Theory of Stochastic Signals/Binomial Distribution - Revision history

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	{{Display}}		{{Display}}
			[[de:Stochastische_Signaltheorie/Binomialverteilung]]

Guenter at 16:32, 7 February 2024

2024-02-07T16:32:23Z

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Guenter at 14:28, 13 December 2022

2022-12-13T14:28:11Z

← Older revision		Revision as of 16:28, 13 December 2022
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	Use the interactive HTML5/JS applet  [[Applets:Binomial_and_Poisson_Distribution_(Applet)\|"Binomial and Poisson distribution"]]  to find the binomial probabilities for any  $I$  and  $p$ .		⇒   Use the interactive HTML5/JS applet  [[Applets:Binomial_and_Poisson_Distribution_(Applet)\|"Binomial and Poisson distribution"]]  to find the binomial probabilities for any  $I$  and  $p$ .


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	*the first order moment   ⇒   "linear mean":		*the first order moment   ⇒   "linear mean":
	:$$m_1 ={\rm E}\big[z\big]= I\cdot p,$$		:$$m_1 ={\rm E}\big[z\big]= I\cdot p,$$
	*the second order moment ~~  ⇒   "quadratic mean"~~:		*the second order moment:
	:$$m_2 ={\rm E}\big[z^2\big]= (I^2-I)\cdot p^2+I\cdot p,$$		:$$m_2 ={\rm E}\big[z^2\big]= (I^2-I)\cdot p^2+I\cdot p,$$
	*the variance by applying "Steiner's theorem":		*the variance by applying "Steiner's theorem":

Hwang at 16:46, 12 December 2022

2022-12-12T16:46:36Z

← Older revision		Revision as of 18:46, 12 December 2022
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	If one transmits blocks each of  $I =10$  binary symbols over a channel		If one transmits blocks each of  $I =10$  binary symbols over a channel
	*with probability  $p = 0.01$  that one symbol is falsified   ⇒   random variable  $e_i = 1$,  and		*with probability  $p = 0.01$  that one symbol is falsified   ⇒   random variable  $e_i = 1$,  and
	*correspondingly with probability  $1 - p = 0.99$  for a unfalsified symbol   ⇒   random variable  $e_i = 0$,		*correspondingly with probability  $1 - p = 0.99$  for an unfalsified symbol   ⇒   random variable  $e_i = 0$,

Hwang at 16:44, 12 December 2022

2022-12-12T16:44:22Z

Hwang at 16:32, 12 December 2022

2022-12-12T16:32:39Z

← Older revision		Revision as of 18:32, 12 December 2022
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	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	$\text{Definition:}$		$\text{Definition:}$
	The  '''binomial distribution'''  represents an important special case for the occurrence probabilities of a discrete random variable.		The  »'''binomial distribution'''«  represents an important special case for the occurrence probabilities of a discrete random variable.

	To derive the binomial distribution,  we assume that  $I$  binary and statistically independent random variables  $b_i$  each can achieve		To derive the binomial distribution,  we assume that  $I$  binary and statistically independent random variables  $b_i$  each can achieve
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	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	$\text{Calculation rule:}$		$\text{Calculation rule:}$
	For the  '''probabilities of the binomial distribution'''  with  $μ = 0, \hspace{0.1cm}\text{...} \hspace{0.1cm}, \ I$:		For the  »'''probabilities of the binomial distribution'''«  with  $μ = 0, \hspace{0.1cm}\text{...} \hspace{0.1cm}, \ I$:
	:$$p_\mu = {\rm Pr}(z=\mu)={I \choose \mu}\cdot p\hspace{0.05cm}^\mu\cdot ({\rm 1}-p)\hspace{0.05cm}^{I-\mu}.$$		:$$p_\mu = {\rm Pr}(z=\mu)={I \choose \mu}\cdot p\hspace{0.05cm}^\mu\cdot ({\rm 1}-p)\hspace{0.05cm}^{I-\mu}.$$
	*The first term here indicates the number of combinations   $($read:  $I\ \text{ over }\ μ)$:		*The first term here indicates the number of combinations   $($read:  $I\ \text{ over }\ μ)$:
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	Additional notes:		Additional notes:
	*For very large values of  $I$,  the binomial distribution can be approximated by the  [[Theory_of_Stochastic_Signals/Poisson_Distribution\|Poisson distribution]]  described in the next section.		*For very large values of  $I$,  the binomial distribution can be approximated by the  [[Theory_of_Stochastic_Signals/Poisson_Distribution\|$\text{Poisson distribution}$]]  described in the next section.
	*If at the same time the product  $I · p \gg 1$,  then according to  [https://en.wikipedia.org/wiki/De_Moivre%E2%80%93Laplace_theorem de Moivre–Laplace's (central limit) theorem],  the Poisson distribution  (and hence the binomial distribution)  transitions to a discrete  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables\|Gaussian distribution]].		*If at the same time the product  $I · p \gg 1$,  then according to  [https://en.wikipedia.org/wiki/De_Moivre%E2%80%93Laplace_theorem $\text{de Moivre–Laplace's (central limit) theorem}$],  the Poisson distribution  (and hence the binomial distribution)  transitions to a discrete  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables\|$\text{Gaussian distribution}$]].


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	Use the interactive HTML5/JS applet  [[Applets:Binomial_and_Poisson_Distribution_(Applet)\|Binomial and Poisson distribution]]  to find the binomial probabilities for any  $I$  and  $p$ .		Use the interactive HTML5/JS applet  [[Applets:Binomial_and_Poisson_Distribution_(Applet)\|"Binomial and Poisson distribution"]]  to find the binomial probabilities for any  $I$  and  $p$ .


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	<br>		<br>
	You can calculate the moments in general using the equations in the chapters		You can calculate the moments in general using the equations in the chapters
	# [[Theory_of_Stochastic_Signals/Moments_of_a_Discrete_Random_Variable\|Moments of a Discrete Random Variable]],		# [[Theory_of_Stochastic_Signals/Moments_of_a_Discrete_Random_Variable\|"Moments of a Discrete Random Variable"]],
	# [[Theory_of_Stochastic_Signals/Binomial_Distribution#Probabilities_of_the_binomial_distribution\|Probabilities of the Binomial Distribution]].		# [[Theory_of_Stochastic_Signals/Binomial_Distribution#Probabilities_of_the_binomial_distribution\|"Probabilities of the Binomial Distribution"]].


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	$\text{Calculation rules:} $		$\text{Calculation rules:} $

	For the  '''$k$-th order moment'''  of a binomially distributed random variable,  the general rule is:		For the  »'''$k$-th order moment'''«  of a binomially distributed random variable,  the general rule is:
	:$$m_k={\rm E}\big[z^k\big]=\sum_{\mu={\rm 0} }^{I}\mu^k\cdot{I \choose \mu}\cdot p\hspace{0.05cm}^\mu\cdot ({\rm 1}-p)\hspace{0.05cm}^{I-\mu}.$$		:$$m_k={\rm E}\big[z^k\big]=\sum_{\mu={\rm 0} }^{I}\mu^k\cdot{I \choose \mu}\cdot p\hspace{0.05cm}^\mu\cdot ({\rm 1}-p)\hspace{0.05cm}^{I-\mu}.$$
	From this,  we obtain for after some transformations		From this,  we obtain for after some transformations

Noah at 18:48, 19 April 2022

2022-04-19T18:48:42Z

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	{{Header		{{Header
	\|Untermenü=~~Dicrete~~ Random Variable		\|Untermenü=Discrete Random Variable
	\|Vorherige Seite=Moments of a Discrete Random Variable		\|Vorherige Seite=Moments of a Discrete Random Variable
	\|Nächste Seite=Poissonverteilung		\|Nächste Seite=Poissonverteilung

Noah at 18:48, 19 April 2022

2022-04-19T18:48:21Z

← Older revision		Revision as of 20:48, 19 April 2022
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	*the second order moment   ⇒   "quadratic mean":		*the second order moment   ⇒   "quadratic mean":
	:$$m_2 ={\rm E}\big[z^2\big]= (I^2-I)\cdot p^2+I\cdot p,$$		:$$m_2 ={\rm E}\big[z^2\big]= (I^2-I)\cdot p^2+I\cdot p,$$
	*the variance by ~~applyings~~ "Steiner's theorem":		*the variance by applying "Steiner's theorem":
	:$$\sigma^2 = {m_2-m_1^2} = {I \cdot p\cdot (1-p)} ,$$		:$$\sigma^2 = {m_2-m_1^2} = {I \cdot p\cdot (1-p)} ,$$
	*the standard deviation:		*the standard deviation:

Noah at 18:38, 19 April 2022

2022-04-19T18:38:15Z

← Older revision		Revision as of 20:38, 19 April 2022
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	$\text{Example 2:}$		$\text{Example 2:}$
	The graph shows the probabilities of the binomial distribution are for  $I =6$  and  $p =0.4$.		The graph shows the probabilities of the binomial distribution are for  $I =6$  and  $p =0.4$.
	[[File:P_ID203__Sto_T_2_3_S2_neu.png \|frame\|Binomial distribution ~~probabilites~~]]		[[File:P_ID203__Sto_T_2_3_S2_neu.png \|frame\|Binomial distribution probabilities]]
	*Thus  $M = I+1=7$  probabilities are different from zero.		*Thus  $M = I+1=7$  probabilities are different from zero.

Noah at 16:33, 15 February 2022

2022-02-15T16:33:55Z

← Older revision		Revision as of 18:33, 15 February 2022
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	*the variance by applyings "Steiner's theorem":		*the variance by applyings "Steiner's theorem":
	:$$\sigma^2 = {m_2-m_1^2} = {I \cdot p\cdot (1-p)} ,$$		:$$\sigma^2 = {m_2-m_1^2} = {I \cdot p\cdot (1-p)} ,$$
	*the standard deviation ~~value  $\rm (rms)$  or  "standard deviation"~~:		*the standard deviation:
	:$$\sigma = \sqrt{I \cdot p\cdot (1-p)}.$$}}		:$$\sigma = \sqrt{I \cdot p\cdot (1-p)}.$$}}

← Older revision		Revision as of 18:44, 12 December 2022
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	If one transmits blocks each of  $I =10$  binary symbols over a channel		If one transmits blocks each of  $I =10$  binary symbols over a channel
	*with probability  $p = 0.01$  that one symbol is ~~corrupted~~   ⇒   random variable  $e_i = 1$,  and		*with probability  $p = 0.01$  that one symbol is falsified   ⇒   random variable  $e_i = 1$,  and
	*correspondingly with probability  $1 - p = 0.99$  for a ~~uncorrupted~~ symbol   ⇒   random variable  $e_i = 0$,		*correspondingly with probability  $1 - p = 0.99$  for a unfalsified symbol   ⇒   random variable  $e_i = 0$,


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	:$$f=\sum_{i=1}^{I}e_i.$$		:$$f=\sum_{i=1}^{I}e_i.$$

	This random variable  $f$  can take all integer values between  $0$  (no symbol is ~~corrupted~~)  and  $I$  (all symbols symbol are ~~corrupted~~) .		This random variable  $f$  can take all integer values between  $0$  (no symbol is falsified)  and  $I$  (all symbols symbol are falsified) .
	*We denote the probabilities for  $\mu$  ~~corruptions~~ by  $p_μ$.		*We denote the probabilities for  $\mu$  falsifications by  $p_μ$.
	*The case where all  $I$  symbols are correctly transmitted occurs with probability  $p_0 = 0.99^{10} ≈ 0.9044$ .		*The case where all  $I$  symbols are correctly transmitted occurs with probability  $p_0 = 0.99^{10} ≈ 0.9044$ .
	*This also follows from the binomial formula for  $μ = 0$  considering the definition  $10\, \text{ over }\, 0 = 1$.		*This also follows from the binomial formula for  $μ = 0$  considering the definition  $10\, \text{ over }\, 0 = 1$.
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	:$$p_1 = \rm 10\cdot 0.01\cdot 0.99^9\approx 0.0914.$$		:$$p_1 = \rm 10\cdot 0.01\cdot 0.99^9\approx 0.0914.$$
	::The first factor considers that there are exactly  $10\, \text{ over }\, 1 = 10$  possibilities for the position of a single error.		::The first factor considers that there are exactly  $10\, \text{ over }\, 1 = 10$  possibilities for the position of a single error.
	::The other two factors take into account that one symbol must be ~~corrupted~~ and nine must be transmitted correctly if  $f =1$  is to hold.		::The other two factors take into account that one symbol must be falsified and nine must be transmitted correctly if  $f =1$  is to hold.
	*For  $f =2$  there are clearly more combinations,  namely  $10\, \text{ over }\, 2 = 45$,   and we get		*For  $f =2$  there are clearly more combinations,  namely  $10\, \text{ over }\, 2 = 45$,   and we get
	:$$p_2 = \rm 45\cdot 0.01^2\cdot 0.99^8\approx 0.0041.$$		:$$p_2 = \rm 45\cdot 0.01^2\cdot 0.99^8\approx 0.0041.$$
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	{{GraueBox\|TEXT=		{{GraueBox\|TEXT=
	$\text{Example 4:}$		$\text{Example 4:}$
	As in  $\text{Example 3}$,  we consider a block of  $I =10$  binary symbols,  each of which is independently ~~corrupted~~ with probability  $p = 0.01$ .		As in  $\text{Example 3}$,  we consider a block of  $I =10$  binary symbols,  each of which is independently falsified with probability  $p = 0.01$ .

	Then holds:		Then holds:
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	In contrast,  in the completely ~~corrupted~~ channel   ⇒   bit error probability  $p = 1/2$  results in the values		In contrast,  in the completely falsified channel   ⇒   bit error probability  $p = 1/2$  results in the values

	*$m_f = 5$   ⇒   on average,  five of the ten bits within a block are wrong,		*$m_f = 5$   ⇒   on average,  five of the ten bits within a block are wrong,