Geometric
Probability mass function
Cumulative distribution function
المتغيرات
0
<
p
≤
1
0
𝑝
1
{\displaystyle{\displaystyle 0<p\leq 1}}
success probability (real )
0
<
p
≤
1
0
𝑝
1
{\displaystyle{\displaystyle 0<p\leq 1}}
success probability (real )
Support
k
∈
{
1
,
2
,
3
,
…
}
𝑘
1
2
3
…
{\displaystyle{\displaystyle k\in\{1,2,3,\dots\}\!}}
k
∈
{
0
,
1
,
2
,
3
,
…
}
𝑘
0
1
2
3
…
{\displaystyle{\displaystyle k\in\{0,1,2,3,\dots\}\!}}
Probability mass function (pmf)
(
1
-
p
)
k
-
1
p
superscript
1
𝑝
𝑘
1
𝑝
{\displaystyle{\displaystyle(1-p)^{k-1}\,p\!}}
(
1
-
p
)
k
p
superscript
1
𝑝
𝑘
𝑝
{\displaystyle{\displaystyle(1-p)^{k}\,p\!}}
Cumulative distribution function (cdf)
1
-
(
1
-
p
)
k
1
superscript
1
𝑝
𝑘
{\displaystyle{\displaystyle 1-(1-p)^{k}\!}}
1
-
(
1
-
p
)
k
+
1
1
superscript
1
𝑝
𝑘
1
{\displaystyle{\displaystyle 1-(1-p)^{k+1}\!}}
Mean
1
p
1
𝑝
{\displaystyle{\displaystyle{\frac{1}{p}}\!}}
1
-
p
p
1
𝑝
𝑝
{\displaystyle{\displaystyle{\frac{1-p}{p}}\!}}
Median
⌈
-
1
log
2
(
1
-
p
)
⌉
1
subscript
2
1
𝑝
{\displaystyle{\displaystyle\left\lceil{\frac{-1}{\log_{2}(1-p)}}\right\rceil%
\!}}
(not unique if
-
1
/
log
2
(
1
-
p
)
1
subscript
2
1
𝑝
{\displaystyle{\displaystyle-1/\log_{2}(1-p)}}
is an integer)
⌈
-
1
log
2
(
1
-
p
)
⌉
-
1
1
subscript
2
1
𝑝
1
{\displaystyle{\displaystyle\left\lceil{\frac{-1}{\log_{2}(1-p)}}\right\rceil%
\!-1}}
(not unique if
-
1
/
log
2
(
1
-
p
)
1
subscript
2
1
𝑝
{\displaystyle{\displaystyle-1/\log_{2}(1-p)}}
is an integer)
Mode
1
1
{\displaystyle{\displaystyle 1}}
0
0
{\displaystyle{\displaystyle 0}}
Variance
1
-
p
p
2
1
𝑝
superscript
𝑝
2
{\displaystyle{\displaystyle{\frac{1-p}{p^{2}}}\!}}
1
-
p
p
2
1
𝑝
superscript
𝑝
2
{\displaystyle{\displaystyle{\frac{1-p}{p^{2}}}\!}}
Skewness
2
-
p
1
-
p
2
𝑝
1
𝑝
{\displaystyle{\displaystyle{\frac{2-p}{\sqrt{1-p}}}\!}}
2
-
p
1
-
p
2
𝑝
1
𝑝
{\displaystyle{\displaystyle{\frac{2-p}{\sqrt{1-p}}}\!}}
Excess kurtosis
6
+
p
2
1
-
p
6
superscript
𝑝
2
1
𝑝
{\displaystyle{\displaystyle 6+{\frac{p^{2}}{1-p}}\!}}
6
+
p
2
1
-
p
6
superscript
𝑝
2
1
𝑝
{\displaystyle{\displaystyle 6+{\frac{p^{2}}{1-p}}\!}}
Entropy
-
(
1
-
p
)
log
2
(
1
-
p
)
-
p
log
2
p
p
1
𝑝
subscript
2
1
𝑝
𝑝
subscript
2
𝑝
𝑝
{\displaystyle{\displaystyle{\tfrac{-(1-p)\log_{2}(1-p)-p\log_{2}p}{p}}\!}}
-
(
1
-
p
)
log
2
(
1
-
p
)
-
p
log
2
p
p
1
𝑝
subscript
2
1
𝑝
𝑝
subscript
2
𝑝
𝑝
{\displaystyle{\displaystyle{\tfrac{-(1-p)\log_{2}(1-p)-p\log_{2}p}{p}}\!}}
Moment-generating function (mgf)
p
e
t
1
-
(
1
-
p
)
e
t
𝑝
superscript
𝑒
𝑡
1
1
𝑝
superscript
𝑒
𝑡
{\displaystyle{\displaystyle{\frac{pe^{t}}{1-(1-p)e^{t}}}\!}}
, for
t
<
-
ln
(
1
-
p
)
𝑡
1
𝑝
{\displaystyle{\displaystyle t<-\ln(1-p)\!}}
p
1
-
(
1
-
p
)
e
t
𝑝
1
1
𝑝
superscript
𝑒
𝑡
{\displaystyle{\displaystyle{\frac{p}{1-(1-p)e^{t}}}\!}}
Characteristic function
p
e
i
t
1
-
(
1
-
p
)
e
i
t
𝑝
superscript
𝑒
𝑖
𝑡
1
1
𝑝
superscript
𝑒
𝑖
𝑡
{\displaystyle{\displaystyle{\frac{pe^{it}}{1-(1-p)\,e^{it}}}\!}}
p
1
-
(
1
-
p
)
e
i
t
𝑝
1
1
𝑝
superscript
𝑒
𝑖
𝑡
{\displaystyle{\displaystyle{\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
𝑃
𝑊
𝑝
superscript
1
𝑝
𝑘
1
{\displaystyle{\displaystyle 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
)
fragments
P
fragments
(
W
)
p
superscript
fragments
(
1
p
)
𝑛
1
p
superscript
𝑞
𝑛
1
fragments
(
n
1
,
2
)
{\displaystyle{\displaystyle P(W)=p(1-p)^{n-1}=pq^{n-1}(n=1,2)}}
p
(
w
)
=
(
1
/
5
)
*
(
5
/
6
)
(
n
-
1
)
fragments
p
fragments
(
w
)
fragments
(
1
5
)
superscript
fragments
(
5
6
)
(
n
1
)
{\displaystyle{\displaystyle 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 )/p 2 :
E
(
X
)
=
1
p
,
var
(
X
)
=
1
-
p
p
2
.
formulae-sequence
E
𝑋
1
𝑝
var
𝑋
1
𝑝
superscript
𝑝
2
{\displaystyle{\displaystyle\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 )/p 2 :
E
(
Y
)
=
1
-
p
p
,
var
(
Y
)
=
1
-
p
p
2
.
formulae-sequence
E
𝑌
1
𝑝
𝑝
var
𝑌
1
𝑝
superscript
𝑝
2
{\displaystyle{\displaystyle\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
κ
n
subscript
𝜅
𝑛
{\displaystyle{\displaystyle\kappa_{n}}}
of the probability distribution of Y satisfy the recursion
κ
n
+
1
=
μ
(
μ
+
1
)
d
κ
n
d
μ
.
subscript
𝜅
𝑛
1
𝜇
𝜇
1
𝑑
subscript
𝜅
𝑛
𝑑
𝜇
{\displaystyle{\displaystyle\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
E
(
Y
)
=
∑
k
=
0
∞
(
1
-
p
)
k
p
⋅
k
=
p
∑
k
=
0
∞
(
1
-
p
)
k
k
=
p
[
d
d
p
(
-
∑
k
=
0
∞
(
1
-
p
)
k
)
]
(
1
-
p
)
=
-
p
(
1
-
p
)
d
d
p
1
p
=
1
-
p
p
.
E
𝑌
absent
superscript
subscript
𝑘
0
⋅
superscript
1
𝑝
𝑘
𝑝
𝑘
missing-subexpression
absent
𝑝
superscript
subscript
𝑘
0
superscript
1
𝑝
𝑘
𝑘
missing-subexpression
absent
𝑝
delimited-[]
𝑑
𝑑
𝑝
superscript
subscript
𝑘
0
superscript
1
𝑝
𝑘
1
𝑝
missing-subexpression
absent
𝑝
1
𝑝
𝑑
𝑑
𝑝
1
𝑝
1
𝑝
𝑝
{\displaystyle{\displaystyle{\begin{aligned} \displaystyle\mathrm{E}(Y)&%
\displaystyle{}=\sum_{k=0}^{\infty}(1-p)^{k}p\cdot k\\
&\displaystyle{}=p\sum_{k=0}^{\infty}(1-p)^{k}k\\
&\displaystyle{}=p\left[{\frac{d}{dp}}\left(-\sum_{k=0}^{\infty}(1-p)^{k}%
\right)\right](1-p)\\
&\displaystyle{}=-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
G
X
(
s
)
=
s
p
1
-
s
(
1
-
p
)
,
G
Y
(
s
)
=
p
1
-
s
(
1
-
p
)
,
|
s
|
<
(
1
-
p
)
-
1
.
subscript
𝐺
𝑋
𝑠
absent
𝑠
𝑝
1
𝑠
1
𝑝
subscript
𝐺
𝑌
𝑠
formulae-sequence
absent
𝑝
1
𝑠
1
𝑝
𝑠
superscript
1
𝑝
1
{\displaystyle{\displaystyle{\begin{aligned} \displaystyle G_{X}(s)&%
\displaystyle={\frac{s\,p}{1-s\,(1-p)}},\\
\displaystyle G_{Y}(s)&\displaystyle={\frac{p}{1-s\,(1-p)}},\quad|s|<(1-p)^{-1%
}.\end{aligned}}}}
توزيعات ذات صلة
The geometric distribution Y is a special case of the negative binomial distribution , with r = 1. More generally, if Y 1 , ..., Y r are independent geometrically distributed variables with parameter p , then the sum
Z
=
∑
m
=
1
r
Y
m
𝑍
superscript
subscript
𝑚
1
𝑟
subscript
𝑌
𝑚
{\displaystyle{\displaystyle Z=\sum_{m=1}^{r}Y_{m}}}
follows a negative binomial distribution with parameters r and '1-'p.
If Y 1 , ..., Y r are independent geometrically distributed variables (with possibly different success parameters p m ), then their minimum
W
=
min
m
∈
1
,
…
,
r
Y
m
𝑊
subscript
𝑚
1
…
𝑟
subscript
𝑌
𝑚
{\displaystyle{\displaystyle W=\min_{m\in 1,\dots,r}Y_{m}\,}}
is also geometrically distributed, with parameter
p
=
1
-
∏
m
(
1
-
p
m
)
.
𝑝
1
subscript
product
𝑚
1
subscript
𝑝
𝑚
{\displaystyle{\displaystyle 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
∑
k
=
1
∞
k
X
k
superscript
subscript
𝑘
1
𝑘
subscript
𝑋
𝑘
{\displaystyle{\displaystyle\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
=
⌊
X
⌋
,
𝑌
𝑋
{\displaystyle{\displaystyle Y=\lfloor X\rfloor,}}
where
⌊
⌋
fragments
⌊
italic-
⌋
{\displaystyle{\displaystyle\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 )[1] ) 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
⌊
ln
(
U
)
/
ln
(
1
-
p
)
⌋
𝑈
1
𝑝
{\displaystyle{\displaystyle\lfloor\ln(U)/\ln(1-p)\rfloor}}
is geometrically distributed with parameter
p
𝑝
{\displaystyle{\displaystyle p}}
, if
U
𝑈
{\displaystyle{\displaystyle U}}
is uniformly distributed in [0,1].
انظر أيضاً
الهامش
وصلات خارجية
Discrete univariate
with finite support with infinite support
Continuous univariate
supported on a bounded interval supported on a semi-infinite interval supported on the whole real line with support whose type varies
Mixed univariate
Multivariate (joint) Directional Degenerate and singular العائلات
قالب:Common univariate probability distributions