The Data Processing Inequality

Markovity

A Markov Chain is a random process that describes a sequence of possible events where the probability of each event depends only on the outcome of the previous event. Thus, we say that $X,Y,Z$ is a Markov chain in this order, denoted as:

$X\rightarrow Y\rightarrow Z$

(1)

If we can write:

$P\left(X=x,Y=y,Z=z\right)=P\left(Z=z\mid Y=y\right)\cdot P\left(Y=y\mid X=x\right)\cdot P\left(X=x\right)$

(2)

Or in a more compact form:

$P\left(x,y,z\right)=P\left(z\mid y\right)\cdot P\left(y\mid x\right)\cdot P\left(x\right)$

(3)

We can use Markov chains to model how a signal is corrupted when passed through noisy channels. For example, if $X$ is a binary signal, it can change with a certain probability, $p$ to $Y$, and it can again be corrupted to produce $Z$.

Consider the joint probability $P\left(x,z\mid y\right)$. We can express this as:

$P\left(x,z\mid y\right)={\frac {P\left(x,y,z\right)}{P\left(y\right)}}$

(4)

And if $X\rightarrow Y\rightarrow Z$, we get:

$P\left(x,z\mid y\right)={\frac {P\left(z\mid y\right)\cdot P\left(y\mid x\right)\cdot P\left(x\right)}{P\left(y\right)}}$

(5)

Since $P\left(y,x\right)=P\left(y\mid x\right)\cdot P\left(x\right)=P\left(x\mid y\right)\cdot P\left(y\right)$, we can write:

$P\left(x,z\mid y\right)={\frac {P\left(z\mid y\right)\cdot P\left(y,x\right)}{P\left(y\right)}}=P\left(z\mid y\right)\cdot P\left(x\mid y\right)$

(6)

Thus, we can say that $X$ and $Z$ are conditionally independent given $Y$. If we think of $X$ as some past event, and $Z$ as some future event, then the past and future events are independent if we know the present event $Y$. Note that this property is good definition of, as well as a useful tool for checking Markovity.

We can rewrite the joint probability $P\left(x,y,z\right)$ as:

${\begin{aligned}P\left(x,y,z\right)&=P\left(z\mid y\right)\cdot P\left(y\mid x\right)\cdot P\left(x\right)\\&={\frac {P\left(z,y\right)}{P\left(y\right)}}\cdot P\left(y,x\right)\\&={\frac {P\left(z,y\right)}{P\left(y\right)}}\cdot P\left(x\mid y\right)\cdot P\left(y\right)\\&=P\left(z,y\right)\cdot P\left(x\mid y\right)\\&=P\left(x\mid y\right)\cdot P\left(y\mid z\right)\cdot P\left(z\right)\\\end{aligned}}$

(7)