Consider a binary memoryless channel characterized by the transition probability diagram shown in the figure.

The channel is

Concept: (i) For lossless channel H(X/Y) = 0

                (ii) For the channel to be deterministic;   H(Y/X) = 0

                (iii) For the channel to be noiseless, (H(x/y) = H(y/x) = 0)

Application: From the given channel, the channel matrix will be:

                                y₁             y₂

$P\left(Y/X\right)=\begin{array}{c}{x}_1\\ x_2\end{array}\left[\begin{array}{cc}0.25& 0.75\\ 0.25& 0.75\end{array}\right]=\left[\begin{array}{cc}\frac{1}{4}& \frac{3}{4}\\ \frac{1}{4}& \frac{3}{4}\end{array}\right]$

Assuming P(x₁) = P(x₂) = 1/2

$\because P\left(X,Y\right)=\left[P\left(X\right)\right]\left[P\left(\frac{Y}{X}\right)\right]$

$P\left(X,Y\right)=\left[\begin{array}{cc}\frac{1}{8}& \frac{3}{8}\\ \frac{1}{8}& \frac{3}{8}\end{array}\right]$

So, $P\left(y_1\right)=\frac{1}{4},P\left(y_2\right)=\frac{3}{4}$

$P\left(\frac{X}{Y}\right)=\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}\end{array}\right]$

(i) For lossless channel H(X/Y) = 0

$H\left(\frac{X}{Y}\right)=\frac{1}{8};{\log }_2\frac{2+3}{8}{\log }_2\frac{2+1}{8}{\log }_2\frac{2+3}{8}{\log }_{2}2$

= 1 ≠ 0 (so, the channel is not lossless)

(ii) For the channel to be deterministic;

H(Y/X) = 0

$H\left(\frac{Y}{X}\right)=\frac{1}{8}{\log }_2\left(4\right)+\frac{3}{8}{\log }_2\left(\frac{4}{3}\right)+\frac{1}{8}{\log }_2\frac{4+3}{8}{\log }_2\left(\frac{4}{3}\right)$

So, H(Y/X) ≠ 0; the channel is not determinant as well.

(iii) For the channel to be noiseless,

(H(x/y) = H(y/x) = 0)

But, H(x/y) ≠ H(y/x) ≠ 0

Say it is not noiseless as well.