Commmon Problems of Entropy
Common problems of Shannon entropy in information theory.
Theorem: Deterministic Distribution Minimizes The Entropy
What is the minimum value of \(H\left(p_1, \ldots, p_n\right)=H(\mathbf{p})\) as \(\mathbf{p}\) ranges over the set of \(n\)-dimensional probability vectors? Find all p's which achieve this minimum.
Solution: We wish to find all probability vectors \(\mathbf{p}=\left(p_1, \ldots, p_n\right)\) which minimize \[ H(\mathbf{p})=-\sum_i p_i \log p_i . \]
Now \(-p_i \log p_i \geq 0\), with equality iff \(p_i=0\) or 1 . Besides we know that \(\sum_i p_i = 1\).
Hence the only possible probability vectors which minimize \(H(\mathbf{p})\) are those with \(p_i=1\) for some \(i\) and \(p_j=0, j \neq i\). There are \(n\) such vectors, i.e., \((1,0, \ldots, 0),(0,1,0, \ldots, 0), \ldots,(0, \ldots, 0,1)\), and the minimum value of \(H(\mathbf{p})\) is 0 .
Theorem: Uniform Distribution Maximizes The Entropy
Source:
- Why is Entropy maximised when the probability distribution is uniform?
- Why does \(\log (1+x)=x+O(x^2)\) when \(𝑥 \rightarrow 0\)?
Proof
Heuristically, the probability density function on \(\left\{x_1, x_2, \ldots, x_n\right\}\) with maximum entropy turns out to be the one that corresponds to the least amount of knowledge of \(\left\{x_1, x_2, \ldots, x_n\right\}\), in other words the Uniform distribution.
Now, for a more formal proof consider the following: A probability density function on \(\left\{x_1, x_2, \ldots, . x_n\right\}\) is a set of nonnegative real numbers \(p_1, \ldots, p_n\) that add up to 1 . Entropy is a continuous function of the \(n\)-tuples \(\left(p_1, \ldots, p_n\right)\), and these points lie in a compact subset of \(\mathbb{R}^n\), so there is an \(n\)-tuple where entropy is maximized. We want to show this occurs at \((1 / n, \ldots, 1 / n)\) and nowhere else.
Suppose the \(p_j\) are not all equal, say \(p_1<p_2\). (Clearly \(n \neq 1\).) We will find a new probability density with higher entropy. It then follows, since entropy is maximized at some \(n\)-tuple, that entropy is uniquely maximized at the \(n\)-tuple with \(p_i=1 / n\) for all \(i\).
Since \(p_1<p_2\), for small positive \(\varepsilon\) we have \(p_1+\varepsilon<p_2-\varepsilon\). The entropy of \(\left\{p_1+\varepsilon, p_2-\varepsilon, p_3, \ldots, p_n\right\}\) minus the entropy of \(\left\{p_1, p_2, p_3, \ldots, p_n\right\}\) equals \[ -p_1 \log \left(\frac{p_1+\varepsilon}{p_1}\right)-\varepsilon \log \left(p_1+\varepsilon\right)-p_2 \log \left(\frac{p_2-\varepsilon}{p_2}\right)+\varepsilon \log \left(p_2-\varepsilon\right) \]
To complete the proof, we want to show this is positive for small enough \(\varepsilon\). Rewrite the above equation as \[ \begin{aligned} -p_1 \log \left(1+\frac{\varepsilon}{p_1}\right) & -\varepsilon\left(\log p_1+\log \left(1+\frac{\varepsilon}{p_1}\right)\right)-p_2 \log \left(1-\frac{\varepsilon}{p_2}\right) \\ & +\varepsilon\left(\log p_2+\log \left(1-\frac{\varepsilon}{p_2}\right)\right) \end{aligned} \]
Recalling that \(\log (1+x)=x+O\left(x^2\right)\) for small \(x\), the above equation is \[ -\varepsilon-\varepsilon \log p_1+\varepsilon+\varepsilon \log p_2+O\left(\varepsilon^2\right)=\varepsilon \log \left(p_2 / p_1\right)+O\left(\varepsilon^2\right) \] which is positive when \(\varepsilon\) is small enough since \(p_1<p_2\).
Simpler Proof
Recalling that \(H(X) \leq \log |\mathcal{X}|\) with equality if and only has a uniform distribution over \(\mathcal X\).
Then for any r.v. \(X\), it's maximum entropy is \(\log |\mathcal X|\), and is achieved by a uniform distribution over \(\mathcal X\).
Drawing with and without replacement
Drawing with and without replacement. An urn contains \(r\) red, \(w\) white, and \(b\) black balls. Which has higher entropy, drawing \(n \geq 2\) balls from the urn with replacement or without replacement?
Solution:
Let random variable \(X_i\) denotes the outcome color of the \(i-\)th ball. The alphabate of all \(X_i\) is the same: \(\mathcal X = \{0,1,2\}\), where
- \(X_i=0\) if the ball is red,
- \(X_i=1\) if the ball is white
- \(X_i=2\) if the ball is black.
Let \(p_{X_i}\) denote the PMF of \(X_i\).
For the case with replacement:
For all \(X_i \in X_1, \cdots, X_n\), Since the total number of red, white, black balls are \(r, w, b\), \[ p(X_i) = \begin{cases} \frac{r}{r + w + b} & \text{if } X = 1 \\ \frac{w}{r + w + b} & \text{if } X = 2 \\ \frac{b}{r + w + b} & \text{if } X = 3 . \end{cases} \] \(p(X_i)\) doesn't change during the drawing process, so all \(p(X_i)\) are independant(because of the replacement).
Due to Theorem: Entropy is additive for independent r.v., \[ H\left(X_1, X_2, \cdots, X_k\right)=H\left(X_1\right)+H\left(X_2\right)+\cdots+H\left(X_k\right) . \] For the case without replacement:
the picks are not independent anymore, due to the chain rule for joint entropy, we have \[ \begin{aligned} H\left(X_1, X_2, \cdots, X_k\right) & =H\left(X_1\right)+H\left(X_2 \mid X_1\right)+\cdots+H\left(X_k \mid X_1, \cdots, X_k-1\right) . \end{aligned} \]
Meanwhile, due to Theorem: Theorem: Conditioning reduces entropy, we have: \[ H\left(X_1\right)+H\left(X_2 \mid X_1\right)+\cdots+H\left(X_k \mid X_1, \cdots, X_k-1\right) \leq H\left(X_1\right)+H\left(X_2\right)+\cdots+H\left(X_k\right) \] Hence, the sample with replacement has higher entropy.