I am trying to derive the concentrated log-likelihood within a limited information maximum likelihood context. The linear model is a compacted instrumental variable regression model and I am researching what heteroskedasticity in the errors does to hypothesis testing problems. I am stuck with concentrating out the first stage regression parameter $\Pi$ however.

I have the following log-likelihood (up to a constant)

\begin{eqnarray} tr(\Omega_1^{-1}A\Pi'Z_1'Y_1)+ tr(\Omega^{-1}_2A\Pi'Z_2'Y_2)-\frac{1}{2} tr(\Omega_1^{-1}A\Pi'Z_1'Z_1\Pi A') -\frac{1}{2} tr(\Omega_2^{-1}A\Pi'Z_2'Z_2\Pi A'). \end{eqnarray}

Now taking the derivative with respect to $\Pi$ and setting that first-order condition to zero results in the following equation, which is where I am stuck: \begin{eqnarray} Z_1'Y_1\Omega^{-1}_1A + Z_2'Y_2 \Omega^{-1}_2A = Z_1'Z_1 \Pi A' \Omega^{-1}_1A + Z_2'Z_2 \Pi A' \Omega^{-1}_2 A \end{eqnarray}

I am confident that there are no algebraic mistakes but what I would like to do next is to re-substitute into the log-likelihood the first-order to condition to continue working with the likelihood expression but I am unsure how to proceed. I have tried finding a general solution to the first-order condition but this seems elusive.

The matrices $A' \Omega^{-1}_i A$ as well as $Z_i'Z_i$ are invertible. Further, $A$ is $1+p \times p$. $Z_i$ is $n_i \times k$ and $\Pi$ is $k \times p$.

Does anybody know how to derive the concentrated log-likelihood with this first-order condition?

Thank you so much!


The equations come from this fairly standard model:

\begin{eqnarray} y_1 & = & y_2\beta+\epsilon \\ y_2 & = & Z\Pi+V, \end{eqnarray} where we have absorbed all dependence on regressors in $W$ into $y_{2}$. For convenience, and without loss of generality, we also treat all regressors in $y_{2}$ as endogenous and hence in need of instrumentation. To derive the likelihood, we compact the model as: \begin{eqnarray} Y = Z \Pi A' + V \end{eqnarray} The individual components are defined as: $Y = (y_{1}~\vdots~y_{2})$ with observations stacked vertically. The matrix $A' = (\beta~\vdots~I_{p})$ contains the only coefficients in our model and $p$ refers to the number of parameters in our coefficient vector. To deduce a likelihood function, we assume that the reduced form error and the first stage error, here written as $V= [ \epsilon + v \beta ~\vdots~ v ]$. This error term has the partitioned covariance matrix $\Omega = \begin{pmatrix} \sigma_{11} & \psi_{12} \\ \psi_{21} & \Omega_{22} \\ \end{pmatrix}$, which we treat as known here. Now, we assume a simple and modeled form of heteroskedasticity such that for observations 1 through $n_1$, we have covariance matrix $\Omega_1$ and for observations $n_1 + 1$ through $n$, we have covariance matrix $\Omega_2$.

  • $\begingroup$ Perhaps you would want to explore the properties of the trace, especially related to the trace of products of matrices, en.wikipedia.org/wiki/… $\endgroup$ Jan 2 '15 at 18:43
  • $\begingroup$ Thanks! @AlecosPapadopoulos However, I am already fairly familiar with the trace operator. So what I ended up deriving is essentially a restriction on the data if one awkwardly resubstitutes it. There must be another way though using some theory about the likelihood in general. $\endgroup$
    – Hirek
    Jan 2 '15 at 21:47
  • $\begingroup$ @AlecosPapadopoulos I think I figured it out. The answer seems to be in "Generalizing weak instrument robust IV statistics towards multiple parameters, unrestricted covariance matrices and identification statistics" by Kleibergen in JOE 2007. The trace operator does not offer more here but the vec notation seems to. Now at it. If I figure it out I shall post an answer here. $\endgroup$
    – Hirek
    Jan 3 '15 at 15:56
  • $\begingroup$ Great. It will be good to see an answer with the $\text {vec}$ operator in action! $\endgroup$ Jan 3 '15 at 17:32
  • 1
    $\begingroup$ I don't think I can answer this anytime soon but I have found the solution to the problem. Using the vectorized notation certainly helps. There is a little more involved here and this is now becoming unpublished research which will be finished in a few short months. Then there will appear a link to a very exhaustive solution. $\endgroup$
    – Hirek
    Jan 11 '15 at 14:48

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