Aufgaben:Exercise 4.09Z: Laplace Distributed Noise

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2D Laplacian PDF

We consider two-dimensional noise  $\boldsymbol{n} = (n_1, n_2)$.

The two noise variables are "independent and identically distributed ", abbreviated i.i.d., and each has a Laplace probability density:

$$p_{n_1}(x) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} K \cdot {\rm e}^{- a \hspace{0.03cm}\cdot \hspace{0.03cm} |x|} \hspace{0.05cm},$$

$$ p_{n_2}(y) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} K \cdot {\rm e}^{- a \hspace{0.03cm}\cdot \hspace{0.03cm} |y|} \hspace{0.05cm}. $$

• The 2D probability density function  $p_{\it \boldsymbol{n}}(x, y)$  is shown in the graph.
• To simplify notation, the realizations of  $n_1$  and  $n_2$  are denoted here by  $x$  and  $y$,  respectively.

Notes:

• The exercise belongs to the chapter  "Approximation of the Error Probability".
• We would like to refer to the interactive applet  "2D Laplace". 
• The integral resulting in subtask (6) must be split into several partial integrals due to the magnitude formation.
• Furthermore, it holds:

$$\int_{0}^{\infty} x^2 \cdot {\rm e}^{-a \hspace{0.03cm}\cdot \hspace{0.03cm} x} \,{\rm d} x = {2}/{a^3} \hspace{0.05cm}.$$

Questions

1

What is the size of the constant  $K$  of the 1D PDF?

	$K = 1$.
	$K = a/2$
	$K = 1/a$.

2

Let  $a = 1$. What are the mean  ${\rm E}\big[n_i \big]$  and variance  $\sigma^2 = {\rm E}\big[n_i^2\big]$  of the two 1D random variables?  $(i = 1, 2)$

${\rm E}\big[n_i\big] \ = \ $

${\rm E}\big[n_i^2\big] \ = \ $

3

What is the shape of the contour lines of the 2D PDF in the first quadrant?

	They are straight lines.
	They are hyperbolas.
	They are circles.

4

Let further  $a = 1$. What is the probability that both  $n_1$  and  $n_2$  are negative?

${\rm Pr}\big[(n_1 < 0) ∩ (n_2 < 0)\big]\ = \ $ $\ \%$

5

What is the probability that  $n_1$  and  $n_2$  are jointly greater than  $1$? 

${\rm Pr}\big[(n_1 > 1) ∩ (n_2 > 1)\big]\ = \ $ $\ \%$

6

What is the probability that the sum is  $n_1 + n_2 > 2$? 

$ {\rm Pr}\big[n_1 + n_2 > 2)\big] \ = \ $ $\ \%$

Solution

Solution

(1)  Solution 2 is correct:

• The area under the PDF must equal $1$:

$$\int_{-\infty}^{+\infty} p_{n_1}(x) \,{\rm d} x = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \int_{0}^{+\infty} p_{n_1}(x) \,{\rm d} x = 0.5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} K \cdot \int_{0}^{\infty} {\rm e}^{- a \hspace{0.03cm}\cdot \hspace{0.03cm}x} \,{\rm d} x = - {K}/{a} \cdot \left [ {\rm e}^{- a \hspace{0.03cm} \cdot \hspace{0.03cm} x} \right ]_{0}^{\infty}= {K}/{a} = 0.5 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} K = {a}/{2}\hspace{0.05cm}.$$

(2)  The linear mean is equal to 0 due to PDF symmetry.

Thus, the variance $\sigma^2$ is actually equal to the root mean square, as already stated in the question:

$$\sigma^2 = {\rm E}[n_1^2] = 2 \cdot \frac{a}{2} \cdot \int_{0}^{\infty} x^2 \cdot {\rm e}^{-a \hspace{0.03cm} \cdot \hspace{0.03cm} x} \,{\rm d} x = a \cdot {2}/{a^3}= {2}/{a^2} \hspace{0.05cm}. \hspace{0.2cm}{\rm Mit}\hspace{0.15cm}a = 1\text{:} \hspace{0.2cm}\hspace{0.1cm}\underline {\sigma^2 = 2 }\hspace{0.05cm}.$$

Contour lines of the 2D Laplace distribution

(3)  Solution 1 is correct:

• In the first quadrant ($x ≥ 0, y ≥ 0$), the magnitude formation can be omitted. Then the 2D PDF is given by:

$$\boldsymbol{ p }_{\boldsymbol{ n }} (x,\hspace{0.15cm} y) = {a^2}/{4} \cdot {\rm e}^{- a \hspace{0.03cm}\cdot \hspace{0.03cm}x} \cdot {\rm e}^{- a \hspace{0.03cm}\cdot \hspace{0.03cm}y }= {a^2}/{4} \cdot {\rm e}^{- a \hspace{0.03cm}\cdot \hspace{0.03cm}(x+y)}\hspace{0.05cm}.$$

• A contour line with factor $\beta$ versus maximum then has the following shape ($0 < \beta < 1$):

$${\rm e}^{- a \hspace{0.03cm}\cdot \hspace{0.03cm}(x+y)} = \beta \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x + y = \frac{{\rm ln}\hspace{0.15cm}1/\beta}{a} \hspace{0.05cm}.$$

• The graph shows the contour lines for $a = 1$ and some values of $\beta$, each giving a square rotated by $45^\circ$  ⇒  the contour lines are thus even.

(4)  The probability event considered here corresponds exactly to the third quadrant of the composite PDF sketched above. Because of symmetry, this probability is:

$${\rm Pr}[(n_1 < 0) ∩ (n_2 < 0)]\hspace{0.15cm}\underline {=25\%}.$$

(5)  For this, the composite PDF can be written:

$${\rm Pr} \left [ (n_1 > 1)\cap (n_2 > 1)\right ] = {1}/{4} \cdot \int_{1}^{\infty} \int_{1}^{\infty}{\rm e}^{- (x+y)} \,{\rm d} x \,{\rm d} y == {1}/{2} \cdot \int_{1}^{\infty} {\rm e}^{- x} \,{\rm d} x \hspace{0.15cm} \cdot \hspace{0.15cm} {1}/{2} \cdot \int_{1}^{\infty} {\rm e}^{- y} \,{\rm d} y $$

$$ \Rightarrow \hspace{0.3cm}{\rm Pr} \left [ (n_1 > 1)\cap (n_2 > 1)\right ] = \left [ {\rm Pr} (n_1 > 1)\right ] \cdot \left [ {\rm Pr} (n_2 > 1)\right ]\hspace{0.05cm}. $$

Considered is the statistical independence between $n_1$ and $n_2$ and the equality $p_{\it n1}(x) = p_{\it n2}(y)$. For $a = 1$ holds:

$${\rm Pr} (n_1 > 1) = {1}/{2} \cdot \int_{1}^{\infty} {\rm e}^{- x} \,{\rm d} x = {1}/({2{\rm e}})\approx 0.184\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm Pr} \left [ (n_1 > 1)\cap (n_2 > 1)\right ] = {1}/({4{\rm e}^2)}\hspace{0.1cm}\hspace{0.15cm}\underline {\approx 3.4\%}\hspace{0.05cm}.$$

Division of the integration domain

(6)  The region considered here is highlighted in color in the following graph.

• However, the regions extend to the right and above to infinity.
• The searched probability results in

$${\rm Pr} [ n_1 \hspace{-0.2cm} \ + \ \hspace{-0.2cm} n_2 > 2 ] =\frac{1}{4} \cdot \int\limits_{-\infty}^{+\infty} {\rm e}^{-|x|} \int\limits_{2-x}^{\infty}{\rm e}^{-|y|} \,{\rm d} y \,{\rm d} x = I_1 + I_2 + I_3 + I_4 \hspace{0.05cm}.$$

Because of the magnitude, a splitting into partial integrals has to be done. Upwards and to the right all areas extend to infinity. Because of the symmetry $I_4 = I_3$ is valid.

$$I_1 = {1}/{4} \cdot \int_{2}^{+\infty} \hspace{-0.15cm}{\rm e}^{-x} \int_{0}^{\infty}\hspace{-0.15cm}{\rm e}^{-y} \,{\rm d} y \,{\rm d} x = {1}/{4} \cdot \int_{2}^{+\infty} {\rm e}^{-x} \,{\rm d} x ={1}/({4{\rm e}^2})\hspace{0.05cm},$$

$$I_2 = {1}/{4} \cdot \hspace{-0.1cm} \int_{0}^{2} \hspace{-0.15cm}{\rm e}^{-x} \int_{2-x}^{\infty}\hspace{-0.15cm}{\rm e}^{-y} \,{\rm d} y \,{\rm d} x = {1}/{4} \cdot \hspace{-0.1cm} \int_{0}^{2} {\rm e}^{-x}\hspace{-0.1cm} \cdot {\rm e}^{x-2} \,{\rm d} x$$

$$\Rightarrow \hspace{0.3cm}I_2 = {1}/{4} \cdot \hspace{-0.1cm}\int_{0}^{2} {\rm e}^{-2} \,{\rm d} x = {1}/({2{\rm e}^2})\hspace{0.05cm},$$

$$I_3 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{4} \cdot \int_{-\infty}^{0} {\rm e}^{x} \int_{2-x}^{\infty}{\rm e}^{-y} \,{\rm d} y \,{\rm d} x = {1}/{4} \cdot \int_{-\infty}^{0} {\rm e}^{x} \cdot {\rm e}^{x-2} \,{\rm d} x = {1}/{4} \cdot \int_{-\infty}^{0} {\rm e}^{2x-2} \,{\rm d} x = \frac{{\rm e}^{-2}}{4} \cdot \int_{0}^{\infty} {\rm e}^{-2x} \,{\rm d} x = {1}/({8{\rm e}^2})\hspace{0.05cm},$$

$$I_4 ={1}/{4} \cdot \int_{-\infty}^{0} {\rm e}^{y} \int_{2-y}^{\infty}{\rm e}^{-x} \,{\rm d} x \,{\rm d} y = ... = {1}/({8{\rm e}^2}) = I_3\hspace{0.05cm}.$$

Thus, the overall result is:

$${\rm Pr} \left [ n_1 + n_2 > 2 \right ] = {\rm e}^{-2} \cdot ({1}/{4} +{1}/{2} +{1}/{8} +{1}/{8})= {\rm e}^{-2} \hspace{0.1cm}\hspace{0.15cm}\underline {\approx 13.5\%}\hspace{0.05cm}.$$