χ^n上の確率質量凾数P(x^n)から求める符号長(符号長の期待値, 平均符号長)
\(\chi^n\)上の確率質量凾数\(P(x^n)\)から求める符号長
$$\begin{array}{rl} I(x^n)=\lceil -\log_2{P(x^n)} \rceil &\quad\dotso \chi^n上の確率質量凾数P(x^n)から求める符号長(I:\chi^n \rightarrow \mathbb{R}^{+})\\ \end{array}$$符号長の期待値
$$\begin{array}{rl} E\left[\mathcal{l}(x^n)\right]=\displaystyle \sum_{x^n\in \chi^n} \mathcal{l}(x^n)P(x^n)&\quad\dotso符号長の期待値, 平均符号長(average\;codeword\;length)\\ \end{array}$$例:\(\chi=\{0,1\},\;n=2\)
| \(y_1\) | \(y_2\) | \(x^2\) \(=y_1y_2\) |
\(P(x^2)\) | \(-\log_2{P(x^2)}\) Shannon情報量 |
\(I(x^2)=\lceil -\log_2{P(x^2)}\rceil\) \(\chi^2\)上の確率質量凾数\(P(x^2)\)から求める符号長 |
\(\pi(x^2)\) | \(\mathcal{l}(x^2)\) \(=|\pi(x^2)|\) |
\(\pi'(x^2)\) | \(\mathcal{l}'(x^2)\) \(=|\pi'(x^2)|\) |
\(\pi''(x^2)\) | \(\mathcal{l}''(x^2)\) \(=|\pi''(x^2)|\) |
|---|---|---|---|---|---|---|---|---|---|---|---|
| 0 | 0 | 00 | \(\frac{1}{8}\) | \(-\log_2{\frac{1}{8}}=3\) | \(\lceil -\log_2{\frac{1}{8}}\rceil=3\) | 000 | 3 | 1 | 1 | 00 | 2 |
| 0 | 1 | 01 | \(\frac{1}{8}\) | \(-\log_2{\frac{1}{8}}=3\) | \(\lceil -\log_2{\frac{1}{8}}\rceil=3\) | 001 | 3 | 01 | 2 | 01 | 2 |
| 1 | 0 | 10 | \(\frac{1}{4}\) | \(-\log_2{\frac{1}{4}}=2\) | \(\lceil -\log_2{\frac{1}{4}}\rceil=2\) | 01 | 2 | 001 | 3 | 10 | 2 |
| 1 | 1 | 11 | \(\frac{1}{2}\) | \(-\log_2{\frac{1}{2}}=1\) | \(\lceil -\log_2{\frac{1}{2}}\rceil=1\) | 1 | 1 | 000 | 3 | 11 | 2 |
上記例で\(\mathcal{l}(x^2)\)が\(I\)と等しい符号化(\(\pi\))の平均符号長
$$\begin{array}{rcl} \displaystyle E\left[\mathcal{l}(x^2)\right] &=&\displaystyle \sum_{x^2\in \chi^2} \mathcal{l}(x^2)P(x^2)\\ &=&\displaystyle 3\times\frac{1}{8}+3\times\frac{1}{8}+2\times\frac{1}{4}+1\times\frac{1}{2}\\ &=&\displaystyle \frac{3+3+4+4}{8} = \frac{14}{8}\\ &=&\displaystyle 1+\frac{3}{4}=1.75\\ \end{array}$$上記例で\(\mathcal{l}'(x^2)\)が\(I\)と等しくない符号化(\(\pi'\))の平均符号長
$$\begin{array}{rcl} \displaystyle E\left[\mathcal{l}'(x^2)\right] &=&\displaystyle \sum_{x^2\in \chi^2} \mathcal{l}'(x^2)P(x^2)\\ &=&\displaystyle 1\times\frac{1}{8}+2\times\frac{1}{8}+3\times\frac{1}{4}+3\times\frac{1}{2}\\ &=&\displaystyle \frac{1+2+6+12}{8} = \frac{21}{8}\\ &=&\displaystyle 2+\frac{5}{8}=2.625\\ \end{array}$$上記例で\(\mathcal{l}(x^n)=2\)と常に一定となる符号化(\(\pi''\))の平均符号長
$$\begin{array}{rcl} \displaystyle E\left[\mathcal{l}''(x^2)\right] &=&\displaystyle \sum_{x^2\in \chi^2} \mathcal{l}''(x^2)P(x^2)\\ &=&\displaystyle 2\times\frac{1}{8}+2\times\frac{1}{8}+2\times\frac{1}{4}+2\times\frac{1}{2}\\ &=&\displaystyle 2\times\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{4}+\frac{1}{2}\right)\\ &=&\displaystyle 2\times 1 \\ &=&\displaystyle 2\\ \end{array}$$
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