Aufgaben:Exercise 3.11: Chebyshev's Inequality

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Exemplary Chebyshev's bound

Values of the  "complementary Gaussian error function"

If nothing else is known about a random variable  $x$  than only

• the mean value  $m_x$  and
• the root mean square  $\sigma_x$,

so the  "Chebyshev's Inequality"  gives an upper bound on the probability that $x$  deviates by more than a value  $\varepsilon$  from its mean.  This bound is:

$${\rm Pr}(|x-m_x|\ge \varepsilon) \le {\sigma_x^{\rm 2}}/{\varepsilon^{\rm 2}}.$$

To explain:

• In the graph,  this upper bound is drawn in red.
• The green curve shows the actual probability for the uniform distribution.
• The blue points are for the exponential distribution.

From this plot it can be seen that the  "Chebyshev's Inequality"  is only a very rough bound. 
It should be used only if really only the mean and the rms are known from the random size.

Hints:

• The exercise belongs to the chapter  "Further Distributions".
• In particular,  reference is made to the section  "Chebyshev's inequality" .

• On the right,  values of the complementary Gaussian error function  ${\rm Q}(x)$  are given.

Questions

1

Which of the following statements are true?

	Conceivably,  a random variable with  ${\rm Pr}(|x -m_x | \ge 3\sigma_x) = 1/4$.
	"Chebyshev"  yields for  $\varepsilon < \sigma_x$  no information.
	${\rm Pr}(|x -m_x | \ge \sigma_x)$  is identically zero for large  $\varepsilon$  if  $x$  is bounded.

2

It holds  $k = 1, \ 2, \ 3, \ 4$.  Give the excess probability  $p_k = {\rm Pr}(|x -m_x | \ge k \cdot \sigma_x)$  for the Gaussian distribution.  How large is  $p_3$?

${\rm Pr}(|x -m_x | \ge 3 \sigma_x) \ = \ $ $\ \%$

3

What are the excess probabilities  $p_k$  for the  exponential distribution.  Here   $m_x = \sigma_x = 1/\lambda$.  What is  $p_3$?

${\rm Pr}(|x -m_x | \ge 3 \sigma_x) \ = \ $ $\ \%$

Solution

Solution

(1)  Correct are the proposed solutions 2 and 3:

• The first statement is false. Chebyshev's inequality provides the bound here  $1/9$.
• For no distribution can the probability considered here be equal  $1/4$ .
• For  $\varepsilon < \sigma_x$  Chebyshev yields a probability greater  $1$.  This information is useless.
• The last statement is true.  For example, in the uniform distribution:

$${\rm Pr}(| x- m_x | \ge \varepsilon)=\left\{ \begin{array}{*{4}{c}} 1-{\varepsilon}/{\varepsilon_{\rm 0}} & \rm f\ddot{u}r\hspace{0.1cm}{\it \varepsilon<\varepsilon_{\rm 0}=\sqrt{\rm 3}\cdot\sigma_x},\rm 0 & \rm else. \end{array} \right. $$

(2)  For the Gaussian distribution holds:

$$p_k={\rm Pr}(| x-m_x| \ge k\cdot\sigma_{x})=\rm 2\cdot \rm Q(\it k).$$

• This results in the following numerical values  (in brackets:   bound according to Chebyshev):

$$k= 1\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge \sigma_{x}) = 31.7 \% \hspace{0.3cm}(100 \%),$$

$$k= 2\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge 2 \cdot \sigma_{x}) = 4.54 \% \hspace{0.3cm}(25 \%),$$

$$k= 3\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge 3 \cdot\sigma_{x})\hspace{0.15cm}\underline{ = 0.26 \%} \hspace{0.3cm}(11.1 \%),$$

$$k= 4\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge 4 \cdot \sigma_{x}) = 0.0064 \% \hspace{0.3cm}(6.25 \%).$$

(3)  Without restricting generality, we set  $\lambda = 1$   ⇒   $m_x = \sigma_x = 1$.  Then holds:

$${\rm Pr}(|x - m_x| \ge k\cdot\sigma_{x}) = {\rm Pr}(| x-1| \ge k).$$

• Since in this special case the random variable is always  $x >0$ , it further holds:

$$p_k= {\rm Pr}( x \ge k+1)=\int_{k+\rm 1}^{\infty}\hspace{-0.15cm} {\rm e}^{-x}\, {\rm d} x={\rm e}^{-( k + 1)}.$$

• This yields the following numerical values for the exponential distribution:

$$k= 1\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge \sigma_{x}) \rm e^{-2}= \rm 13.53\%,$$

$$k= 2\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge 2 \cdot \sigma_{x})= \rm \rm e^{-3}=\rm 4.97\% ,$$

$$k= 3\text\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge 3 \cdot\sigma_{x})= \rm \rm e^{-4}\hspace{0.15cm}\underline{ =\rm 1.83\% },$$

$$k= 4\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge 4 \cdot \sigma_{x}) = \rm e^{-5}= \rm 0.67\%.$$