Solving a difficult equation for a variable?

Question

I'm trying to obtain the maximum likelihood estimate of the parameters for a model I'm building. I have constants $\sigma$, $\mu$, and $q_0$; a boolean matrix $\alpha$; and vectors $A, \beta, r, d,$ and $Q$, all quantities non-negative. I set the derivative of the log likelihood function equal to zero with

$$0=\sum_j \alpha_{kj} \Bigg[ \beta_j r_j \frac{\ln (A_j - r_j\sum_i \alpha_{ij} Q_i) + \sigma^2 - \mu}{\sigma^2 (A_j - r_j\sum_i \alpha_{ij} Q_i)} + (1-\beta_j)\frac{\ln(1-d_j)}{q_0} \Bigg]$$

How can I solve for $Q$?

Answer 1

I suspect that the question you have asked may not actually be the one that you wanted to ask. However, I will answer it anyway. Your equation contains the expression $$ \alpha_{kj}Q_k + \sum_{i\neq k} \alpha_{ij} Q_i $$ but this is just the same as $\sum_i\alpha_{ij}Q_j$. It therefore does not depend on $k$; I will call it $u_j$. I will then put $$ v_j = \frac{\ln(A_j-r_ju_j)+\sigma^2-\mu}{\sigma^2(A_j-r_ju_j)} $$ and $$ w_j = \beta_jv_j + (1-\beta_j)\frac{\ln(1-d_j)}{q_0}. $$ Your equation is then $\sum_j\alpha_{kj}w_j=0$. Presumably this is supposed to hold for all $k$. It is now a purely linear problem to find the possible vectors $w$, and another linear problem with the same matrix to recover the vector $Q$ from the vector $u$. It sounds like you believe that there should be a unique solution for $Q$. That can only happen if the matrix $\alpha$ has trivial kernel, so $w$ would have to be zero. From there it is trivial to find $v$, then $u_j$ can be expressed in terms of $v_j$ using the Lambert W function, and finally it is a linear problem to find $Q$.