0
$\begingroup$

I am wondering why stochastic encoder and decoder can not help the Shannon source coding? I know the achievability scheme of source coding, which is based on typicality, is deterministic, and hence we dont need any further randomization, but I want to know what is the reason why further private randomization does not help.

Thanks,

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Could you elaborate on your wisdom on "I know the achievability scheme of source coding, which is based on typicality, is deterministic"? $\endgroup$
    – Turbo
    Commented Apr 30, 2014 at 23:15

1 Answer 1

Reset to default
1
$\begingroup$

The question, as it stands, is unclear.

In any case there is data processing inequality, which proves that any further randomization does not increase mutual information.

See e.g. section 6. from this pdf. Or:

  • Sec. 2.8 in Thomas M. Cover, Joy A. Thomas, "Elements of Information Theory", Second Edition (1991).
Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .