<p>Since the interaction information can be <a href="https://en.wikipedia.org/wiki/Interaction_information" rel="nofollow noreferrer">either negative or positive</a>, and a Markov chain that erases everything can be used to take the interaction information to <span class="math-container">$0$</span>, without further conditions the answer is &quot;no&quot;.</p>

CC BY-SA 4.0
2020-07-02T23:56:35.623Z
2020-07-02T23:56:35.623Z
Peter Shor 
<p>The interaction information is defined as <span class="math-container">$I(X;Y)-I(X;Y|Z)$</span>. Let <span class="math-container">$Z-(X, Y) -(X', Y')$</span>  be a Markov chain. Is there an inequality similar to the data processing inequality, relating <span class="math-container">$I(X';Y')-I(X';Y'|Z)$</span> to <span class="math-container">$I(X;Y)-I(X;Y|Z)$</span>? Thanks in advance.</p>

<it.information-theory>
Data processing inequality for interaction information
Dina Abdelhadi
# Data processing inequality for interaction information

(Asked by: username_0 on 2020-07-01)

The interaction information is defined as $I(X;Y)-I(X;Y|Z)$. Let $Z-(X, Y) -(X', Y')$ be a Markov chain. Is there an inequality similar to the data processing inequality, relating $I(X';Y')-I(X';Y'|Z)$ to $I(X;Y)-I(X;Y|Z)$? Thanks in advance.

**Tags:** it.information-theory

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## 1 answers

### Answer by username_1 on 2020-07-02. Score: 1 | Accepted answer

Since the interaction information can be [either negative or positive](https://en.wikipedia.org/wiki/Interaction_information), and a Markov chain that erases everything can be used to take the interaction information to $0$, without further conditions the answer is "no".


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