Maximum Entropy in Gaussian Setting

Click For Summary
SUMMARY

The discussion centers on maximizing differential entropy within a set of inequalities in a Gaussian setting, specifically using mutual information concepts from Information Theory. The Normal distribution is established as the maximizer of differential entropy. The inequalities involve random variables U, V, X1, and X2, with Gaussian noise terms N1, N2, and N3. The user seeks confirmation on whether the jointly Gaussian distribution of these variables maximizes the conditional entropy h(b*X1+X2+N3|V) given previous assumptions about the distributions.

PREREQUISITES
  • Understanding of differential entropy and its maximization
  • Familiarity with mutual information and the Entropy Power Inequality (EPI)
  • Knowledge of Gaussian distributions and properties
  • Basic concepts in Information Theory, particularly rate regions
NEXT STEPS
  • Study the properties of Gaussian distributions in the context of entropy maximization
  • Explore the Entropy Power Inequality (EPI) and its applications in Information Theory
  • Investigate the implications of jointly Gaussian distributions on conditional entropies
  • Review the concept of rate regions and their significance in communication theory
USEFUL FOR

Researchers and practitioners in Information Theory, statisticians, and engineers working on communication systems who are interested in entropy maximization and Gaussian distributions.

ferreat
Messages
2
Reaction score
0
Hello,
I have a doubt about the distribution of random variables that maximize the differential entropy in a set of inequalities. It is well known that the Normal distribution maximizes the differential entropy. I have the following set of inequalities:

T1 < I(V;Y1|U)
T2 < I(U;Y2)
T3 < I(X1,X2;Y3|V)
T4 < I(X1,X2;Y3)

where, Y1=X1+N1, Y2=a*X1+N2, Y3=b*X1+X2+N3. N1,N2,N3 are Gaussian ~ N(0,1). The lower case a and b are positive real numbers a < b. U, V, X1 and X2 are random variables. I want to maximize that set of inequalities. I know the following:

(i) From T4, h(Y3) maximum is when Y3 is Gaussian then X1 and X2 are Gaussian.

(ii) From T2 we maximize it by having h(Y2) or h(a*X1+N2) maximum. From this by the Entropy Power Inequality (EPI) we bound -h(a*X1+N2|U) and have X1|U Gaussian.

(iii) From T1 we maximize it by having h(Y1|U) or h(X1+N1|U) maximum which we can do as -h(a*X1+N2|U) in the part ii can be bounded having Y1 Gaussian (satisfying the maximum entropy theorem).

The Question:

From T3, can I assume that jointly Gaussian distribution will maximize h(Y3|V) or h(b*X1+X2+N3) having the assumptions i,ii,iii ?

My aim is to show that jointly Gaussian distribution of U, V, X1 and X2 maximizes the set of inequalities. I hope anyone can help me out with this.
 
Physics news on Phys.org
I don't understand your notation. Is "T4" a number or is it only a designator for an expression? Are the vertical bars "|" to denote absolute values? - conditional probabilities?
 
Thanks Stephen for your reply. Basically the set of inequalities is what is known in Information Theory as a rate region:
T1 < I(V;Y1|U)
T2 < I(U;Y2)
T3 < I(X1,X2;Y3|V)
T1+T2+T3 < I(X1,X2;Y3).
T1, T2 adn T3 are the rates obtained when transmitting messages 1, 2 and 3. The I's are Mutual Informations and the vertical bars "|" indicate conditioning. For instance I(V;Y1|U) = h(Y1|U) - h(Y1|U,V) where h(x) is the differential entropy.
My question is basically is after having assumed h(X1+N1|U) maximum implies (X1+N1|U) Gaussian in (iii), could I assume h(b*X1+X2+N3|V) maximum implies (b*X1+X2+N3|V) Gaussian? I know if I hadn't assumed (i,ii,iii) this last question would be affirmative, but having (i,ii,iii) is it still true?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
8K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K