I have searched high and low for examples like this online but I cannot find it anywhere. This question is pulled from a PhD course in econometrics.

Given $(y^*_1,y^*_2)$~ bivariate normal with the following parameters:

$\mu_1=x_1'\beta, \mu_2=x_2'\beta, \sigma^2_1=1, \sigma^2_2=2$ and covariance $\sigma_{12}$, consider a Tobit model with observations on $y_2, x_1, x_2$ in the data.

The censoring rule is:

$y_2=y_2^*$ if $x_1'\beta+u_1>0$ and $x_2'\beta+u_2>0$ $y_2=0$ otherwise

What is the likelihood function?

I have found all sorts of results for a univariate case or a MV case with different truncation points, but I can't seem to map those answers into what I need.


Secondly, the model:

$y^*_{1i}=x'_{1i}\beta_1+u_{1i}$ $y^*_{2i}=x'_{2i}\beta_2+u_{2i}$

But now all we know is that the error terms are kid with density functions $f_1, f_2$ and

$y_{1i}=y^*_{1i}*\mathbf{1}_{y^*_{2i}>0}$ $y_{2i}=\mathbf{1}_{y^*_{2i}>0}$

In the first model, the following are observed in the data:

$y_{1i}, y^*_{2i}, x_{1i}, x_{2i}$

In the second model, these are observed:

$y_{1i}, y_{2i}, x_{1i}, x_{2i}$

I am beyond stuck here. Any help on any of these three iterations of the Tobit specification would be greatly appreciated.


1 Answer 1


Let $d_i=I\{y_{i1}>0\;and\;y_{i2}>0\}$ be equal to 1 if the statement inside the brackets is true and 0 otherwise. Then I believe the likelihood would be $$ \mathscr{L} = \prod_{i=1}^n \boldsymbol{\phi}_{\Sigma}\bigg( \begin{matrix} y_{i1}-x'_{i1}\beta\\ y_{i2}-x'_{i2}\beta \end{matrix}\; \bigg)^{d_i}\bigg[\Phi_{\Sigma}\bigg( \begin{matrix} -x'_{i1}\beta\\ -x'_{i2}\beta \end{matrix}\; \bigg)\frac{\phi_{\sigma_1}(y_{i1}-x'_{i1}\beta)}{\Phi{\sigma_1}(-x'_{i1}\beta)}\bigg]^{(1-d_i)} $$

Where $\Sigma$ is the entire covariance matrix, $\boldsymbol{\phi}_{\Sigma}$ the centered bi-variate normal pdf of covariance $\Sigma$ ,$\Phi_{\sigma_2}$ the centered normal cdf of variance $\sigma_2^2$, and $\phi_{\sigma_1}$ the centered normal pdf of variance $\sigma_1^2$. We can also write it as $$ \mathscr{L} = \prod_{i=1}^n \boldsymbol{\phi}_{\Sigma}\bigg( \begin{matrix} y_{i1}-x'_{i1}\beta\\ y_{i2}-x'_{i2}\beta \end{matrix}\; \bigg)^{d_i}\bigg[\Phi_{\Sigma}\bigg( \begin{matrix} -x'_{i1}\beta\\ -x'_{i2}\beta \end{matrix}\; \bigg)\frac{1}{\sigma_1}\frac{\phi_{1}(\frac{y_{i1}-x'_{i1}\beta}{\sigma_1})}{\Phi_{1}(\frac{-x'_{i1}\beta}{\sigma_1})}\bigg]^{(1-d_i)} $$ The term $\Phi_{\Sigma}\bigg( \begin{matrix} -x'_{i1}\beta\\ -x'_{i2}\beta \end{matrix}\; \bigg)$ is the joint probability that both $y_{i1}$ and $y_{i2}$ are less than 0, which is the mass point creating bias in MV least squared regression. Note that if the covariance $\sigma_{12}=0$,
$$\Phi_{\Sigma}\bigg( \begin{matrix} -x'_{i1}\beta\\ -x'_{i2}\beta \end{matrix}\; \bigg)=\Phi{\sigma_1}(-x'_{i1}\beta)\Phi{\sigma_2}(-x'_{i2}\beta)$$ and $$\boldsymbol{\phi}_{\Sigma}\bigg( \begin{matrix} y_{i1}-x'_{i1}\beta\\ y_{i2}-x'_{i2}\beta \end{matrix}\; \bigg)=\phi_{\sigma_1}(y_{i1}-x'_{i1}\beta)\phi_{\sigma_2}(y_{i2}-x'_{i2}\beta)$$ Unfortunately I do not have a source for you. Let me know if there are any problems or if further explanation is needed. Note that if the question goes beyond just how to write the likelihood to actually maximizing it via computational methods, things get much more complicated.

I am not so sure about the last two cases. I am assuming you mean that the error terms in the latent model are iid $f_1,f_2$. First I restate $y_{1i}$ and $y_{2i}$ as piece-wise functions of $y^*_{1i}$ and $y^*_{2i}$.

$$ y_{1i} = \left\{ \begin{array}{lr} y^*_{1i} & if\;\; y^*_{2i}>0\\ 0 & if\;\; y^*_{2i} \leq 0 \end{array} \right.\;\;\;\;\;\;\;\; y_{2i} = \left\{ \begin{array}{lr} 1 & if\;\; y^*_{2i}>0\\ 0 & if\;\; y^*_{2i} \leq 0 \end{array} \right. $$

(or at least that's my interpretation of your notation). Then I create some indicator function. $d_i=I\{y_{2i}>0\}$.

  1. We observe $(y_{1},y^*_{2})$:

In this case notice that when $d_i=1$, the conditional probability of observing any pair $(y_{1i},y^*_{2i})$ is simply the conditional joint pdf, which I will call $f_{1,2}$, as a function of the $(y_{1i},y^*_{2i})$,$(x_{1i},x_{2i})$ and $(\beta_{1},\beta_{2})$.

When $d_i=0$ the joint probability would be $$ Pr(Y_{1i}=0\,|\,Y^*_{2i}\leq0)\frac{f_2(Y^*_{2i}-x'_{2i}\beta_2)}{Pr(Y^*_{2i}\leq 0)}=Pr(Y^*_{2i}\leq 0)\frac{f_2(y^*_{2i}-x'_{2i}\beta_2)}{Pr(Y^*_{2i}\leq 0)}=f_2(y^*_{2i}-x'_{2i}\beta_2) $$

The term $\frac{f_2(y^*_{2i}-x'_{2i}\beta_2)}{Pr(y^*_{2i}\leq 0)}$ is the left-of-zero truncated distribution of $u_{2i}$. So we write the likelihood: $$ \mathscr{L} = \prod_{i=1}^n f_{1,2}\bigg( \begin{matrix} y_{i1}-x'_{i1}\beta\\ y_{i2}-x'_{i2}\beta \end{matrix}\; \bigg)^{d_i}f_2(y^*_{2i}-x'_{2i}\beta_2)^{(1-d_i)} $$

  1. We observe $(y_{1},y_{2})$:

So when $d_i=1$ the joint probability is $$ Pr(Y^*_{2i}>0\cap Y_{1i}=y_{1i} )=Pr(Y^*_{2i}>0\,|\, Y_{1i}=y_{1i})Pr(Y_{1i}=y_{1i})= $$ $$ Pr(U_{2i}>-x'_{2i}\beta_2\,|\, Y_{1i}=y_{1i})Pr(Y_{1i}=y_{1i})= $$ $$ \bigg(\int_{-x'_{2i}\beta_2}^{\eta}\frac{f_{12}(y_{1i}-x'_{1i},t)}{f_1(y_{1i}-x'_{1i}\beta_1)}dt\bigg)f_1(y_{1i}-x'_{1i}\beta_1)=\int_{-x'_{2i}\beta_2}^{\eta}f_{12}(y_{1i}-x'_{1i}\beta_1,t)dt $$ for some upper bound $\eta$ which may or may not be $\infty$.

When $d_i=0$ the only information we have to construct a probability is $$ Pr(y_i \leq 0)=F_2(-x'_{2i}\beta_2) $$ So the Likelihood will be $$ \mathscr{L} = \prod_{i=1}^n \bigg[\int_{-x'_{2i}\beta_2}^{\eta}f_{12}\bigg(\begin{matrix} y_{i1}-x'_{i1}\beta\\ t \end{matrix}\; \bigg)dt\bigg] ^{d_i}F_2(-x'_{2i}\beta_2)^{(1-d_i)} $$

  • $\begingroup$ Thank you very much! I had some pieces that looked like this, but I just couldn't see how to fit them together. I do not need to do any computational maximization. It's all theory at this point. Can you comment on the second/third model, where the errors only have a generic distributions, f? $\endgroup$
    – Justin
    Feb 22, 2015 at 14:20
  • $\begingroup$ I made some edits. I tried to answer the last 2 model likelihood functions. I also edited my previous answer because it disregarded possible correlation between $y_1$ and $y_2$ when both where below zero. Hope that helps $\endgroup$ Feb 22, 2015 at 17:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.