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43 | 43 | Let $I$ be the of all $x$ for which $p(x)$ is non-zero. Then, proving \eqref{eq:Gibbs-ineq} requires to show that |
44 | 44 |
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45 | 45 | $$ \label{eq:Gibbs-ineq-s1} |
46 | | -\sum_{x \in I} p(x) \, \frac{\ln p(x)}{\ln q(x)} \geq 0 \; . |
| 46 | +\sum_{x \in I} p(x) \, \ln \frac{p(x)}{q(x)} \geq 0 \; . |
47 | 47 | $$ |
48 | 48 |
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49 | 49 | Because $\ln x \leq x - 1$, i.e. $-\ln x \geq 1 - x$, for all $x > 0$, with equality only if $x = 1$, we can say about the left-hand side that |
50 | 50 |
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51 | 51 | $$ \label{eq:Gibbs-ineq-s2} |
52 | 52 | \begin{split} |
53 | | -\sum_{x \in I} p(x) \, \frac{\ln p(x)}{\ln q(x)} &\geq \sum_{x \in I} p(x) \left( 1 - \frac{p(x)}{q(x)} \right) \\ |
| 53 | +\sum_{x \in I} p(x) \, \ln \frac{p(x)}{q(x)} &\geq \sum_{x \in I} p(x) \left( 1 - \frac{p(x)}{q(x)} \right) \\ |
54 | 54 | &= \sum_{x \in I} p(x) - \sum_{x \in I} q(x) \; . |
55 | 55 | \end{split} |
56 | 56 | $$ |
57 | 57 |
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58 | | -Finally, since $p(x)$ and $q(x)$ are [probability mass functions](/D/pmf), it holds that $0 \leq p(x),q(x) \leq 1$ and we also have |
| 58 | +Finally, since $p(x)$ and $q(x)$ are [probability mass functions](/D/pmf), we have |
59 | 59 |
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60 | 60 | $$ \label{eq:p-q-pmf} |
61 | 61 | \begin{split} |
62 | | -\sum_{x \in I} p(x) &= 1 \\ |
63 | | -\sum_{x \in I} q(x) &\leq 1 \; , |
| 62 | +0 \leq p(x) \leq 1, \quad \sum_{x \in I} p(x) &= 1 \quad \text{and} \\ |
| 63 | +0 \leq q(x) \leq 1, \quad \sum_{x \in I} q(x) &\leq 1 \; , |
64 | 64 | \end{split} |
65 | 65 | $$ |
66 | 66 |
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67 | 67 | such that it follows from \eqref{eq:Gibbs-ineq-s2} that |
68 | 68 |
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69 | 69 | $$ \label{eq:Gibbs-ineq-s3} |
70 | 70 | \begin{split} |
71 | | -\sum_{x \in I} p(x) \, \frac{\ln p(x)}{\ln q(x)} &\geq \sum_{x \in I} p(x) - \sum_{x \in I} q(x) \\ |
| 71 | +\sum_{x \in I} p(x) \, \ln \frac{p(x)}{q(x)} &\geq \sum_{x \in I} p(x) - \sum_{x \in I} q(x) \\ |
72 | 72 | &= 1 - \sum_{x \in I} q(x) \geq 0 \; . |
73 | 73 | \end{split} |
74 | 74 | $$ |
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