توزيع هندسي

Geometric

Probability mass function
Cumulative distribution function
المتغيرات	$0<p\le 1$ success probability (real)	$0<p\le 1$ 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	$\lceil \frac{-1}{{\log }_2(1-p)}\rceil$ (not unique if $-1/{\log }_2(1-p)$ is an integer)	$\lceil \frac{-1}{{\log }_2(1-p)}\rceil -1$ (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	$\frac{-(1-p){\log }_2(1-p)-p{\log }_{2}p}{p}$	$\frac{-(1-p){\log }_2(1-p)-p{\log }_{2}p}{p}$
Moment-generating function (mgf)	$\frac{p{e}^t}{1-(1-p)e^t}$, for $t<-\ln (1-p)$	$\frac{p}{1-(1-p)e^t}$
Characteristic function	$\frac{p{e}^{it}}{1-(1-p)e^{it}}$	$\frac{p}{1-(1-p)e^{it}}$

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

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

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

 العزوم والتراكمات

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

$\mathrm{E}(X)=\frac{1}{p},\mathrm{v}\mathrm{a}\mathrm{r}(X)=\frac{1-p}{p^2}.$

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

$\mathrm{E}(Y)=\frac{1-p}{p},\mathrm{v}\mathrm{a}\mathrm{r}(Y)=\frac{1-p}{p^2}.$

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

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

(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,

$\begin{array}{ll}{G}_X(s)& =\frac{sp}{1-s(1-p)},\\ [10pt]G_Y(s)& =\frac{p}{1-s(1-p)},\left|s\right|<(1-p{)}^{-1}.\end{array}$

• 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

$\sum _{k=1}^{\mathrm{\infty }}k{X}_k$

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 $\lfloor \rfloor$ 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