# 4 equations, 4 variables

 P: 39 How do you solve the system of equations $$e^{x_{1}+y_{1}}+e^{x_{1}+y_{2}}=a_{1}$$ $$e^{x_{2}+y_{1}}+e^{x_{2}+y_{2}}=a_{2}$$ $$e^{x_{1}+y_{1}}+e^{x_{2}+y_{1}}=b_{1}$$ $$e^{x_{1}+y_{2}}+e^{x_{2}+y_{2}}=b_{2}$$ x1, x2, y1, y2 are the variables for which I want to solve the equations, a1, a2, b1, b2 are known. Context: I need to solve this in order to get the unknown maximum entropy joint probabilities $$p_{ij}=e^{-1-x_{i}-y_{j}}$$ $$\mbox{for the known marginal probabilities (}a_{i}\mbox{ and }b_{j}\mbox{).}$$ i know there is way to do this in information theory, but I need to solve it algebraically.
 P: 1,403 4 equations, 4 variables There's a problem here. First subsitute $u_1 = e^{x1}$ $v_1 = e^{y1}$ etc. Than the four equations become: $$u_1 v_1 + u_1 v_2 = a_1$$ $$u_2 v_1 + u_2 v_2 = a_2$$ $$u_1 v_1 + u_2 v_1 = b_1$$ $$u_1 v_2 + u_2 v_2 = b_2$$ No if you add the first two you get: $$u_1 v_1 + u_1 v_2 + u_2 v_1 + u_2 v_2 = a_1 + a_2$$ and if you add the last two you get $$u_1 v_1 + u_1 v_2 + u_2 v_1 + u_2 v_2 = b_1 + b_2$$ These can't both be true unless $a_1 + a_2 = b_1 + b_2$ And if that is the case, you have only 3 equations left for 4 unknowns, so there won't be an unique solution.
 P: 39 Thank you! Quick reply here: yes, a1+a2=b1+b2 because they are marginal probabilities and sum to 1. Also, $$u_{1}v_{1}+u_{1}v_{2}+u_{2}v_{1}+u_{2}v_{2}=1$$ because these are the joint probabilities. Sorry! I should have mentioned that. I will be back in half an hour to report if this gives me enough information to solve the system.
 P: 1,403 If you write the equations as: $$u_1 (v_1 + v_2) = a_1$$ $$v_1 (u_1 + u_2) = b_1$$ $$(v_1 + v_2)(u_1 + u_2) = 1$$ it's easy to see if $(u_1,u_2,v_1,v_2)$ is a solution, so is $(c u_1, c u_2, \frac {v_1}{c}, \frac {v_2}{c} )$ to get a solution you can set u1 + u2 = 1 so v1 this gets you $$u_1 = c a_1$$ $$u_2 = c a_2$$ $$v_1 = \frac {b_1}{c}$$ $$v_2 = \frac {b_2}{c}$$ as the complete solution.