Example: Random Bit String

Consider a random sequence of 1s and 0s, i.e., the probability of a 0 or 1 is always $P(0)=P(1)=1/2$ . The corresponding probability density function is

$$p_b(x) = \frac{1}{2}\delta(x) + \frac{1}{2}\delta(x-1)$$

and the entropy is

$$h(p_b) = \frac{1}{2}$$lg$$(2) + \frac{1}{2}$$lg$$(2) =$$   lg$$(2) = 1$$

Thus, 1 bit is required for each bit of the sequence. In other words, the sequence cannot be compressed. There is no redundancy.

If instead the probability of a 0 is 1/4 and that of a 1 is 3/4, we get

$$\begin{eqnarray*} p_b(x) &=& \frac{1}{4}\delta(x) + \frac{3}{4}\delta(x-1)\\ h(p_b) &=& \frac{1}{4}\mbox{lg}(4) + \frac{3}{4}\mbox{lg}\left(\frac{4}{3}\right) = 0.81128\ldots \end{eqnarray*}$$

and the sequence can be compressed about $19\%$ .

In the degenerate case for which the probability of a 0 is 0 and that of a 1 is 1, we get

$$\begin{eqnarray*} p_b(x) &=& \lim_{\epsilon \to0}\left[\epsilon \delta(x) + (1-\epsilon )\delta(x-1)\right]\\ h(p_b) &=& \lim_{\epsilon \to0}\epsilon \cdot\mbox{lg}\left(\frac{1}{\epsilon }\right) + 1\cdot\mbox{lg}(1) = 0 \end{eqnarray*}$$