Exercise 4.09Z: Laplace Distributed Noise

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:

The exercise belongs to the chapter "Approximation of the Error Probability".

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

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

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

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

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

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

$\ \%$

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

$\ \%$

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

$\ \%$

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:

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