توزيع هندسي

Geometric

Probability mass function
Cumulative distribution function
المتغيرات	$0<p\leq 1$ success probability (real)	$0<p\leq 1$ success probability (real)
Support	$k\in \{1,2,3,\dots \}\!$	$k\in \{0,1,2,3,\dots \}\!$
Probability mass function (pmf)	$(1-p)^{k-1}\,p\!$	$(1-p)^{k}\,p\!$
Cumulative distribution function (cdf)	$1-(1-p)^{k}\!$	$1-(1-p)^{k+1}\!$
Mean	${\frac {1}{p}}\!$	${\frac {1-p}{p}}\!$
Median	$\left\lceil {\frac {-1}{\log _{2}(1-p)}}\right\rceil \!$ (not unique if $-1/\log _{2}(1-p)$ is an integer)	$\left\lceil {\frac {-1}{\log _{2}(1-p)}}\right\rceil \!-1$ (not unique if $-1/\log _{2}(1-p)$ is an integer)
Mode	$1$	$0$
Variance	${\frac {1-p}{p^{2}}}\!$	${\frac {1-p}{p^{2}}}\!$
Skewness	${\frac {2-p}{\sqrt {1-p}}}\!$	${\frac {2-p}{\sqrt {1-p}}}\!$
Excess kurtosis	$6+{\frac {p^{2}}{1-p}}\!$	$6+{\frac {p^{2}}{1-p}}\!$
Entropy	${\tfrac {-(1-p)\log _{2}(1-p)-p\log _{2}p}{p}}\!$	${\tfrac {-(1-p)\log _{2}(1-p)-p\log _{2}p}{p}}\!$
Moment-generating function (mgf)	${\frac {pe^{t}}{1-(1-p)e^{t}}}\!$, for $t<-\ln(1-p)\!$	${\frac {p}{1-(1-p)e^{t}}}\!$
Characteristic function	${\frac {pe^{it}}{1-(1-p)\,e^{it}}}\!$	${\frac {p}{1-(1-p)\,e^{it}}}\!$

التوزيع الهندسي Geometric distribution وهو جزء من التوزيع الاحتمالي الغير متعلق بمحاولات برنولي Bernoulli trials. ويستخدم التوزيع الهندسي كم عدد المحاولات التي نحتاجها للحصول على النتيجة المطلوبة $P(W)=p*(1-p)^{k-1}$

مثال ليكن لدينا نرد متجانس (1,2,3,4,5,6,) كم عدد المحاولات (n) التي نحتاجها للحصول على الرقم 6 الحل :

الاحتمال الصحيح P = 1/6

الاحتمالات الخاطئة q=1-P = 5/6

$P(W)=p(1-p)^{n-1}=pq^{n-1}(n=1,2)$

$p(w)=(1/5)*(5/6)^{(}n-1)$

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 العزوم والتراكمات

The expected value of a geometrically distributed random variable X is 1/p and the variance is (1 − p)/p²:

$\mathrm {E} (X)={\frac {1}{p}},\qquad \mathrm {var} (X)={\frac {1-p}{p^{2}}}.$

Similarly, the expected value of the geometrically distributed random variable Y is (1 − p)/p, and its variance is (1 − p)/p²:

$\mathrm {E} (Y)={\frac {1-p}{p}},\qquad \mathrm {var} (Y)={\frac {1-p}{p^{2}}}.$

Let μ = (1 − p)/p be the expected value of Y. Then the cumulants $\kappa _{n}$ of the probability distribution of Y satisfy the recursion

$\kappa _{n+1}=\mu (\mu +1){\frac {d\kappa _{n}}{d\mu }}.$

Outline of proof: That the expected value is (1 − p)/p can be shown in the following way. Let Y be as above. Then

${\begin{aligned}\mathrm {E} (Y)&{}=\sum _{k=0}^{\infty }(1-p)^{k}p\cdot k\\&{}=p\sum _{k=0}^{\infty }(1-p)^{k}k\\&{}=p\left[{\frac {d}{dp}}\left(-\sum _{k=0}^{\infty }(1-p)^{k}\right)\right](1-p)\\&{}=-p(1-p){\frac {d}{dp}}{\frac {1}{p}}={\frac {1-p}{p}}.\end{aligned}}$

(The interchange of summation and differentiation is justified by the fact that convergent power series converge uniformly on compact subsets of the set of points where they converge.)

 خواص أخرى

• The probability-generating functions of X and Y are, respectively,

${\begin{aligned}G_{X}(s)&={\frac {s\,p}{1-s\,(1-p)}},\\[10pt]G_{Y}(s)&={\frac {p}{1-s\,(1-p)}},\quad |s|<(1-p)^{-1}.\end{aligned}}$

• Golomb coding is the optimal prefix code for the geometric discrete distribution.

 توزيعات ذات صلة

• The geometric distribution Y is a special case of the negative binomial distribution, with r = 1. More generally, if Y₁, ..., Y_r are independent geometrically distributed variables with parameter p, then the sum

$Z=\sum _{m=1}^{r}Y_{m}$

follows a negative binomial distribution with parameters r and '1-'p.

• If Y₁, ..., Y_r are independent geometrically distributed variables (with possibly different success parameters p_m), then their minimum

$W=\min _{m\in 1,\dots ,r}Y_{m}\,$

is also geometrically distributed, with parameter $p=1-\prod _{m}(1-p_{m}).$

• Suppose 0 < r < 1, and for k = 1, 2, 3, ... the random variable X_k has a Poisson distribution with expected value r ^k/k. Then

$\sum _{k=1}^{\infty }k\,X_{k}$

has a geometric distribution taking values in the set {0, 1, 2, ...}, with expected value r/(1 − r).

• The exponential distribution is the continuous analogue of the geometric distribution. If X is an exponentially distributed random variable with parameter λ, then

$Y=\lfloor X\rfloor ,$

where $\lfloor \quad \rfloor$ is the floor (or greatest integer) function, is a geometrically distributed random variable with parameter p = 1 − e^−λ (thus λ = −ln(1 − p)) and taking values in the set {0, 1, 2, ...}. This can be used to generate geometrically distributed pseudorandom numbers by first generating exponentially distributed pseudorandom numbers from a uniform pseudorandom number generator: then $\lfloor \ln(U)/\ln(1-p)\rfloor$ is geometrically distributed with parameter $p$, if $U$ is uniformly distributed in [0,1].

 انظر أيضاً

• توزيع هندسي زائدي
• مشكلة جامع الكوبونات

 وصلات خارجية

• Geometric distribution على بلانيت ماث
• Geometric distribution on MathWorld.
• Online geometric distribution calculator