Exercise 4.3Z: Exponential and Laplace Distribution
We consider here the probability density functions $\rm (PDF)$ of two continuous random variables:
- The random variable $X$ is exponentially distributed (see top plot): For $x<0$ ⇒ $f_X(x) = 0$, and for positive $x$–values:
- $$f_X(x) = \lambda \cdot {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\hspace{0.05cm}. $$
- On the other hand, for the Laplace distributed random variable $Y$ in the whole range $ - \infty < y < + \infty$ holds (lower sketch):
- $$f_Y(y) = \lambda/2 \cdot {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}|\hspace{0.05cm}y\hspace{0.05cm}|}\hspace{0.05cm}.$$
To be calculated are the differential entropies $h(X)$ and $h(Y)$ depending on the PDF parameter $\it \lambda$. For example:
- $$h(X) = -\hspace{-0.7cm} \int\limits_{x \hspace{0.05cm}\in \hspace{0.05cm}{\rm supp} \hspace{0.03cm}(\hspace{-0.03cm}f_X)} \hspace{-0.55cm} f_X(x) \cdot {\rm log} \hspace{0.1cm} \big [f_X(x) \big ] \hspace{0.1cm}{\rm d}x \hspace{0.05cm}.$$
If $\log_2$ is used, add the pseudo-unit "bit".
In subtasks (2) and (4) specify the differential entropy in the following form:
- $$h(X) = {1}/{2} \cdot {\rm log} \hspace{0.1cm} ({\it \Gamma}_{{\hspace{-0.01cm} \rm L}}^{\hspace{0.08cm}(X)} \cdot \sigma^2), \hspace{0.8cm}h(Y) = {1}/{2} \cdot {\rm log} \hspace{0.1cm} ({\it \Gamma}_{{\hspace{-0.05cm} \rm L}}^{\hspace{0.08cm}(Y)} \cdot \sigma^2) \hspace{0.05cm}.$$
Determine by which factor ${\it \Gamma}_{{\hspace{-0.05cm} \rm L}}^{\hspace{0.08cm}(X)}$ the exponential PDF is characterized and which factor ${\it \Gamma}_{{\hspace{-0.01cm} \rm L}}^{\hspace{0.08cm}(Y)}$ results for the Laplace PDF.
Hints:
- The exercise belongs to the chapter Differential Entropy.
- Useful hints for solving this task can be found in particular on the page Differential entropy of some power-constrained random variables.
- For the variance of the exponentially distributed random variable $X$ holds, as derived in Exercise 4.1Z: $\sigma^2 = 1/\lambda^2$.
- The variance of the Laplace distributed random variable $Y$ is twice as large for the same $\it \lambda$: $\sigma^2 = 2/\lambda^2$.
Questions
Solution
- Then the differential entropy is:
- $$h(X) = -\hspace{-0.7cm} \int\limits_{x \hspace{0.05cm}\in \hspace{0.05cm}{\rm supp} \hspace{0.03cm}(\hspace{-0.03cm}f_X)} \hspace{-0.35cm} f_X(x) \cdot {\rm ln} \hspace{0.1cm} \big [f_X(x)\big] \hspace{0.1cm}{\rm d}x \hspace{0.05cm}.$$
- For the exponential distribution, the integration limits are $0$ and $+∞$. In this range, the PDF $f_X(x)$ according to the specification sheet is used:
- $$h(X) =- \int_{0}^{\infty} \hspace{-0.15cm} \lambda \cdot {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x} \hspace{0.05cm} \cdot \hspace{0.05cm} \left [ {\rm ln} \hspace{0.1cm} (\lambda) + {\rm ln} \hspace{0.1cm} ({\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x})\right ]\hspace{0.1cm}{\rm d}x - \hspace{0.05cm} {\rm ln} \hspace{0.1cm} (\lambda) \cdot \int_{0}^{\infty} \hspace{-0.15cm} \lambda \cdot {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\hspace{0.1cm}{\rm d}x \hspace{0.1cm} + \hspace{0.1cm} \lambda \cdot \int_{0}^{\infty} \hspace{-0.15cm} \lambda \cdot x \cdot {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\hspace{0.1cm}{\rm d}x \hspace{0.05cm}.$$
We can see:
- The first integrand is identical to the PDF $f_X(x)$ considered here. Thus, the integral over the entire integration domain yields $1$.
- The second integral corresponds exactly to the definition of the mean value $m_1$ (moment of first order). For the exponential PDF, $m_1 = 1/λ$ holds. From this follows:
- $$h(X) = - \hspace{0.05cm} {\rm ln} \hspace{0.1cm} (\lambda) + 1 = - \hspace{0.05cm} {\rm ln} \hspace{0.1cm} (\lambda) + \hspace{0.05cm} {\rm ln} \hspace{0.1cm} ({\rm e}) = {\rm ln} \hspace{0.1cm} ({\rm e}/\lambda) \hspace{0.05cm}.$$
- This result is to be given the additional unit "nat". Using $\log_2$ instead of $\ln$, we obtain the differential entropy in "bit":
- $$h(X) = {\rm log}_2 \hspace{0.1cm} ({\rm e}/\lambda) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \lambda = 1{\rm :} \hspace{0.3cm} h(X) = {\rm log}_2 \hspace{0.1cm} ({\rm e}) = \frac{{\rm ln} \hspace{0.1cm} ({\rm e})}{{\rm ln} \hspace{0.1cm} (2)} \hspace{0.15cm}\underline{= 1.443\,{\rm bit}} \hspace{0.05cm}.$$
(2) Considering the equation $\sigma^2 = 1/\lambda^2$ valid for the exponential distribution, we can transform the result found in (1) as follows:
- $$h(X) = {\rm log}_2 \hspace{0.1cm} ({\rm e}/\lambda) = {1}/{2}\cdot {\rm log}_2 \hspace{0.1cm} ({\rm e}^2/\lambda^2) = {1}/{2} \cdot {\rm log}_2 \hspace{0.1cm} ({\rm e}^2 \cdot \sigma^2) \hspace{0.05cm}.$$
- A comparison with the required basic form $h(X) = {1}/{2} \cdot {\rm log}_2 \hspace{0.1cm} ({\it \Gamma}_{\hspace{-0.05cm} \rm L}^{\hspace{0.08cm}(X)} \cdot \sigma^2)$ leads to the result:
- $${\it \Gamma}_{{\hspace{-0.05cm} \rm L}}^{\hspace{0.08cm}(X)} = {\rm e}^2 \hspace{0.15cm}\underline{\approx 7.39} \hspace{0.05cm}.$$
(3) For the Laplace distribution, we divide the integration domain into two subdomains:
- $Y$ negative ⇒ proportion $h_{\rm neg}(Y)$,
- $Y$ positive ⇒ proportion $h_{\rm pos}(Y)$.
The total differential entropy, taking into account $h_{\rm neg}(Y) = h_{\rm pos}(Y)$ is given by
- $$h(Y) = h_{\rm neg}(Y) + h_{\rm pos}(Y) = 2 \cdot h_{\rm pos}(Y) $$
- $$\Rightarrow \hspace{0.3cm} h(Y) = - 2 \cdot \int_{0}^{\infty} \hspace{-0.15cm} \lambda/2 \cdot {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}y} \hspace{0.05cm} \cdot \hspace{0.05cm} \left [ {\rm ln} \hspace{0.1cm} (\lambda/2) + {\rm ln} \hspace{0.1cm} ({\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}y})\right ]\hspace{0.1cm}{\rm d}y = - \hspace{0.05cm} {\rm ln} \hspace{0.1cm} (\lambda/2) \cdot \int_{0}^{\infty} \hspace{-0.15cm} \lambda \cdot {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}y}\hspace{0.1cm}{\rm d}y \hspace{0.1cm} + \hspace{0.1cm} \lambda \cdot \int_{0}^{\infty} \hspace{-0.15cm} \lambda \cdot y \cdot {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}y}\hspace{0.1cm}{\rm d}y \hspace{0.05cm}.$$
If we again consider that the first integral gives the value $1$ (PDF area) and the second integral gives the mean value $m_1 = 1/\lambda$ we obtain:
- $$h(Y) = - \hspace{0.05cm} {\rm ln} \hspace{0.1cm} (\lambda/2) + 1 = - \hspace{0.05cm} {\rm ln} \hspace{0.1cm} (\lambda/2) + \hspace{0.05cm} {\rm ln} \hspace{0.1cm} ({\rm e}) = {\rm ln} \hspace{0.1cm} (2{\rm e}/\lambda) \hspace{0.05cm}.$$
- Since the result is required in "bit", we still need to replace "$\ln$" by "$\log_2$":
- $$h(Y) = {\rm log}_2 \hspace{0.1cm} (2{\rm e}/\lambda) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \lambda = 1{\rm :} \hspace{0.3cm} h(Y) = {\rm log}_2 \hspace{0.1cm} (2{\rm e}) \hspace{0.15cm}\underline{= 2.443\,{\rm bit}} \hspace{0.05cm}.$$
(4) For the Laplace distribution, the relation $\sigma^2 = 2/\lambda^2$ holds. Thus, we obtain:
- $$h(X) = {\rm log}_2 \hspace{0.1cm} (\frac{2{\rm e}}{\lambda}) = {1}/{2} \cdot {\rm log}_2 \hspace{0.1cm} (\frac{4{\rm e}^2}{\lambda^2}) = {1}/{2} \cdot {\rm log}_2 \hspace{0.1cm} (2 {\rm e}^2 \cdot \sigma^2) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\it \Gamma}_{{\hspace{-0.05cm} \rm L}}^{\hspace{0.08cm}(Y)} = 2 \cdot {\rm e}^2 \hspace{0.15cm}\underline{\approx 14.78} \hspace{0.05cm}.$$
- Consequently, the ${\it \Gamma}_{{\hspace{-0.05cm} \rm L}}$ value is twice as large for the Laplace distribution as for the exponential distribution.
- Thus, the Laplace PDF is better than the exponential PDF in terms of differential entropy when power-limited signals are assumed.
- Under the constraint of peak limitation, both the exponential and Laplace PDF are completely unsuitable, as is the Gaussian PDF. These all extend to infinity.