# 幾何分佈

• 伯努利試驗中，得到一次成功所需要的試驗次数${\displaystyle X}$${\displaystyle X}$的值域是{ 1, 2, 3, ... }
參數 支撑集 概率質量函數 累積分布函數 ${\displaystyle 0成功概率（实） ${\displaystyle 0成功概率（实） ${\displaystyle k\in \{1,2,3,\dots \}\!}$ ${\displaystyle k\in \{0,1,2,3,\dots \}\!}$ ${\displaystyle (1-p)^{k-1}\,p\!}$ ${\displaystyle (1-p)^{k}\,p\!}$ ${\displaystyle 1-(1-p)^{k}\!}$ ${\displaystyle 1-(1-p)^{k+1}\!}$ ${\displaystyle {\frac {1}{p}}\!}$ ${\displaystyle {\frac {1-p}{p}}\!}$ ${\displaystyle \left\lceil {\frac {-1}{\log _{2}(1-p)}}\right\rceil \!}$（如果${\displaystyle -1/\log _{2}(1-p)}$是整数，则中位数不唯一） ${\displaystyle \left\lceil {\frac {-1}{\log _{2}(1-p)}}\right\rceil \!-1}$（如果${\displaystyle -1/\log _{2}(1-p)}$是整数，则中位数不唯一） ${\displaystyle 1}$ ${\displaystyle 0}$ ${\displaystyle {\frac {1-p}{p^{2}}}\!}$ ${\displaystyle {\frac {1-p}{p^{2}}}\!}$ ${\displaystyle {\frac {2-p}{\sqrt {1-p}}}\!}$ ${\displaystyle {\frac {2-p}{\sqrt {1-p}}}\!}$ ${\displaystyle 6+{\frac {p^{2}}{1-p}}\!}$ ${\displaystyle 6+{\frac {p^{2}}{1-p}}\!}$ ${\displaystyle {\tfrac {-(1-p)\log _{2}(1-p)-p\log _{2}p}{p}}\!}$ ${\displaystyle {\tfrac {-(1-p)\log _{2}(1-p)-p\log _{2}p}{p}}\!}$ ${\displaystyle {\frac {pe^{t}}{1-(1-p)e^{t}}}\!}$, for ${\displaystyle t<-\ln(1-p)\!}$ ${\displaystyle {\frac {p}{1-(1-p)e^{t}}}\!}$ ${\displaystyle {\frac {pe^{it}}{1-(1-p)\,e^{it}}}\!}$ ${\displaystyle {\frac {p}{1-(1-p)\,e^{it}}}\!}$
• 在得到第一次成功之前所经历的失败次数${\displaystyle Y=X-1}$Y的值域是{ 0, 1, 2, 3, ... }

${\displaystyle \Pr(X=k)=(1-p)^{k-1}\,p\,}$

${\displaystyle \Pr(Y=k)=(1-p)^{k}\,p\,}$

## 性质

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

${\displaystyle P(T>s+t\;|\;T>t)=P(T>s)\;\;{\hbox{for all}}\ }$ s, t ∈ℕ.