Show that typical sets satisfy $|A_\epsilon^{(n)}(p)|>|A_\epsilon^{(n)}(q)|$ if $0.5\ge p>q$

Question

I'm struggling with the following question: For typical sets in source coding, decide if it is true or false and provide a detailed proof.

My ideas:

a. true, since by definition we get 2^log(p) > 2^log(q), 2^(nH(p)+eps)>? 2^(nH(q)+eps) 2^(n*(-1/nlog(p)+eps) > 2^(n(-1/n*log(q)+eps) p > q, as stated in the question.

b. false, p and q may be not equal, but take p close to q by epsilon, and we might get the same typical set, which yields overlapping.

c. true, since p=1-q, if we take p=1/2 => q=1/2. and if all values of An(p) are close to An(q), then it's the same set, which means they are the whole {0,1}n

I'm not sure about my answers, would like to hear your thoughts, and proofs if possible.

Answer 1

By the AEP, we know that the size of a typical set satisfies $$(1-\epsilon) 2^{n(H(X)-\epsilon)} \le |A_\epsilon^{(n)}|\le 2^{n(H(X)+\epsilon)},$$ for large enough $n$. Roughly speaking, we can think this as saying that $|A_\epsilon^{(n)}|\sim 2^{n H(X)}$.

Furthermore, if $0.5\ge p>q$, then $H(p)<H(q)$. The conclusion follows immediately.