Prof. Matthieu Bloch
Wednesday, September 13, 2023
Let \(\set{X_i}_{i=1}^n\) be i.i.d. random variables \(\sim P_X\). Then, \(\frac{1}{n}\sum_{i=1}^nX_i\) converges in probability to \(\E{X}\). Specifically,
\[\begin{align*} \forall\epsilon>0\qquad \lim_{n\to \infty}\P{\abs{\frac{1}{n}\sum_{i=1}^nX_i-\E{X}}>\epsilon}=0. \end{align*}\]
Let \(\set{X_i}_{i=1}^n\) be i.i.d. random variables \(\sim P_X\). Then,
\[\begin{align*} \frac{-1}{n}\log P_{X}^{\otimes n}(X_1,\cdots,X_n)\mathop{\longrightarrow}_{n\to\infty}\bbH(X)\textsf{ in probability.} \end{align*}\]
The typical set \(\calA_\epsilon^n(P_X)\) is
\[\begin{align*} \calA_\epsilon^n(P_X) \eqdef \set{x^n\in\calX^n:2^{-n(\bbH(X)+\epsilon)}\leq P_X^{\otimes n}(x^n)\leq2^{-n(\bbH(X)-\epsilon)}} \end{align*}\]
For \(x^n\in\calA_\epsilon^n(P_X)\)
\[\begin{align*} \bbH(X)-\epsilon \leq -\frac{1}{n}\log P_X^{\otimes n}(x^n) \leq \bbH(X)+\epsilon \end{align*}\]
\(\P{\calA_\epsilon^n(P_X)}\to 1\) as \(n\to\infty\)
\(\card{\calA_\epsilon^n(P_X)} \leq 2^{n(\bbH(X)+\epsilon)}\)
\(\card{\calA_\epsilon^n(P_X)} \geq (1-\epsilon)2^{n(\bbH(X)-\epsilon)}\) for \(n\) large enough
A fixed-lengh \((n,M_n)\) code for a discrete memoryless source \(p_X\) consists of:
The rate of the code \(R\eqdef \frac{\log_2 M_n}{n}\) (bits/source symbol)
The average probability of error
\[\begin{align*} P_e^{(n)}(\calC) \eqdef \P{\widehat{X}^n\neq X^n} \eqdef \sum_{x^n\in\calX^n}P_X^{\otimes n}(x^n)\indic{g_n(f_n(x^n))\neq x^n}. \end{align*}\]
Question 1: for a fixed \(n\in\bbN\), what is the minimum \(R\) required to achieve \(P_e^{(n)}\leq \epsilon\)?
Question 2: as \(n\to\infty\), what is the minimum \(R\) that ensures that \(\lim_{n\to \infty} P_e^{(n)}=0\) \[ C_{\textsf{source}}\eqdef \inf\left\{R: \exists {(f_n,g_n)}_{n\geq 1} with \lim_{n\to\infty}\frac{\log M_n}{n}\leq R \text{ and } \lim_{n\to\infty}P_e^{(n)}=0\right\} \]
For a discrete memoryless source \(p_X\), \(C_{\textsf{source}} = \bbH(X)\)
Consider a sequence of fixed length \((n,M_n)\) source codes \(\set{(f_n,g_n)}_{n\geq1}\) such that \[ \lim_{n\to\infty}\frac{\log M_n}{n}\leq R \text{ and } \lim_{n\to\infty}P_e^{(n)}=0 \] Then \(R\geq \bbH(X)\).
On average (over the random binning to create \(F\)), \[ \E[C]{P_e} \leq \P[P_{U}]{U\notin \calB_\gamma} + \frac{2^{\gamma}}{M}. \]