# Exercise 4.09Z: Laplace Distributed Noise

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Two-dimensional 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 function  $\rm (PDF)$:

$$p_{n_1}(x) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} K \cdot {\rm e}^{- a \hspace{0.03cm}\cdot \hspace{0.03cm} |x|} \hspace{0.05cm},$$
$$p_{n_2}(y) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} K \cdot {\rm e}^{- a \hspace{0.03cm}\cdot \hspace{0.03cm} |y|} \hspace{0.05cm}.$$
• The two-dimensional 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:

• We would like to refer to the  (German language)  interactive SWF 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 one-dimensional PDF?

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

2

Let  $a = 1$.  What are the mean  ${\rm E}\big[n_i \big]$  and the variance  $\sigma^2 = {\rm E}\big[n_i^2\big]$  of the two one-dimensional 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 two-dimensional 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  $n_1 + n_2 > 2$?

 ${\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 second moment,  as already stated in the question:
Contour lines of the two-dimensional Laplace PDF
$$\sigma^2 = {\rm E}[n_1^2] = 2 \cdot \frac{a}{2} \cdot \int_{0}^{\infty} x^2 \cdot {\rm e}^{-a \hspace{0.03cm} \cdot \hspace{0.03cm} x} \,{\rm d} x = a \cdot {2}/{a^3}= {2}/{a^2} \hspace{0.05cm}. \hspace{0.2cm}{\rm With}\hspace{0.15cm}a = 1\text{:} \hspace{0.2cm}\hspace{0.1cm}\underline {\sigma^2 = 2 }\hspace{0.05cm}.$$

(3)  Solution 1  is correct:

• In the first quadrant  $(x ≥ 0, y ≥ 0)$,  the magnitude formation can be omitted.  Then the two-dimensional 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:
$${\rm Pr} \left [ n_1 + n_2 > 2 \right ] = {\rm e}^{-2} \cdot ({1}/{4} +{1}/{2} +{1}/{8} +{1}/{8})= {\rm e}^{-2} \hspace{0.1cm}\hspace{0.15cm}\underline {\approx 13.5\%}\hspace{0.05cm}.$$