Exercise 4.4: Conventional Entropy and Differential Entropy

Two uniform distributions

We consider the two continuous random variables $X$ and $Y$ with probability density functions $f_X(x)$ and $f_Y(y)$. For these random variables one can

not specify the conventional entropies $H(X)$ and $H(Y)$ , respectively,
but the differential entropies $h(X)$ and $h(Y)$.

We also consider two value-discrete random variables:

The variable $Z_{X,\hspace{0.05cm}M}$ is obtained by (suitably) quantizing the random quantity $X$ with the quantization level number $M$
⇒ quantization interval width ${\it \Delta} = 0.5/M$.
The variable $Z_{Y,\hspace{0.05cm}M}$ is obtained after quantization of the random quantity $Y$ with the quantization level number $M$
⇒ quantization interval width ${\it \Delta} = 2/M$.

The probability density functions $\rm (PDF)$ of these discrete random variables are each composed of $M$ Dirac functions whose momentum weights are given by the interval areas of the associated value-continuous random variables.

From this, the entropies $H(Z_{X,\hspace{0.05cm}M})$ and $H(Z_{Y,\hspace{0.05cm}M})$ can be determined in the conventional way according to the page Probability mass function and entropy .

In the section Entropy of value-ontinuous random variables after quantization, an approximation was also given. For example:

$$H(Z_{X, \hspace{0.05cm}M}) \approx -{\rm log}_2 \hspace{0.1cm} ({\it \Delta}) + h(X)\hspace{0.05cm}. $$

In the course of the task it will be shown that in the case of rectangular PDF ⇒ uniform distribution this "approximation" gives the same result as the direct calculation.
But in the general case – so in $\text{Example 2}$ with triangular PDF – this equation is in fact only an approximation, which agrees with the actual entropy $H(Z_{X,\hspace{0.05cm}M})$ only in the limiting case ${\it \Delta} \to 0$ .

Hints:

The exercise belongs to the chapter Differential Entropy.
Useful hints for solving this task can be found in particular on the page Entropy of value-continuous random variables after quantization.

Questions

$ h(X) \ = \ $

$\ \rm bit$

$ h(Y) \ = \ $

$\ \rm bit$

$H(Z_{X,\hspace{0.05cm}M=4})\ = \ $

$\ \rm bit$

$H(Z_{X,\hspace{0.05cm}M=4})\ = \ $

$\ \rm bit$

$H(Z_{Y,\hspace{0.05cm}M=8})\ = \ $

$\ \rm bit$

	The entropy of a discrete value random variable $Z$ is always $H(Z) \ge 0$.
	The differential entropy of a continuous value random variable $X$ is always $h(X) \ge 0$.

Solution

(1) According to the corresponding theory page, with $x_{\rm min} = 0$ and $x_{\rm max} = 1/2$:

$$h(X) = {\rm log}_2 \hspace{0.1cm} (x_{\rm max} - x_{\rm min}) = {\rm log}_2 \hspace{0.1cm} (1/2) \hspace{0.15cm}\underline{= - 1\,{\rm bit}}\hspace{0.05cm}.$$

(2) On the other hand, with $y_{\rm min} = -1$ and $y_{\rm max} = +1$ the differential entropy of the random variable $Y$ is given by:

$$h(Y) = {\rm log}_2 \hspace{0.1cm} (y_{\rm max} - y_{\rm min}) = {\rm log}_2 \hspace{0.1cm} (2) \hspace{0.15cm}\underline{= + 1\,{\rm bit}}\hspace{0.05cm}. $$

Quantized random variable $Z_{X, \ M = 4}$

(3) The adjacent graph illustrates the best possible quantization of random variable $X$ with quantization level number $M = 4$ ⇒ random variable $Z_{X, \ M = 4}$:

The interval width here is equal to ${\it \Delta} = 0.5/4 = 1/8$.
The possible values (at the center of the interval, respectively) are $z \in \{0.0625,\ 0.1875,\ 0.3125,\ 0.4375\}$.

Using the probability mass function, the direct entropy calculation gives $P_Z(Z) = \big [1/4,\ \text{...} , \ 1/4 \big]$:

$$H(Z_{X, \ M = 4}) = {\rm log}_2 \hspace{0.1cm} (4) \hspace{0.15cm}\underline{= 2\,{\rm bit}} \hspace{0.05cm}.$$

With the approximation, considering the result of (1), we obtain:

$$H(Z_{X,\hspace{0.05cm} M = 4}) \approx -{\rm log}_2 \hspace{0.1cm} ({\it \Delta}) + h(X) = 3\,{\rm bit} +(- 1\,{\rm bit})\hspace{0.15cm}\underline{= 2\,{\rm bit}}\hspace{0.05cm}. $$

Note: Only in the case of uniform distribution, the approximation gives exactly the same result as the direct calculation, i.e. the actual entropy.

Quantized random variable $Z_{Y, \ M = 4}$

(4) From the second graph, one can see the similarities / differences to subtask (3):

The quantization parameter is now ${\it \Delta} = 2/4 = 1/2$.
The possible values are now $z \in \{\pm 0.75,\ \pm 0.25\}$.
Thus, here the "approximation" (as well as the direct calculation) gives the result:

$$H(Z_{Y,\hspace{0.05cm} M = 4}) \approx -{\rm log}_2 \hspace{0.1cm} ({\it \Delta}) + h(Y) = 1\,{\rm bit} + 1\,{\rm bit}\hspace{0.15cm}\underline{= 2\,{\rm bit}}\hspace{0.05cm}.$$

Quantized random variable $Z_{Y, \ M = 8}$

(5) In contrast to subtask (5): ${\it \Delta} = 1/4$ is now valid. From this follows for the "approximation":

$$H(Z_{Y,\hspace{0.05cm} M = 8}) \approx -{\rm log}_2 \hspace{0.1cm} ({\it \Delta}) + h(Y) = 2\,{\rm bit} + 1\,{\rm bit}\hspace{0.15cm}\underline{= 3\,{\rm bit}}\hspace{0.05cm}.$$

Again, one gets the same result as in the direct calculation.

(6) Only statement 1 is correct:

The entropy $H(Z)$ of a value-discrete random variable $Z = \{z_1, \ \text{...} \ , z_M\}$ is never negative.
For example, the limiting case $H(Z) = 0$ results for ${\rm Pr}(Z = z_1) = 1$ and ${\rm Pr}(Z = z_\mu) = 0$ for $2 \le \mu \le M$.

In contrast, the differential entropy $h(X)$ of a continuous value random variable $X$ can be as follows:
- $h(X) < 0$ $($subtask 1$)$,
- $h(X) > 0$ $($subtask 2$)$, or even
- $h(X) = 0$ $($for example fo $x_{\rm min} = 0$ and $x_{\rm max} = 1)$.