4 equations, 4 variables


by stlukits
Tags: equations, joint probabilities, system of equations, variables
stlukits
stlukits is offline
#1
Feb9-14, 02:37 AM
P: 35
How do you solve the system of equations

[tex]e^{x_{1}+y_{1}}+e^{x_{1}+y_{2}}=a_{1}[/tex]
[tex]e^{x_{2}+y_{1}}+e^{x_{2}+y_{2}}=a_{2}[/tex]
[tex]e^{x_{1}+y_{1}}+e^{x_{2}+y_{1}}=b_{1}[/tex]
[tex]e^{x_{1}+y_{2}}+e^{x_{2}+y_{2}}=b_{2}[/tex]

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

[tex]p_{ij}=e^{-1-x_{i}-y_{j}}[/tex]

[tex]\mbox{for the known marginal probabilities (}a_{i}\mbox{ and }b_{j}\mbox{).}[/tex]

i know there is way to do this in information theory, but I need to solve it algebraically.
Phys.Org News Partner Mathematics news on Phys.org
Researchers help Boston Marathon organizers plan for 2014 race
'Math detective' analyzes odds for suspicious lottery wins
Pseudo-mathematics and financial charlatanism
chiro
chiro is offline
#2
Feb9-14, 02:45 AM
P: 4,570
Hey stlukits.

This is basically a multi-equation root finding problem which involves multi-variable calculus and optimization.

Take a look at this for more information:

http://homes.soic.indiana.edu/classe...ons_method.pdf
stlukits
stlukits is offline
#3
Feb9-14, 09:03 AM
P: 35
I don't need a numeric solution or an approximation. I want to solve the equation for x1, x2, y1, y2, if possible. If not, I'll have to live with it.

willem2
willem2 is offline
#4
Feb9-14, 09:36 AM
P: 1,351

4 equations, 4 variables


There's a problem here. First subsitute [itex] u_1 = e^{x1} [/itex] [itex] v_1 = e^{y1} [/itex] etc.

Than the four equations become:

[tex] u_1 v_1 + u_1 v_2 = a_1 [/tex]
[tex] u_2 v_1 + u_2 v_2 = a_2 [/tex]
[tex] u_1 v_1 + u_2 v_1 = b_1 [/tex]
[tex] u_1 v_2 + u_2 v_2 = b_2 [/tex]

No if you add the first two you get:

[tex] u_1 v_1 + u_1 v_2 + u_2 v_1 + u_2 v_2 = a_1 + a_2 [/tex]

and if you add the last two you get

[tex] u_1 v_1 + u_1 v_2 + u_2 v_1 + u_2 v_2 = b_1 + b_2 [/tex]

These can't both be true unless [itex] a_1 + a_2 = b_1 + b_2 [/itex]

And if that is the case, you have only 3 equations left for 4 unknowns, so there won't be an unique solution.
stlukits
stlukits is offline
#5
Feb9-14, 04:57 PM
P: 35
Thank you! Quick reply here: yes, a1+a2=b1+b2 because they are marginal probabilities and sum to 1. Also,

[tex]u_{1}v_{1}+u_{1}v_{2}+u_{2}v_{1}+u_{2}v_{2}=1[/tex]

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.
willem2
willem2 is offline
#6
Feb9-14, 06:01 PM
P: 1,351
If you write the equations as:

[tex] u_1 (v_1 + v_2) = a_1 [/tex]
[tex] v_1 (u_1 + u_2) = b_1 [/tex]
[tex] (v_1 + v_2)(u_1 + u_2) = 1 [/tex]

it's easy to see if [itex] (u_1,u_2,v_1,v_2) [/itex] is a solution, so is [itex] (c u_1, c u_2, \frac {v_1}{c}, \frac {v_2}{c} ) [/itex]

to get a solution you can set u1 + u2 = 1 so v1

this gets you

[tex] u_1 = c a_1 [/tex]
[tex] u_2 = c a_2 [/tex]
[tex] v_1 = \frac {b_1}{c} [/tex]
[tex] v_2 = \frac {b_2}{c} [/tex]

as the complete solution.
stlukits
stlukits is offline
#7
Feb9-14, 07:13 PM
P: 35
Got it. Thank you, willem2. There was information hiding here that I didn't take into account (basically, that it is sufficient to know p_ij for i=1,...,m-1 and j=1,...,n-1 in order to know the m x n dimensional joint probability matrix). Be that as it may, problem solved!


Register to reply

Related Discussions
System of three equations and four variables Precalculus Mathematics Homework 25
Moving Variables in Equations General Math 6
Two Equations, Three variables General Math 2
Solving for variables using 3 different equations (simultaneous equations) Precalculus Mathematics Homework 1
Complex conjugate variables as independent variables in polynomial equations Linear & Abstract Algebra 0