Difference between revisions of "Aufgaben:Exercise 4.1Z: Calculation of Moments"

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{{quiz-Header|Buchseite=Informationstheorie/Differentielle Entropie
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{{quiz-Header|Buchseite=Information_Theory/Differential_Entropy
 
}}
 
}}
  
[[File:EN_Inf_Z_4_1_vers2.png|right|frame|Exponential distribution (top) and Laplace distribution (bottom)]]
+
[[File:EN_Inf_Z_4_1_vers2.png|right|frame|Exponential PDF (top), <br>Laplace PDF (bottom)]]
The graph above shows the probability density function (PDF) of the&nbsp; [[Theory_of_Stochastic_Signals/Exponentialverteilte_Zufallsgrößen|exponential distribution]]:
+
The upper graph shows the probability density function&nbsp; $\rm (PDF)$&nbsp; of the&nbsp; [[Theory_of_Stochastic_Signals/Exponentialverteilte_Zufallsgrößen|exponential distribution]]:
 
:$$f_X(x) = \left\{ \begin{array}{c} A_{ X} \cdot {\rm e}^{-\lambda \hspace{0.05cm} \cdot \hspace{0.05cm}x} \\ A_{ X}/2 \\ 0 \\  \end{array} \right. \begin{array}{*{20}c}  {\rm{f\ddot{u}r}} \hspace{0.1cm}x>0, \\ {\rm{f\ddot{u}r}} \hspace{0.1cm}x=0,  \\    {\rm{f\ddot{u}r}} \hspace{0.1cm}x<0. \\ \end{array}$$
 
:$$f_X(x) = \left\{ \begin{array}{c} A_{ X} \cdot {\rm e}^{-\lambda \hspace{0.05cm} \cdot \hspace{0.05cm}x} \\ A_{ X}/2 \\ 0 \\  \end{array} \right. \begin{array}{*{20}c}  {\rm{f\ddot{u}r}} \hspace{0.1cm}x>0, \\ {\rm{f\ddot{u}r}} \hspace{0.1cm}x=0,  \\    {\rm{f\ddot{u}r}} \hspace{0.1cm}x<0. \\ \end{array}$$
Drawn below is the WDF of the&nbsp; [[Theory_of_Stochastic_Signals/Exponentialverteilte_Zufallsgrößen#Zweiseitige_Exponentialverteilung_.E2.80.93_Laplaceverteilung|Laplace distribution]], which can be specified for all&nbsp; $y$&ndash;values as follows:
+
Drawn below is the PDF of the&nbsp; [[Theory_of_Stochastic_Signals/Exponentialverteilte_Zufallsgrößen#Zweiseitige_Exponentialverteilung_.E2.80.93_Laplaceverteilung|Laplace distribution]], which can be specified for all&nbsp; $y$&ndash;values as follows:
 
:$$f_Y(y) =  A_{ Y} \cdot {\rm e}^{-\lambda \hspace{0.05cm} \cdot \hspace{0.05cm} |\hspace{0.03cm}y\hspace{0.03cm}|}\hspace{0.05cm}.$$
 
:$$f_Y(y) =  A_{ Y} \cdot {\rm e}^{-\lambda \hspace{0.05cm} \cdot \hspace{0.05cm} |\hspace{0.03cm}y\hspace{0.03cm}|}\hspace{0.05cm}.$$
  
The two continuous value random variables&nbsp; $X$&nbsp; and&nbsp; $Y$&nbsp; are to be compared with respect to the following characteristics:
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The two continuous random variables&nbsp; $X$&nbsp; and&nbsp; $Y$&nbsp; are to be compared with respect to the following characteristics:
*the linear mean&nbsp; $m_1$&nbsp; (first order moment),
+
*The linear mean&nbsp; $m_1$&nbsp; (first order moment),
 
*the second order moment &nbsp; &#8658; &nbsp; $m_2$,
 
*the second order moment &nbsp; &#8658; &nbsp; $m_2$,
 
*the variance&nbsp; $\sigma^2 = m_2 - m_1^2$ &nbsp; &#8658; &nbsp; Steiner's theorem,  
 
*the variance&nbsp; $\sigma^2 = m_2 - m_1^2$ &nbsp; &#8658; &nbsp; Steiner's theorem,  
*the dispersion&nbsp; $\sigma$.
+
*the standard deviation&nbsp; $\sigma$.
  
  
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Hints:
 
Hints:
 
*The task belongs to the chapter&nbsp; [[Information_Theory/Differentielle_Entropie|Differential Entropy]].
 
*The task belongs to the chapter&nbsp; [[Information_Theory/Differentielle_Entropie|Differential Entropy]].
*Useful hints for solving this task and further information on continuous-valued random variables can be found in the third chapter  "Continuous Random Variables" of the book&nbsp;  [[Theory of Stochastic Signals]].
+
*Useful hints for solving this task and further information on continuous random variables can be found in the third chapter  "Continuous Random Variables" of the book&nbsp;  [[Theory of Stochastic Signals]].
 
   
 
   
 
*Also given are the two indefinite integrals:
 
*Also given are the two indefinite integrals:
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|type="[]"}
 
|type="[]"}
 
+ The linear mean is&nbsp; $m_1 = 1/\lambda$.
 
+ The linear mean is&nbsp; $m_1 = 1/\lambda$.
+ The root mean square is&nbsp; $m_2 = 2/\lambda^2$.
+
+ The second order moment is&nbsp; $m_2 = 2/\lambda^2$.
 
+ The variance is&nbsp; $\sigma^2 = 1/\lambda^2$.
 
+ The variance is&nbsp; $\sigma^2 = 1/\lambda^2$.
  
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|type="[]"}
 
|type="[]"}
 
- The linear mean is&nbsp; $m_1 = 1/\lambda$.
 
- The linear mean is&nbsp; $m_1 = 1/\lambda$.
+ The root mean square is&nbsp; $m_2 = 2/\lambda^2$.
+
+ The second order moment is&nbsp; $m_2 = 2/\lambda^2$.
 
- The variance is&nbsp; $\sigma^2 = 1/\lambda^2$.
 
- The variance is&nbsp; $\sigma^2 = 1/\lambda^2$.
  
{With what probabilities does the random variable&nbsp; $(X$&nbsp; or &nbsp; $Y)$&nbsp; differ from the respective mean in amount by more than the dispersion &nbsp;$\sigma$?
+
{With what probabilities does the random variable&nbsp; $(X$&nbsp; or &nbsp; $Y)$&nbsp; differ from the respective mean in magnitude by more than the dispersion &nbsp;$\sigma$?
 
|type="{}"}
 
|type="{}"}
 
$\text{Exponential:}\; \;{\rm Pr}( |X  -  m_X| > \sigma_X) \ = \ $  { 0.135 3% }
 
$\text{Exponential:}\; \;{\rm Pr}( |X  -  m_X| > \sigma_X) \ = \ $  { 0.135 3% }
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:$$A_{X} \cdot\int_{0}^{\infty} \hspace{-0.01cm}  {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\hspace{0.1cm}{\rm d}x = A_{X} \cdot (-1/\lambda)\cdot\big [{\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\big ]_{0}^{\infty} = A_{X} \cdot (1/\lambda) \stackrel{!}{=} 1
 
:$$A_{X} \cdot\int_{0}^{\infty} \hspace{-0.01cm}  {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\hspace{0.1cm}{\rm d}x = A_{X} \cdot (-1/\lambda)\cdot\big [{\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\big ]_{0}^{\infty} = A_{X} \cdot (1/\lambda) \stackrel{!}{=} 1
 
  \hspace{0.3cm} \Rightarrow\hspace{0.3cm} A_{X} = \lambda \hspace{0.05cm}. $$
 
  \hspace{0.3cm} \Rightarrow\hspace{0.3cm} A_{X} = \lambda \hspace{0.05cm}. $$
 
  
  
 
'''(2)'''&nbsp; <u>Proposed solution 1</u> is correct:
 
'''(2)'''&nbsp; <u>Proposed solution 1</u> is correct:
*From the graph on the information page, we can see that the heighz&nbsp; $A_Y$&nbsp; of the Laplace distribution is only half as large as the maximum of the exponential distribution:
+
*From the graph on the information page, we can see that the height&nbsp; $A_Y$&nbsp; of the Laplace PDF is only half as large as the maximum of the exponential PDF:
 
:$$A_Y = \lambda/2.$$  
 
:$$A_Y = \lambda/2.$$  
  
  
 
+
'''(3)'''&nbsp; Correct is <u>YES</u>,&nbsp; although for&nbsp; $z \ne 0$&nbsp; always &nbsp;$f_X(z) = f_Y(z)$. &nbsp;  Let us now consider the special case&nbsp; $z= 0$:
'''(3)'''&nbsp; Correct is <u>YES</u>, although for&nbsp; $z \ne 0$&nbsp; always &nbsp;$f_X(z) = f_Y(z)$&nbsp;. Let us now consider the special case&nbsp; $z= 0$:
+
* For the Laplace PDF:&nbsp; $f_Y(y = 0) = \lambda/2$.
* For the Laplace distribution,&nbsp; $f_Y(y = 0) = \lambda/2$.
+
* For the exponential PDF,&nbsp; the left-hand and right-hand limits differ for&nbsp; $x \to 0$.  
* For the exponential distribution, the left-hand and right-hand limits differ for&nbsp; $x \to 0$.  
+
*The PDF value at point&nbsp; $x= 0$&nbsp; is the average of these two limits:
*The PDF&ndash;value at the point&nbsp; $x= 0$&nbsp; is the average of these two limits:
 
 
:$$f_X(0) = \frac{1}{2} \cdot \big [ 0 + \lambda \big] = \lambda/2 =  f_Y(0)\hspace{0.05cm}.$$
 
:$$f_X(0) = \frac{1}{2} \cdot \big [ 0 + \lambda \big] = \lambda/2 =  f_Y(0)\hspace{0.05cm}.$$
 
  
  
 
'''(4)'''&nbsp; <u>All proposed solutions</u> are correct.&nbsp;  
 
'''(4)'''&nbsp; <u>All proposed solutions</u> are correct.&nbsp;  
  
For the exponential distribution, the&nbsp; $k$th order moment is generally calculated to be
+
For the exponential distribution, the&nbsp; $k$&ndash;th order moment is generally calculated to be
 
:$$m_k = \frac{k!}{\lambda^k} \hspace{0.3cm}\Rightarrow\hspace{0.3cm} m_1 = \frac{1}{\lambda}, \hspace{0.3cm} m_2 = \frac{2}{\lambda^2}, \hspace{0.3cm} m_3 = \frac{6}{\lambda^3}, \ \text{...}$$
 
:$$m_k = \frac{k!}{\lambda^k} \hspace{0.3cm}\Rightarrow\hspace{0.3cm} m_1 = \frac{1}{\lambda}, \hspace{0.3cm} m_2 = \frac{2}{\lambda^2}, \hspace{0.3cm} m_3 = \frac{6}{\lambda^3}, \ \text{...}$$
 
Thus one obtains for
 
Thus one obtains for
 
* the linear mean (first order moment):
 
* the linear mean (first order moment):
 
:$$m_1 = \lambda \cdot\int_{0}^{\infty} \hspace{-0.01cm} x \cdot  {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\hspace{0.1cm}{\rm d}x = \lambda \cdot \left [\frac{{\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}}{(-\lambda)^2}\cdot(-\lambda \cdot x-1)\right ]_{0}^{\infty}= {1}/{\lambda} \hspace{0.05cm},$$
 
:$$m_1 = \lambda \cdot\int_{0}^{\infty} \hspace{-0.01cm} x \cdot  {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\hspace{0.1cm}{\rm d}x = \lambda \cdot \left [\frac{{\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}}{(-\lambda)^2}\cdot(-\lambda \cdot x-1)\right ]_{0}^{\infty}= {1}/{\lambda} \hspace{0.05cm},$$
* the root mean square (second order moment):
+
* the second order moment:
 
:$$m_2 = \lambda \cdot\int_{0}^{\infty} \hspace{-0.01cm} x^2 \cdot  {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\hspace{0.1cm}{\rm d}x = \lambda \cdot\left [ {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\cdot
 
:$$m_2 = \lambda \cdot\int_{0}^{\infty} \hspace{-0.01cm} x^2 \cdot  {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\hspace{0.1cm}{\rm d}x = \lambda \cdot\left [ {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\cdot
 
(\frac{x^2}{-\lambda} - \frac{2x}{\lambda^2} + \frac{2}{\lambda^3})
 
(\frac{x^2}{-\lambda} - \frac{2x}{\lambda^2} + \frac{2}{\lambda^3})
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+
[[File:EN_Inf_A_4_3.png|right|frame|To illustrate the sample solution to problem&nbsp; '''(5)''']]
 
'''(5)'''&nbsp; Only the <u>proposed solution 2</u> is correct:  
 
'''(5)'''&nbsp; Only the <u>proposed solution 2</u> is correct:  
*The quadratic mean of the Laplace distribution is exactly the same as for the exponential distribution because of the symmetric PDF:
+
*The second moment of&nbsp; "Laplace"&nbsp; is the same as for the exponential distribution because of the symmetric PDF:
[[File:EN_Inf_A_4_3.png|right|frame|To illustrate the sample solution to problem&nbsp; '''(5)''']]
 
 
:$$m_2 = \frac{\lambda}{2} \cdot \int_{-\infty}^{\infty} \hspace{-0.01cm} y^2 \cdot  {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}|y|}\hspace{0.1cm}{\rm d}y = \lambda \cdot\int_{0}^{\infty} \hspace{-0.01cm} y^2 \cdot  {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}y}\hspace{0.1cm}{\rm d}y = {2}/{\lambda^2} \hspace{0.05cm}.$$
 
:$$m_2 = \frac{\lambda}{2} \cdot \int_{-\infty}^{\infty} \hspace{-0.01cm} y^2 \cdot  {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}|y|}\hspace{0.1cm}{\rm d}y = \lambda \cdot\int_{0}^{\infty} \hspace{-0.01cm} y^2 \cdot  {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}y}\hspace{0.1cm}{\rm d}y = {2}/{\lambda^2} \hspace{0.05cm}.$$
  
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\hspace{0.3cm} \Rightarrow\hspace{0.3cm}  
 
\hspace{0.3cm} \Rightarrow\hspace{0.3cm}  
 
\sigma = {\sqrt{2}}/{\lambda}\hspace{0.05cm}.$$
 
\sigma = {\sqrt{2}}/{\lambda}\hspace{0.05cm}.$$
 
 
  
  
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\right ]_{2/\lambda}^{\infty}
 
\right ]_{2/\lambda}^{\infty}
 
   =  {\rm e}^{-2} \hspace{0.15cm}\underline {\approx 0.135}.$$
 
   =  {\rm e}^{-2} \hspace{0.15cm}\underline {\approx 0.135}.$$
For the Laplace distribution (lower graph), with &nbsp;$m_Y = 0$&nbsp; and &nbsp;$\sigma_Y = \sqrt{2}/\lambda$ we obtain::
+
For the Laplace distribution (lower graph),&nbsp; with &nbsp;$m_Y = 0$&nbsp; and &nbsp;$\sigma_Y = \sqrt{2}/\lambda$ we obtain::
 
:$${\rm Pr}( |Y  -  m_Y| > \sigma_Y) =
 
:$${\rm Pr}( |Y  -  m_Y| > \sigma_Y) =
 
2 \cdot {\rm Pr}( Y > \sqrt{2}/\lambda)  
 
2 \cdot {\rm Pr}( Y > \sqrt{2}/\lambda)  
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  = -  {\rm e}^{-\sqrt{2}} \hspace{0.15cm}\underline {\approx 0.243}\hspace{0.05cm}.$$
 
  = -  {\rm e}^{-\sqrt{2}} \hspace{0.15cm}\underline {\approx 0.243}\hspace{0.05cm}.$$
  
A comparison of the shaded areas in the accompanying graph qualitatively confirms the result: <br> &nbsp; &nbsp;  The blue areas together are slightly larger than the red area.
+
A comparison of the shaded areas in the accompanying graph qualitatively confirms the result: <br> &rArr; &nbsp;  The blue areas together are slightly larger than the red area.
  
  

Latest revision as of 13:19, 18 January 2023

Exponential PDF (top),
Laplace PDF (bottom)

The upper graph shows the probability density function  $\rm (PDF)$  of the  exponential distribution:

$$f_X(x) = \left\{ \begin{array}{c} A_{ X} \cdot {\rm e}^{-\lambda \hspace{0.05cm} \cdot \hspace{0.05cm}x} \\ A_{ X}/2 \\ 0 \\ \end{array} \right. \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \hspace{0.1cm}x>0, \\ {\rm{f\ddot{u}r}} \hspace{0.1cm}x=0, \\ {\rm{f\ddot{u}r}} \hspace{0.1cm}x<0. \\ \end{array}$$

Drawn below is the PDF of the  Laplace distribution, which can be specified for all  $y$–values as follows:

$$f_Y(y) = A_{ Y} \cdot {\rm e}^{-\lambda \hspace{0.05cm} \cdot \hspace{0.05cm} |\hspace{0.03cm}y\hspace{0.03cm}|}\hspace{0.05cm}.$$

The two continuous random variables  $X$  and  $Y$  are to be compared with respect to the following characteristics:

  • The linear mean  $m_1$  (first order moment),
  • the second order moment   ⇒   $m_2$,
  • the variance  $\sigma^2 = m_2 - m_1^2$   ⇒   Steiner's theorem,
  • the standard deviation  $\sigma$.





Hints:

  • The task belongs to the chapter  Differential Entropy.
  • Useful hints for solving this task and further information on continuous random variables can be found in the third chapter "Continuous Random Variables" of the book  Theory of Stochastic Signals.
  • Also given are the two indefinite integrals:
$$\int \hspace{-0.01cm} x \cdot {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\hspace{0.1cm}{\rm d}x = \frac{{\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}}{(-\lambda)^2}\cdot(-\lambda \cdot x-1)\hspace{0.05cm}, $$
$$\int \hspace{-0.01cm} x^2 \cdot {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\hspace{0.1cm}{\rm d}x = {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\cdot (\frac{x^2}{-\lambda} - \frac{2x}{\lambda^2} + \frac{2}{\lambda^3}) \hspace{0.05cm}. $$


Questions

1

What is the maximum value  $A_X$  of the PDF  $f_X(x)$?

$A_X = \lambda/2$,
$A_X = \lambda$,
$A_X = 1/\lambda$.

2

What is the maximum value  $A_Y$  of the PDF  $f_Y(y)$?

$A_Y = \lambda/2$,
$A_Y = \lambda$,
$A_Y = 1/\lambda$.

3

Is there an argument  $z$, such that  $f_X(z) = f_Y(z)$ ?

Yes.
No.

4

Which statements are true about the characteristics of the exponential distribution?

The linear mean is  $m_1 = 1/\lambda$.
The second order moment is  $m_2 = 2/\lambda^2$.
The variance is  $\sigma^2 = 1/\lambda^2$.

5

Which statements are true about the characteristics of the Laplace distribution?

The linear mean is  $m_1 = 1/\lambda$.
The second order moment is  $m_2 = 2/\lambda^2$.
The variance is  $\sigma^2 = 1/\lambda^2$.

6

With what probabilities does the random variable  $(X$  or   $Y)$  differ from the respective mean in magnitude by more than the dispersion  $\sigma$?

$\text{Exponential:}\; \;{\rm Pr}( |X - m_X| > \sigma_X) \ = \ $

$\text{Laplace:}\; \;{\rm Pr}( |Y - m_Y| > \sigma_Y) \ = \ $


Solution

(1)  Proposed solution 2 is correct:

  • The area under the PDF must always be  $1$ .  It follows for the exponential distribution:
$$A_{X} \cdot\int_{0}^{\infty} \hspace{-0.01cm} {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\hspace{0.1cm}{\rm d}x = A_{X} \cdot (-1/\lambda)\cdot\big [{\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\big ]_{0}^{\infty} = A_{X} \cdot (1/\lambda) \stackrel{!}{=} 1 \hspace{0.3cm} \Rightarrow\hspace{0.3cm} A_{X} = \lambda \hspace{0.05cm}. $$


(2)  Proposed solution 1 is correct:

  • From the graph on the information page, we can see that the height  $A_Y$  of the Laplace PDF is only half as large as the maximum of the exponential PDF:
$$A_Y = \lambda/2.$$


(3)  Correct is YES,  although for  $z \ne 0$  always  $f_X(z) = f_Y(z)$.   Let us now consider the special case  $z= 0$:

  • For the Laplace PDF:  $f_Y(y = 0) = \lambda/2$.
  • For the exponential PDF,  the left-hand and right-hand limits differ for  $x \to 0$.
  • The PDF value at point  $x= 0$  is the average of these two limits:
$$f_X(0) = \frac{1}{2} \cdot \big [ 0 + \lambda \big] = \lambda/2 = f_Y(0)\hspace{0.05cm}.$$


(4)  All proposed solutions are correct. 

For the exponential distribution, the  $k$–th order moment is generally calculated to be

$$m_k = \frac{k!}{\lambda^k} \hspace{0.3cm}\Rightarrow\hspace{0.3cm} m_1 = \frac{1}{\lambda}, \hspace{0.3cm} m_2 = \frac{2}{\lambda^2}, \hspace{0.3cm} m_3 = \frac{6}{\lambda^3}, \ \text{...}$$

Thus one obtains for

  • the linear mean (first order moment):
$$m_1 = \lambda \cdot\int_{0}^{\infty} \hspace{-0.01cm} x \cdot {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\hspace{0.1cm}{\rm d}x = \lambda \cdot \left [\frac{{\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}}{(-\lambda)^2}\cdot(-\lambda \cdot x-1)\right ]_{0}^{\infty}= {1}/{\lambda} \hspace{0.05cm},$$
  • the second order moment:
$$m_2 = \lambda \cdot\int_{0}^{\infty} \hspace{-0.01cm} x^2 \cdot {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\hspace{0.1cm}{\rm d}x = \lambda \cdot\left [ {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\cdot (\frac{x^2}{-\lambda} - \frac{2x}{\lambda^2} + \frac{2}{\lambda^3}) \right ]_{0}^{\infty} ={2}/{\lambda^2} \hspace{0.05cm}.$$

From this, using Steiner's theorem for the variance of the exponential distribution, we get:

$$\sigma^2 = m_2 - m_1^2 = {2}/{\lambda^2} -{1}/{\lambda^2} = {1}/{\lambda^2} \hspace{0.3cm} \Rightarrow\hspace{0.3cm} \sigma = {1}/{\lambda}\hspace{0.05cm}.$$


To illustrate the sample solution to problem  (5)

(5)  Only the proposed solution 2 is correct:

  • The second moment of  "Laplace"  is the same as for the exponential distribution because of the symmetric PDF:
$$m_2 = \frac{\lambda}{2} \cdot \int_{-\infty}^{\infty} \hspace{-0.01cm} y^2 \cdot {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}|y|}\hspace{0.1cm}{\rm d}y = \lambda \cdot\int_{0}^{\infty} \hspace{-0.01cm} y^2 \cdot {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}y}\hspace{0.1cm}{\rm d}y = {2}/{\lambda^2} \hspace{0.05cm}.$$
  • In contrast, the mean of the Laplace distribution is  $m_1 = 0$.
  • Thus, the variance of the Laplace distribution is twice that of the exponential distribution:
$$\sigma^2 = m_2 - m_1^2 = {2}/{\lambda^2} - 0 ={2}/{\lambda^2} \hspace{0.3cm} \Rightarrow\hspace{0.3cm} \sigma = {\sqrt{2}}/{\lambda}\hspace{0.05cm}.$$


(6)  For the exponential distribution, according to the upper graph with  $m_X = \sigma_X = 1/\lambda$:

$${\rm Pr}( |X - m_X| > \sigma_X) \hspace{-0.05cm} = \hspace{-0.05cm} {\rm Pr}( X > 2/\lambda) \hspace{-0.05cm} = \hspace{-0.05cm} \lambda \cdot\int_{2/\lambda}^{\infty} \hspace{-0.08cm} {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\hspace{0.1cm}{\rm d}x = -\left [ {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x} \right ]_{2/\lambda}^{\infty} = {\rm e}^{-2} \hspace{0.15cm}\underline {\approx 0.135}.$$

For the Laplace distribution (lower graph),  with  $m_Y = 0$  and  $\sigma_Y = \sqrt{2}/\lambda$ we obtain::

$${\rm Pr}( |Y - m_Y| > \sigma_Y) = 2 \cdot {\rm Pr}( Y > \sqrt{2}/\lambda) = 2 \cdot \frac{\lambda}{2} \cdot\int_{\sqrt{2}/\lambda}^{\infty} \hspace{-0.01cm} {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\hspace{0.1cm}{\rm d}x $$
$$\Rightarrow \hspace{0.3cm}{\rm Pr}( |Y - m_Y| > \sigma_Y) = \left [ {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x} \right ]_{\sqrt{2}/\lambda}^{\infty} = - {\rm e}^{-\sqrt{2}} \hspace{0.15cm}\underline {\approx 0.243}\hspace{0.05cm}.$$

A comparison of the shaded areas in the accompanying graph qualitatively confirms the result:
⇒   The blue areas together are slightly larger than the red area.