finding the maximum likelihood estimator: conditional generalised linear model Find the maximum likelihood estimator of $\alpha$,$\beta$ and $\lambda$  given the model
\begin{equation*}P(Z=z|X=x)=\exp\bigg\{\sum_{j=1}^m(\alpha_j+\beta_j{x})z_j+\sum_{j<k}\lambda_{jk}z_jz_k-\Psi(\alpha+\beta{x},\lambda)\bigg\}\end{equation*}
where
\begin{equation*}\Psi
(\alpha+\beta{x},\lambda)=\log\bigg\{\sum_{z\in\Omega}\exp\big\{\sum_{j=1}^m\big(\alpha_j+\beta_jx\big)z_j+\sum_{j<k}\lambda_{jk}z_jz_k\big\}\bigg\}\end{equation*} 
\begin{equation*}z_j=\begin{cases}
 1 & \text{if success}\\
 0 & \text{if failure}
 \end{cases}\end{equation*}
$x$ is the vector of covariates, $\alpha$,$\beta$ and $\lambda$ are the unknown parameters
$\sum_{z\in\Omega}$ is the summation over all possible values $z$ can take. i was able to find the likelihood function and the log likelihood functions as follows: 
Denote $L$ the likelihood function of the model for a random
 sample of size $N$ so that we have the following\
 \begin{equation*}L(\theta;z)=\prod_{i=1}^N\exp\bigg\{\sum_{j=1}^m(\alpha_j+\beta_j{x_i})z_{ij}+\sum_{j<k}\lambda_{jk}z_{ij}z_{ik}-\Psi(\alpha+\beta{x_i},\lambda)\bigg\}\end{equation*} where $\theta=(\alpha,\beta,\lambda)$
so that the log likelihood function $l$ becomes
 \begin{equation}l=\sum_{j=1}^m\alpha_j\sum_{i=1}^Nz_{ij}+\sum_{j=1}^m\beta_j\sum_{i=1}^Nx_iz_{ij}+
 \sum_{j<k}\lambda_{jk}\sum_{i=1}^Nz_{ij}z_{ik}-\sum_{i=1}^N\Psi(\alpha+\beta{x_i},\lambda)\end{equation}
my main problem is to find the maximum likelihood estimator of  $\alpha$,$\beta$ and $\lambda$. I had a problem in finding the first and second partial of the log likelihood function more especially finding the partial of the $\Psi$ function. Help me find the score and fisher information.
 A: Well, that's not an answer, I'll just try to help with matrix nonation.
First of all let's stack all column-vectors $z_i=(z_{i1}\dots z_{im})^T$ in rows of a matrix, so that
$$Z=(z_1^T,z_2^T\dots z_N^T)^T=\begin{bmatrix}
z_{11}      & \cdots & z_{1m}      \\
\vdots & \ddots & \vdots \\ 
z_{N1}     & \cdots & z_{Nm}
\end{bmatrix}
$$
Then let's put all $\lambda_{jk}$ in a matrix $\Lambda$ (note that it will be upper triangular) and all $\alpha, \beta, x$ will be combined in corresponding vectors $\vec{\alpha}, \vec{\beta}, \vec{x}$.
Then your log likelihood function will look like:
$$l=\vec{1}^TZ\vec{\alpha}+\vec{x}^TZ\vec{\beta}+\mathrm{Tr}(Z\Lambda Z^T)-\vec{1}^T\vec{\Psi}$$
Here $\vec{1}^T=(1,1\dots 1)$ and $\vec{\Psi}=(\Psi(\alpha+\beta{x_1},\lambda)\dots \Psi(\alpha+\beta{x_N},\lambda))^T$.
And $$\Psi(\alpha+\beta{x_i},\lambda)=\log\bigg\{\sum_{z\in\Omega}\exp\big\{\big(\vec{\alpha}^T+\vec{\beta}^Tx_i\big)z_i+\mathrm{Tr}(Z\Lambda Z^T)\big\}\bigg\}$$
Either way the last term looks very nasty, especially the summation over all possible $z$.
In order to find ML estimates $\hat{\theta}=(\hat{\vec{\alpha}}, \hat{\vec{\beta}}, \hat{\vec{\lambda}})$ of $\theta=(\vec{\alpha}, \vec{\beta}, \vec{\lambda})$ one has to compute the derivatives with the respect to $\theta$, set the obtained equations to zero and solve them. So the obtained system will look like:
$$\begin{cases} 
\frac{\partial \ l}{\partial \vec{\alpha}}=\vec{1}^TZ- \vec{1}^T\frac{\partial \vec{\Psi}}{\partial \vec{\alpha}}=0\\
\frac{\partial \ l}{\partial \vec{\beta}}=\vec{x}^TZ- \vec{1}^T\frac{\partial \vec{\Psi}}{\partial \vec{\beta}}=0\\
\frac{\partial \ l}{\partial \Lambda}=Z^TZ- \frac{\partial \  \mathrm{Tr}(\vec{1}\!\otimes\!\vec{\Psi})}{\partial \Lambda}=0
 \end{cases}$$
here $\otimes$ - is the outer product.
So the solution of those equations will yield you ML estimates.
Do I understand you correctly that $\Omega=(0,1)$?
