累積分布と逆CDF

Rによる多変量確率分布

Surajit Ray

Professor, University of Glasgow

CDFと逆CDFの計算が必要な場面

$\mu=210$、$\sigma=10$の正規密度

$x < 200$における曲線下面積

pnorm(200, mean = 210, sd = 10)

[1] 0.159

累積確率が0.95となる$x_0$は？


qnorm( p = 0.95, mean = 210, sd = 10)


[1] 226.45

$\Rightarrow$ コーヒー瓶の95%は226.45グラム未満

x = 2、y = 4における二変量CDFと正規分布 ${\mu}={\begin{pmatrix} 1 \\ 2 \end{pmatrix}},\quad { \Sigma }={\begin{pmatrix} 1 & .5 \\ .5 & 2 \end{pmatrix}}$

x = 2、y = 4における二変量CDFと正規分布 ${\mu}={\begin{pmatrix} 1 \\ 2 \end{pmatrix}},\quad { \Sigma }={\begin{pmatrix} 1 & 0.5 \\ 0.5 & 2 \end{pmatrix}}$

mu1 <- c(1, 2)
sigma1 <- matrix(c(1, 0.5, 0.5, 2), 2)

pmvnorm(upper = c(2, 4), mean = mu1, sigma = sigma1)

[1] 0.79
attr(,"error")
[1] 1e-15
attr(,"msg")
[1] "Normal Completion"

$ 1 < x <2$、$2 < y < 4$の確率

pmvnorm(lower = c(1, 2), 
          upper = c(2, 4), 
             mean = mu1, 
              sigma = sigma1)

$ 1 < x <2$、$2 < y < 4$の確率

pmvnorm(lower = c(1, 2), 
          upper = c(2, 4), 
             mean = mu1, 
              sigma = sigma1)

[1] 0.163

濃い赤の楕円は0.95分位点

qmvnorm

sigma1 <- diag(2)
sigma1

     [,1] [,2]
[1,]    1    0
[2,]    0    1

qmvnorm(p = 0.95, sigma = sigma1, tail = "both")

$quantile
[1] 2.24
$f.quantile
[1] -1.31e-06
attr(,"message")
[1] "Normal Completion"

半径2.24の赤い円は確率0.95を含む

Rによる多変量確率分布