Difference between revisions of "Aufgaben:Exercise 3.11: Chebyshev's Inequality"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Further_Distributions |
}} | }} | ||
− | [[File:EN_Sto_A_3_11.png|right|frame| | + | [[File:EN_Sto_A_3_11.png|right|frame|Exemplary Chebyshevsch barrier]] |
− | + | If nothing else is known about a random size $x$ than only | |
− | * | + | *the mean value $m_x$ and |
− | * | + | *the rms $\sigma_x$, |
− | so | + | so the <i>Chebyshev's Inequality</i> an upper bound on the probability that $x$ deviates by more than one value from its mean value $\varepsilon$ is given. |
− | + | This bound is: | |
− | :$${\rm Pr}(|x-m_x|\ge \varepsilon) \le | + | :$${\rm Pr}(|x-m_x|\ge \varepsilon) \le {\sigma_x^{\rm 2}}/{\varepsilon^{\rm 2}}.$$ |
− | + | To explain: | |
− | *In | + | *In the graph, this upper bound is drawn in red. |
− | * | + | *The larger curve shows the actual probability in the uniform distribution. |
− | * | + | *The blue points are for the exponential distribution. |
− | + | From this plot it can be seen that the <i>Chebyshev's Inequality</i> is only a very rough bound. <br>It should be used only if really only the mean and the rms are known from the random size. | |
− | [[File:P_ID921__Sto_A_3_11_b.png|frame| | + | [[File:P_ID921__Sto_A_3_11_b.png|frame|Values of the complementary Gaussian error function]] |
<br> | <br> | ||
− | + | Hints: | |
− | * | + | *The exercise belongs to the chapter [[Theory_of_Stochastic_Signals/Further_Distributions|Further Distributions]]. |
− | * | + | *In particular, reference is made to the page [[Theory_of_Stochastic_Signals/Further_Distributions#Chebyshev.27s_inequality|Chebyshev's inequality]] . |
− | * | + | *On the right, values of the complementary Gaussian error function ${\rm Q}(x)$ are given. |
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− | === | + | ===Questions=== |
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {Which of the following statements are true? |
|type="[]"} | |type="[]"} | ||
− | - | + | - Conceivably, a random variable with ${\rm Pr}(|x -m_x> | \ge 3\sigma_x) = 1/4$. |
− | + " | + | + "Chebyshev" yields für $\varepsilon < \sigma_x$ no information. |
− | + ${\rm Pr}(|x -m_x> | \ge \sigma_x)$ | + | + ${\rm Pr}(|x -m_x> | \ge \sigma_x)$ is identically zero for large $\varepsilon$ if $x$ is bounded. |
− | { | + | {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 <u>Gaussian distribution</u>. How large is $p_3$? |
|type="{}"} | |type="{}"} | ||
${\rm Pr}(|x -m_x | \ge 3 \sigma_x) \ = \ $ { 0.26 3% } $\ \%$ | ${\rm Pr}(|x -m_x | \ge 3 \sigma_x) \ = \ $ { 0.26 3% } $\ \%$ | ||
− | { | + | {What are the excess probabilities $p_k$ for the <u>exponential distribution</u>. Here $m_x = \sigma_x = 1/\lambda$. What is $p_3$? |
|type="{}"} | |type="{}"} | ||
${\rm Pr}(|x -m_x | \ge 3 \sigma_x) \ = \ $ { 1.83 3% } $\ \%$ | ${\rm Pr}(|x -m_x | \ge 3 \sigma_x) \ = \ $ { 1.83 3% } $\ \%$ | ||
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</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' Correct are <u>the proposed solutions 2 and 3</u>: |
− | * | + | *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}, | + | :$${\rm Pr}(| x- m_x | \ge \varepsilon)=\left\{ \begin{array}{*{4}{c}} 1-{\varepsilon}/{\varepsilon_{\rm 0}} & \rm f\ddot{u}r\hspace{0.1cm}{\it \varepsilon<\varepsilon_{\rm 0}=\sqrt{\rm 3}\cdot\sigma_x},\rm 0 & \rm else. \end{array} \right. $$ |
− | '''(2)''' | + | '''(2)''' For the Gaussian distribution holds: |
:$$p_k={\rm Pr}(| x-m_x| \ge k\cdot\sigma_{x})=\rm 2\cdot \rm Q(\it k).$$ | :$$p_k={\rm Pr}(| x-m_x| \ge k\cdot\sigma_{x})=\rm 2\cdot \rm Q(\it k).$$ | ||
− | * | + | *This results in the following numerical values (in brackets: bound according to Chebyshev): |
:$$k= 1\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge \sigma_{x}) = 31.7 \% \hspace{0.3cm}(100 \%),$$ | :$$k= 1\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge \sigma_{x}) = 31.7 \% \hspace{0.3cm}(100 \%),$$ | ||
:$$k= 2\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge 2 \cdot \sigma_{x}) = 4.54 \% \hspace{0.3cm}(25 \%),$$ | :$$k= 2\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge 2 \cdot \sigma_{x}) = 4.54 \% \hspace{0.3cm}(25 \%),$$ | ||
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:$$k= 4\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge 4 \cdot \sigma_{x}) = 0.0064 \% \hspace{0.3cm}(6.25 \%).$$ | :$$k= 4\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge 4 \cdot \sigma_{x}) = 0.0064 \% \hspace{0.3cm}(6.25 \%).$$ | ||
− | '''(3)''' | + | '''(3)''' Without restricting generality, we set $\lambda = 1$ |
− | ⇒ $m_x = \sigma_x = 1$. | + | ⇒ $m_x = \sigma_x = 1$. Then holds: |
− | :$${\rm Pr}(|x - m_x| \ge | + | :$${\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} | :$$p_k= {\rm Pr}( x \ge k+1)=\int_{k+\rm 1}^{\infty}\hspace{-0.15cm} | ||
{\rm e}^{-x}\, {\rm d} x={\rm e}^{-( k + 1)}.$$ | {\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}) | + | :$$k= 1\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge \sigma_{x}) \rm e^{-2}= \rm 13.53\%,$$ |
:$$k= 2\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge 2 \cdot \sigma_{x})= \rm \rm e^{-3}=\rm 4.97\% ,$$ | :$$k= 2\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge 2 \cdot \sigma_{x})= \rm \rm e^{-3}=\rm 4.97\% ,$$ | ||
:$$k= 3\text\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge 3 \cdot\sigma_{x})= \rm \rm e^{-4}\hspace{0.15cm}\underline{ =\rm 1.83\% },$$ | :$$k= 3\text\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge 3 \cdot\sigma_{x})= \rm \rm e^{-4}\hspace{0.15cm}\underline{ =\rm 1.83\% },$$ |
Revision as of 22:41, 12 January 2022
If nothing else is known about a random size $x$ than only
- the mean value $m_x$ and
- the rms $\sigma_x$,
so the Chebyshev's Inequality an upper bound on the probability that $x$ deviates by more than one value from its mean value $\varepsilon$ is given.
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 larger curve shows the actual probability in 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 page Chebyshev's inequality .
- On the right, values of the complementary Gaussian error function ${\rm Q}(x)$ are given.
Questions
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:
- $$k= 1\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge \sigma_{x}) \rm e^{-2}= \rm 13.53\%,$$
- $$k= 2\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge 2 \cdot \sigma_{x})= \rm \rm e^{-3}=\rm 4.97\% ,$$
- $$k= 3\text\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge 3 \cdot\sigma_{x})= \rm \rm e^{-4}\hspace{0.15cm}\underline{ =\rm 1.83\% },$$
- $$k= 4\text{:}\hspace{0.5cm} {\rm Pr}(|x-m_x| \ge 4 \cdot \sigma_{x}) = \rm e^{-5}= \rm 0.67\%.$$