ν λ²μ μνμμ μ±κ³΅νλ₯ μ΄ p p p μΈ κ²½μ°, 첫 λ²μ§Έ μ±κ³΅μ΄ λ°μν λκΉμ§ μννλ λ
립μνμ νμ λ₯Ό X X X λΌ νλ©΄, νλ₯ λ³μ X X X λ λ€μκ³Ό κ°μ κΈ°νλΆν¬λ₯Ό λ°λ₯Έλ€[X βΌ G ( p ) X \sim G(p) X βΌ G ( p ) ]. κ·Έ μ΄μ λ x x x λ²μ§Έ μνμμ 첫 λ²μ§Έ μ±κ³΅μ΄ λ°μνλ €λ©΄ κ·Έ μ΄μ μ ( x β 1 ) (x-1) ( x β 1 ) λ²μ μνμμ μ°μν΄μ 'μ€ν¨'κ° λμμΌ νκ³ , λ§μ§λ§ μνμμ 'μ±κ³΅'μ΄ λμμΌ νκΈ° λλ¬Έμ΄λ€.
κΈ°νλΆν¬(Geometric Distribution)
μ±κ³΅νλ₯ μ΄ μΌμ ν μνμμ 첫 λ²μ§Έ μ±κ³΅μ΄ λ°μν λκΉμ§ μνν νμμ νλ₯ λΆν¬λ λ€μκ³Ό κ°λ€.
f ( x ) = P ( X = x ) = ( 1 β p ) x β 1 p , f(x) = P(X=x) = (1-p)^{x-1}p , f ( x ) = P ( X = x ) = ( 1 β p ) x β 1 p , x = 1 , 2 , 3 , . . . x = 1, 2, 3,... x = 1 , 2 , 3 , ...
κΈ°ν νλ₯ λ³μλ μ²μ μ±κ³΅μ΄ λμ¬ λ κΉμ§ μΌλ§λ§νΌ μλνλμ§κ° μ€μνλ€.
κΈ°ννλ₯ λ³μμ λμ λΆν¬ν¨μ
F ( x ) = β y = 1 x ( 1 β p ) y β 1 p = p [ 1 β ( 1 β p ) x ] 1 β ( 1 β p ) ) F(x) = \sum_{y=1}^{x} (1-p)^{y-1}p = \frac{p[1-(1-p)^x]}{1-(1-p))} F ( x ) = β y = 1 x β ( 1 β p ) y β 1 p = 1 β ( 1 β p )) p [ 1 β ( 1 β p ) x ] β
= 1 β ( 1 β p ) x , = 1-(1-p)^x, = 1 β ( 1 β p ) x , x = 1 , 2 , 3 , . . . x = 1, 2, 3,... x = 1 , 2 , 3 , ...
Expected Value and Variance of Geometric Distribution
E ( X ) = 1 p , E(X) = \frac{1}{p}, E ( X ) = p 1 β , V a r ( X ) = 1 β p p 2 Var(X) = \frac{1-p}{p^2} Va r ( X ) = p 2 1 β p β
EXAMPLE 25. μ±κ³΅νλ₯ μ΄ κ°κ° 0.1, 0.3, 0.5μΈ λ¬΄νλͺ¨μ§λ¨μμ 첫 λ²μ§Έ μ±κ³΅μ μννμμ λ, νλ₯ λΆν¬ κ·Έλνλ₯Ό μμ±νκ³ , κΈ°λκ°κ³Ό λΆμ°μ ꡬνμ¬ λΉκ΅νλΌ.
Using Rstat Package : dgeom()
, disc.mexp()
R Source E(X) and Var(X) Plot
Copy library(Rstat)
# success : p, Range of X
p <- c(0.1, 0.3, 0.5); x <- 1:200
len <- length(p)
# Compute the Probability : dgeom() uses the number of failure.
fx <- list()
for ( i in 1:len) fx[[i]] <- dgeom(x-1, p[i])
# Sum(P(X)) == 1 ?
sapply(fx, sum)
# E(X) and Var(X)
disc.mexp(x, fx, plot=F)
# Plotting
mt <- paste0("Geometric(", p, ")")
win.graph(9,3); par(mfrow=c(1,3)); par(mar=c(3,4,4,2))
for (k in 1:len) plot(1:50, fx[[k]][1:50], type="h", main=mt[k],
ylab="f(x)", xlab="", lwd=3, col=2)
Copy > # E(X) and Var(X)
> disc.mexp(x, fx, plot=F)
## E(X) = 10 V(X) = 190 - 10Β² = 90 D(X) = β(90) = 9.487
## E(X) = 3.333 V(X) = 18.889 - 3.333Β² = 7.778 D(X) = β(7.778) = 2.789
## E(X) = 2 V(X) = 6 - 2Β² = 2 D(X) = β(2) = 1.414
EXAMPLE 26. μ£Όμ¬μ 1κ°λ₯Ό '6'μ΄ λμ¬ λκΉμ§ λ°λ³΅ν΄μ ꡴리λ μ€νμμ μννμλ₯Ό XλΌ ν λ, λ€μμ ꡬνλΌ.
3νμ μν μ΄λ΄μ '6'μ΄ λμ¬ νλ₯
[ Solution ]
R Source E(X) and Var(X) F(3)
Copy library(Rstat)
p <- 1/6; x <- 1:100
fx <- dgeom(x-1, p)
# E(X) and Var(X)
disc.exp(x, fx, prt=TRUE)
# Within 0, 1, 2 failures, P(Success)
f <- c(0, 1, 2)
dgeom(f, p)
sum(dgeom(0:2, p))
pgeom(2, p)
Copy > # E(X) and Var(X)
> disc.exp(x, fx, prt=TRUE)
## E(X) = 6
## V(X) = 66 - 6Β² = 30
## D(X) = β(30) = 5.477
Copy > dgeom(f, p)
## [1] 0.1666667 0.1388889 0.1157407
>
> sum(dgeom(0:2, p))
## [1] 0.4212963
>
> pgeom(2, p)
## [1] 0.4212963