Source Coding Theorem (Achievability) (Scribe: LIANG Tong)
We now derive and use properties of the to obtain a source code.
`As shown in the previous class/notes,
(1)
To remind ourselves, this bound arises by first proving that the probability that a sequence is atypical is , where
is defined as the relative entropy between two probability mass functions p(x) and q(x) (also called the Kullback-Leibler divergence), where the denotes the sample space of the random variable X.
Second, as asked to prove in PS2, if we have . Combining these two results gives (1).
For a "reasonable" source code we would expect that be very small, or in fact asymptotically negligible in n. This is possible if we let be a function of n, and set . In all that follows, we condition on the assumption that is indeed typical, and that
(2)
Bound on the size of the typical set
The number of typical
(here we assume that p < 1/2; if p>1/2, a similar argument yields an upper bound of )
Question :
By the Taylor series expansion of H(p), for some p' in . For fixed p distinct from 0 and 1, the maximum value of can be bounded by some c (that is a function of p, but not of )
Therefore, the size of the typical set is at most .
Source coding scheme
1. Use the first bit to describe whether the sequence is typical or not
2(a). If the sequence is atypical, describe the entire sequence uncompressed using n bits.
2(b). If the sequence is typical, describe the index number of the sequence in the set of typical sequences, using bits.
Expected total # bits transmitted (one bit to describe to whether typical or not, and then bits to describe the sequence (typical/atypical) )
.
Note that the size of the typical set decreases as epsilon decreases, but corresponding the probability of atypicality increases.
Discussion of typicality
p <1/2 and n =10 in the following figures
Fig1: the relation between #of heads and probability,
Fig2: between # of heads and size of T(k,n)
Note that the value of k for which the elements of T(k,n) have the largest (individual) probabilities is k=0 (Figure 1), and the value of k for which the size of the set T(k,n) is largest is k =n/2 (Figure 2).
However, as we saw via the analysis before, the most likely T(k,n) (the type-class with the largest probability of occuring) is T(np, n), i.e., when k=np.
Binary source coding theorem
(a) such that for all n>N, , code with expected rate and 0-error.
(b) code with and expected rate < (we shall prove this "converse" in a later class).
The above theorem can be generalized to general i.i.d. sources over finite alphabets, with the only difference being in how the corresponding entropy function is defined -- the achievability is left as an excercise for the reader.
Comments (10)
Cho Yiu Ng said
at 11:23 am on Jan 29, 2009
No comments?
sidjaggi said
at 12:37 am on Feb 2, 2009
Hmm, I really would like the class to join in and comment a bit on this, before I jump in and edit. Open for comments until Friday the 6th of Feb.
Mak said
at 2:45 pm on Feb 2, 2009
You wrote: "the typical set decrease, as the epsilon decrease." What does that mean? Thanks.
Silas said
at 10:13 am on Feb 4, 2009
I cannot follow the logical flow easily in this scribe note. May you add some connectives between statements?
Thanks a lot.
Silas said
at 10:18 am on Feb 4, 2009
Since H(p) is not an increasing function, H(p+\epsilon) >= H(k/n) does not hold in general. Therefore, I don't understand why it is true in the middle of the scribe note. Please correct me if I am wrong.
Tong said
at 11:38 am on Feb 4, 2009
"Since H(p) is not an increasing function, H(p+\epsilon) >= H(k/n) does not hold in general"
Here, p< 1/2. and p+\epsilon < 1/2 (i think it is because the epsilon is a small number)
Cho Yiu Ng said
at 11:14 am on Feb 5, 2009
What is the script X in your definition of divergence?
Please make sure all the symbols are defined and state the assumptions explicitly.
Tong said
at 11:51 am on Feb 5, 2009
the set of the value which x can take
Cho Yiu Ng said
at 5:03 pm on Feb 5, 2009
Please put this in the scribe note.
sidjaggi said
at 8:10 am on Feb 10, 2009
Finished my revision -- if there are any comments, speak now, or forever hold your peace :)
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