Distribució binomial: diferència entre les revisions
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* Una partícula es mou unidimensionalment amb probabilitat '' q '' de moure's cap enrere i '' 1-q '' de moure's cap endavant
La probabilitat d'obtenir exactament <math> k\, \! </math> èxits en <math> n \, \! </math> assaigs(proves) independents de Bernouilli és donada per la seva [[funció de probabilitat]]:
▲La seva [[funció de probabilitat]] és:
<math> \! f (x) ={n \choose x}p^x (1-p)^{n-x}\, \! </math>▼
== Mitjana i Variància d'una distribució binomial <math> X \sim B (n, p) \, </math> ==
: <math> \mathbb{E}[X] = np \, </math>
: <math> \text{Var}[X] = np (1-p) \, </math>
Cumulative distribution function===
The [[cumulative distribution function]] can be expressed as:
▲:<math>F(k;n,p) = \
where <math>\lfloor k\rfloor</math> is the "floor" under ''k'', i.e. the [[greatest integer]] less than or equal to ''k''.
It can also be represented in terms of the [[regularized incomplete beta function]], as follows:<ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref>
:<math>\begin{align}
F(k;n,p) & = \Pr(X \le k) \\
&= I_{1-p}(n-k, k+1) \\
& = (n-k) {n \choose k} \int_0^{1-p} t^{n-k-1} (1-t)^k \, dt.
\end{align}</math>
which is equivalent to the [[cumulative distribution function]] of the [[F-distribution|{{mvar|F}}-distribution]]:<ref>Jowett G H (1963), The Relationship Between the Binomial and F Distributions, Journal of the Royal Statistical Society D, 13, 55-57.</ref>
:<math>F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).</math>
Some closed-form bounds for the cumulative distribution function are given [[#Tail bounds|below]].
===Example===
Suppose a [[fair coin|biased coin]] comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is
:<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math>
== Properties ==
===Expected value and variance===
If ''X'' ~ ''B''(''n'', ''p''), that is, ''X'' is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the [[expected value]] of ''X'' is:<ref>See [https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution Proof Wiki]</ref>
:<math> \operatorname{E}[X] = np.</math>
This follows from the linearity of the expected value along with fact that {{mvar|X}} is the sum of {{mvar|n}} identical Bernoulli random variables, each with expected value {{mvar|p}}. In other words, if <math>X_1, \ldots, X_n</math> are identical (and independent) Bernoulli random variables with parameter {{mvar|p}}, then <math>X = X_1 + \cdots + X_n</math> and
:<math>\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.</math>
The [[variance]] is:
:<math> \operatorname{Var}(X) = np(1 - p).</math>
This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.
===Higher moments===
The first 6 central moments are given by
:<math>\begin{align}
\mu_1 &= 0, \\
\mu_2 &= np(1-p),\\
\mu_3 &= np(1-p)(1-2p),\\
\mu_4 &= np(1-p)(1+(3n-6)p(1-p)),\\
\mu_5 &= np(1-p)(1-2p)(1+(10n-12)p(1-p)),\\
\mu_6 &= np(1-p)(1-30p(1-p)(1-4p(1-p))+5np(1-p)(5-26p(1-p))+15n^2 p^2 (1-p)^2).
\end{align}</math>
===Mode===
== Relacions amb altres variables aleatòries ==
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