on the conditional density <span class="math-container">$$g(b|X,u,N,n)$$</span> to derive the likelihood function whose parameters have to be estimated using observed explanatory variables only. The following unconditional likelihood function is of interest:
<span class="math-container">$$L(b|X,N,n)= \int_0^\infty g(b|X,u,N,n)h(u|X,N,n) du,$$</span>
with <span class="math-container">$h$</span> being the density of <span class="math-container">$u$</span>.
We obtain the Weibull density for <span class="math-container">$b$</span> from the cdf by <span class="math-container">$g=G'$</span> which yields
<span class="math-container">$$ g(b|X,u,N,n)= \frac{u\rho(n)}{\lambda(X,N,n)}\left(\frac{b}{\lambda(X,N,n)}\right)^{\rho(n)-1} \exp\left(-u\left(\frac{b}{\lambda(X,N,n)}\right)^{\rho(n)}\right). $$</span>
For the random term <span class="math-container">$u$</span>, the authors consider a &quot;Gamma distribution with unit mean and variance <span class="math-container">$\theta$</span>, and which is independent of <span class="math-container">$X$</span>, <span class="math-container">$N$</span>, and <span class="math-container">$n$</span>&quot;, that is
<span class="math-container">$$ h(u|X,N,n)=h(u)= \left(\frac{1}{\theta}\right)^{1/\theta} u^{1/\theta-1} \frac{\exp(- u/\theta)}{\Gamma(1/\theta)}. $$</span>
Once inserted into the likelihood function we have (omitting the conditioning variables):
<span class="math-container">\begin{align*}
     L(b|X,N,n) =  &amp; \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 \\
     =  &amp;  \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}} \\
     =  &amp;  \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*}</span>
The second equality is achieved after use of the following result:
<span class="math-container">\begin{align*}
&amp; \frac{\Gamma(\beta+1)}{\alpha^{\beta+1}} = \int_0^\infty \exp(-\alpha u)u^\beta du. \\ 
\end{align*}</span>
For one observation, the log-likelihood is:
<span class="math-container">\begin{align*}
\ln L(b|X,N,n) &amp;= \ln \left( \rho/\lambda (b/ \lambda)^{\rho-1} \right)
- 1/\theta\ln \theta - 1/\theta\ln( (b/\lambda)^{\rho} + 1/\theta) \\
&amp;= \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) \\
&amp;= \ln \left( \rho/\lambda (b/ \lambda)^{\rho-1} \right) - 1/\theta \ln( 1+ \theta(b/\lambda)^{\rho} ).
\end{align*}</span></p>
<p>Summation over <span class="math-container">$n$</span> yields the sample log-likelihood. There are obvious differences between this expression and the one expected to be found... Where is the error ?</p>

CC BY-SA 4.0
2022-02-14T07:46:35.333Z
2022-02-14T07:46:35.333Z
Bertrand
<p>I am reading <a href="https://www.nber.org/system/files/working_papers/w14590/w14590.pdf" rel="nofollow noreferrer">Comparing Open and Sealed Bid Auctions: Evidence From Timber Auctions</a> (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:</p>
<p><span class="math-container">$$G(b|X,u,N,n)=1-\exp\left(-u\left(\frac{b}{\lambda(X,N,n)}\right)^{\rho(n)}\right)$$</span></p>
<p>where <span class="math-container">$\lambda(X,N,n) = \exp(X\beta)$</span>, <span class="math-container">$\rho(n)=\exp(\gamma n)$</span> and <span class="math-container">$u$</span> has a Gamma distribution with unit mean and variance <span class="math-container">$\theta$</span>.
While I can understand the log-likelihood without <span class="math-container">$u$</span>, the log-likelihood for auction <span class="math-container">$t$</span> would be the following with <span class="math-container">$u$</span> (simplified version of p.32):</p>
<p><span class="math-container">$$\ln L_t = n\ln\theta + \ln\Gamma\left(\frac{1}{\theta}+n\right)-\ln\Gamma\left(\frac{1}{\theta}\right) + \sum_{i=1}^n\ln\left(\rho_{it}\lambda_{it}\left(\frac{b_{it}}{\lambda_{it}}\right)^{\rho_{it}-1}\right) + \left(\frac{1}{\theta} + n\right)\ln\left(1+\theta\sum^n_{i=1}\left(\frac{b_{it}}{\lambda_{it}}\right)^{\rho_{it}}\right)$$</span></p>
<p>How do I derive this equation?</p>

<statistics><auctions>
A Gamma-Weibull distribution for empirical auction model
hrkshr
# A Gamma-Weibull distribution for empirical auction model

(Asked by: username_0 on 2022-02-11)

I am reading [Comparing Open and Sealed Bid Auctions: Evidence From Timber Auctions](https://www.nber.org/system/files/working_papers/w14590/w14590.pdf) (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?

**Tags:** statistics, auctions


---

## 1 answers

### Answer by username_1 on 2022-02-14. Score: 2

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 ?


---

