ThmDex – An index of mathematical definitions, results, and conjectures.

Result R4817 on D4422: Mutual information
Subresult of R3503: Data processing inequality

Data processing inequality for transformed endpoint

Formulation 0

Let $X$ and $Y$ each be a D202: Random variable such that

(i)	$f$ is a D201: Measurable map

Then

(1)	\begin{equation} I(X; Y) \geq I(X; f(Y)) \end{equation}
(2)	\begin{equation} I(X; Y) = I(X; f(Y)) \quad \iff \quad I(X; Y \mid f(Y)) = 0 \end{equation}

Proofs

Proof 0

Let $X$ and $Y$ each be a D202: Random variable such that

(i)	$f$ is a D201: Measurable map

Since $f$ is measurable, result R4816: Markov chain triple if endpoint is a transformation of the midpoint shows that $X \to Y \to f(Y)$ is a Markov chain triple, whence this result is a special case of R3503: Data processing inequality. $\square$