演習問題の解     Last modified: Aug 26, 2015

表 2' のように $x\cdot f ( x )$ ,$x^{2}\cdot f ( x )$ ,$x^{3}\cdot f ( x )$ ,$x^{4}\cdot f ( x )$ を計算する。

> x <- 0:7
> f <- c(0.13533528, 0.27067057, 0.27067057, 0.18044704, 0.09022352, 0.03608941, 0.01202980, 0.00453381)
> x.f <- x*f
> x2.f <- x^2*f
> x3.f <- x^3*f
> x4.f <- x^4*f
> old <- options(digits=9)
> ( d <- data.frame(x, f, x.f, x2.f, x3.f, x4.f) )
  x          f        x.f       x2.f       x3.f        x4.f
1 0 0.13533528 0.00000000 0.00000000 0.00000000  0.00000000
2 1 0.27067057 0.27067057 0.27067057 0.27067057  0.27067057
3 2 0.27067057 0.54134114 1.08268228 2.16536456  4.33072912
4 3 0.18044704 0.54134112 1.62402336 4.87207008 14.61621024
5 4 0.09022352 0.36089408 1.44357632 5.77430528 23.09722112
6 5 0.03608941 0.18044705 0.90223525 4.51117625 22.55588125
7 6 0.01202980 0.07217880 0.43307280 2.59843680 15.59062080
8 7 0.00453381 0.03173667 0.22215669 1.55509683 10.88567781
それぞれの合計 $\nu_{1}$,$\nu_{2}$,$\nu_{3}$,$\nu_{4}$ を求める。

> ( nu <- colSums(d)[3:6] )
        x.f        x2.f        x3.f        x4.f 
 1.99860943  5.97841727 21.74712037 91.34701091
これらより,

母平均値 $= \mu = \nu_{1} = 1.99860943$

母分散 $= \sigma^{2} = \nu_{2}' = \nu_{2} - \nu_{1}^{2} = 5.97841727 - 1.99860943^{2} = 1.98397762$

> nu <- unname(nu)
> ( mu <- nu[1] )
[1] 1.99860943
> ( sigma2 <- nu2.prime <- nu[2]-nu[1]^2 )
[1] 1.98397762
また,

$\nu_{3}' = \nu_{3} - 3 \nu_{2} \nu_{1} + 2 \nu_{1}^{3} = 21.74712037 - 3 \cdot 5.97841727 \cdot 1.99860943 + 2 \cdot 1.99860943^{3} = 1.86820649$ $\nu_{4}' = \nu_{4} - 4 \nu_{3} \nu_{1} + 6 \nu_{2} \nu_{1}^{2} - 3 \nu_{1}^{4} = 91.34701091 - 4 \cdot 21.74712037 \cdot 1.99860943 + 6 \cdot 5.97841727 \cdot 1.99860943^{2} - 3 \cdot 1.99860943^{4} = 12.90692914$

> ( nu3.prime <- nu[3]-3*nu[2]*nu[1]+2*nu[1]^3 )
[1] 1.86820649
> ( nu4.prime <- nu[4]-4*nu[3]*nu[1]+6*nu[2]*nu[1]^2-3*nu[1]^4 )
[1] 12.9069291
より,

母歪度 $= \alpha_{3} = \nu_{3}' / \sigma^{3} = 1.86820649 / 1.98397762^{1.5} = 0.66852819$

母尖度 $= \alpha_{4} = \nu_{4}' / \sigma^{4} = 12.90692914 / 1.98397762^{2} = 3.27906020$

> ( alpha3 <- nu3.prime/sigma2^1.5 )
[1] 0.66852819
> ( alpha4 <- nu4.prime/sigma2^2 )
[1] 3.2790602
> options(old)

表 2'.表 2 に基づく計算
$x$ $f(x)$ $x\ f(x)$ $x^{2}\ f(x)$ $x^{3}\ f(x)$ $x^{4}\ f(x)$
0 0.13533528 0.00000000 0.00000000 0.00000000 0.00000000
1 0.27067057 0.27067057 0.27067057 0.27067057 0.27067057
2 0.27067057 0.54134114 1.08268228 2.16536456 4.33072912
3 0.18044704 0.54134112 1.62402336 4.87207008 14.61621024
4 0.09022352 0.36089408 1.44357632 5.77430528 23.09722112
5 0.03608941 0.18044705 0.90223525 4.51117625 22.55588125
6 0.01202980 0.07217880 0.43307280 2.59843680 15.59062080
7 0.00453381 0.03173667 0.22215669 1.55509683 10.88567781
合計 1.00000000 1.99860943 5.97841727 21.74712037 91.34701091


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