*Why*

*I*(

*X*;

*Y*

^{3}) >

*I*(

*X*;

*Y*

^{2}) ?

*X*emitting symbols

*x*

_{0},

*x*

_{1}, treating both as a single symbol ⟨

*x*

_{0},

*x*

_{1}⟩ into an encoder

*E*emitting encoded sequence ⟨

*x*

_{0},

*x*

_{1}, ∏⟩ where parity ∏ = sum of bits ≜

*x*

_{0}⊕

*x*

_{1}, is an "addition modulo–2" or "exclusive–or" function.

Hence the total information for the DMS is

*H*(

*x*

_{0},

*x*

_{1}, ∏) =

*H*(

*x*

_{0}) +

*H*(

*x*

_{1}|

*x*

_{0}) +

*H*(∏|

*x*

_{0},

*x*

_{1})

=

*H*(

*X*) +

*H*(

*X*) + 0

= 2

*H*(

*X*)

*H*(

*X*) × 1 ∕ 3 is called the

**coded information rate**. ❶

Let us now consider the above case with some numbers. Assume that the compound symbols (*C*) are *x*_{0}, *x*_{1}, ∏ all ∈ {0, 1} with *p*(*x*_{0} = 0) = 0.4, *p*(*x*_{1} = 1) = 0.6. Thus *p*(*C*) = *p*(*x*_{0}) *p*(*x*_{1}). Then for four decoding rules (four cases)

In each respective case only four possibilities are noted. This is because 0 (from source) is given out as 0 and similarly 1→1. But when e occurs 0↛1 or 1↛0. In other words, when e occurs all other possibilities of *Y*^{3} are declared **"error detected and re-transmit"**. This is therefore also called an Automatic Repeat reQuest (ARQ) system.

Because of the e output it is not possible for either *x*_{0} or *x*_{1} to be confounded. We can therefore transmit *without any information loss* through the erasure channel *at the cost* of sending more data symbols.
❷

*Next:*

Summary (p:5) ➽