- #1

noowutah

- 57

- 3

It is well-known that with known marginal probabilities a_{i} and

b_{j} the joint probability distribution maximizing the entropy

[tex]H(P)=-\sum_{i=1}^{m}\sum_{j=1}^{n}p_{ij}\log{}p_{ij}[/tex]

is [tex]p_{ij}=a_{i}b_{j}[/tex]

For m=3 and n=3, a=(0.2,0.3,0.5), b=(0.1,0.6,0.3), for example,

[tex]\begin{equation}

\label{eq:r1}

P=\left(

\begin{array}{rrr}

0.02 & 0.12 & 0.06 \\

0.03 & 0.18 & 0.09 \\

0.05 & 0.30 & 0.15

\end{array}

\right)

\end{equation}[/tex]

Now, I have a problem where the joint probability distribution is

constrained (much like a random walk where the path from one node to

another is blocked). For example, let m, n, a, and b be as above with

the constraint that

[tex]\begin{equation}

\label{eq:r2}

P=\left(

\begin{array}{rrr}

p_{11} & 0 & p_{13} \\

p_{21} & p_{22} & p_{23} \\

p_{31} & p_{32} & p_{33} \\

\end{array}

\right)

\end{equation}[/tex]

Because a and b are known, it suffices to find out any 2x2 matrix

contained in the 3x3 matrix, for example (x,y,z the variables by which

w_{1}, w_{2}, v_{1}, v_{2}, and sigma are expressible, given a and b)

[tex]\begin{equation}

\label{eq:r3}

P=\left(

\begin{array}{rrr}

x & 0 & w_{1} \\

y & z & w_{2} \\

v_{1} & v_{2} & \sigma \\

\end{array}

\right)

\end{equation}[/tex]

I use this to write out the entropy and differentiate with respect to

x, y, and z to find out that the maximum will be where

[tex]\begin{equation}

\label{eq:r4}

\det\left(

\begin{array}{ll}

x & w_{1} \\

v_{1} & \sigma

\end{array}

\right)=0

\end{equation}[/tex]

[tex]\begin{equation}

\label{eq:r5}

\det\left(

\begin{array}{ll}

y & w_{2} \\

v_{1} & \sigma

\end{array}

\right)=0

\end{equation}[/tex]

[tex]\begin{equation}

\label{eq:r6}

\det\left(

\begin{array}{ll}

z & w_{2} \\

v_{2} & \sigma

\end{array}

\right)=0

\end{equation}[/tex]

This is a system of 3 non-linear equations which are awkward to solve

algebraically. In the end, I am interested to know which property a

and b need to have to ensure that P is a proper probability

distribution (i.e. no negative elements). For now, however, I'd be

thrilled if anybody could give me a hint how I could find the

solutions for x, y, and z algebraically without these non-linear

equations.

The numeric solution for this is

[tex]\begin{equation}

\label{eq:r7}

P=\left(

\begin{array}{rrr}

0.093 & 0.000 & 0.107 \\

0.001 & 0.113 & 0.186 \\

0.006 & 0.487 & 0.007

\end{array}

\right)

\end{equation}[/tex]

But I have definitely seen systems where all solutions were complex

and some of the probabilities ended up <0, especially when m and n are

larger than 3 and there are more than one zero constraint in the joint

probability matrix.

b_{j} the joint probability distribution maximizing the entropy

[tex]H(P)=-\sum_{i=1}^{m}\sum_{j=1}^{n}p_{ij}\log{}p_{ij}[/tex]

is [tex]p_{ij}=a_{i}b_{j}[/tex]

For m=3 and n=3, a=(0.2,0.3,0.5), b=(0.1,0.6,0.3), for example,

[tex]\begin{equation}

\label{eq:r1}

P=\left(

\begin{array}{rrr}

0.02 & 0.12 & 0.06 \\

0.03 & 0.18 & 0.09 \\

0.05 & 0.30 & 0.15

\end{array}

\right)

\end{equation}[/tex]

Now, I have a problem where the joint probability distribution is

constrained (much like a random walk where the path from one node to

another is blocked). For example, let m, n, a, and b be as above with

the constraint that

[tex]\begin{equation}

\label{eq:r2}

P=\left(

\begin{array}{rrr}

p_{11} & 0 & p_{13} \\

p_{21} & p_{22} & p_{23} \\

p_{31} & p_{32} & p_{33} \\

\end{array}

\right)

\end{equation}[/tex]

Because a and b are known, it suffices to find out any 2x2 matrix

contained in the 3x3 matrix, for example (x,y,z the variables by which

w_{1}, w_{2}, v_{1}, v_{2}, and sigma are expressible, given a and b)

[tex]\begin{equation}

\label{eq:r3}

P=\left(

\begin{array}{rrr}

x & 0 & w_{1} \\

y & z & w_{2} \\

v_{1} & v_{2} & \sigma \\

\end{array}

\right)

\end{equation}[/tex]

I use this to write out the entropy and differentiate with respect to

x, y, and z to find out that the maximum will be where

[tex]\begin{equation}

\label{eq:r4}

\det\left(

\begin{array}{ll}

x & w_{1} \\

v_{1} & \sigma

\end{array}

\right)=0

\end{equation}[/tex]

[tex]\begin{equation}

\label{eq:r5}

\det\left(

\begin{array}{ll}

y & w_{2} \\

v_{1} & \sigma

\end{array}

\right)=0

\end{equation}[/tex]

[tex]\begin{equation}

\label{eq:r6}

\det\left(

\begin{array}{ll}

z & w_{2} \\

v_{2} & \sigma

\end{array}

\right)=0

\end{equation}[/tex]

This is a system of 3 non-linear equations which are awkward to solve

algebraically. In the end, I am interested to know which property a

and b need to have to ensure that P is a proper probability

distribution (i.e. no negative elements). For now, however, I'd be

thrilled if anybody could give me a hint how I could find the

solutions for x, y, and z algebraically without these non-linear

equations.

The numeric solution for this is

[tex]\begin{equation}

\label{eq:r7}

P=\left(

\begin{array}{rrr}

0.093 & 0.000 & 0.107 \\

0.001 & 0.113 & 0.186 \\

0.006 & 0.487 & 0.007

\end{array}

\right)

\end{equation}[/tex]

But I have definitely seen systems where all solutions were complex

and some of the probabilities ended up <0, especially when m and n are

larger than 3 and there are more than one zero constraint in the joint

probability matrix.

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