정보량과 로그

ㄱ. $\displaystyle {\mathrm{P}} ( E ) = \frac{1}{2}$이므로 $I ( E ) = - \log _{2} {\mathrm{P}} ( E )$$\displaystyle = - \log _{2} \frac{1}{2}$$= 1$ (참) ㄴ. 두 사건 $\mathrm{A} , B$가 서로 독립이므로 $\mathrm{P} ( A \cap B ) = P ( A ) P ( B )$ ∴ $I ( A \cap B ) = - \log _{2} {\mathrm{P}} ( A \cap B )$$= - \log _{2} {\mathrm{P}} ( A ) {\mathrm{P}} ( B )$ $= - \left\{ \log _{2} {\mathrm{P}} ( A ) + \log _{2} {\mathrm{P}} ( B ) \right\}$ $= - \log _{2} {\mathrm{P}} ( A ) - \log _{2} {\mathrm{P}} ( B )$ $= I ( A ) + I ( B )$ (참) ㄷ. $2 I ( A \cup B ) = - 2 \log _{2} {\mathrm{P}} ( A \cup B )$$= - \log _{2} \left\{ {\mathrm{P}} ( A \cup B ) \right\} ^{2}$ $I ( A ) + I ( B ) = - \log _{2} {\mathrm{P}} ( A ) - \log _{2} {\mathrm{P}} ( B )$$= - \log _{2} {\mathrm{P}} ( A ) {\mathrm{P}} ( B )$ $\mathrm{P} \mathit{(} A \cup B ) \geq \mathrm{P} \mathit{(} A ) > 0$, $\mathrm{P} \mathit{(} A \cup B ) \geq \mathrm{P} \mathit{(} B ) > 0$이므로 $\left\{ \mathrm{P} \mathit{(} A \cup B ) \right\} ^{2} \geq \mathrm{P} \mathit{(} A ) \mathrm{P} \mathit{(} B )$ ∴ $\log _{2} \left\{ \mathrm{P} \mathit{(} A \cup B ) \right\} ^{2} \geq \log _{2} \mathrm{P} \mathit{(} A ) \mathrm{P} \mathit{(} B )$ $- \log _{2} \left\{ \mathrm{P} \mathit{(} A \cup B ) \right\} ^{2} \leq - \log _{2} \mathrm{P} \mathit{(} A ) \mathrm{P} \mathit{(} B )$ ∴ $2 I ( A \cup B ) \leq I ( A ) + I ( B )$ (참) 따라서 보기 중 옳은 것은 ㄱ, ㄴ, ㄷ이다.