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标题: 關於handbook裡面3.1.3的例 [打印本页]

作者: jazlynn1031    时间: 2015-5-8 19:14     标题: 關於handbook裡面3.1.3的例

1. 關於handbook裡面3.1.3的例
......To be more precise, we could use the chi-square distribution for the variance
with k = T − 1 = 239. For the X^2(239), the 2.5% lower and 2.5% higher quantiles are q2.5% = 198.1 and q97.5% = 283.7
The exact confidence band is then sqrt(198.1/239) × 3.24% to sqrt(283.7/239) × 3.24%, or [2.949%, 3.530%]

想請問
a. q2.5%=198.1 怎麼算出來的?
b. exact confidence band = sqrt(198.1/239) × 3.24% 是用哪一個公式得出?


2. A confidence interval can be constructed using the 2.5 percent lower and 2.5 percent higher cutoff values from the chi-square distribution.
Define the first cutoff value as q2.5% = q[2.5%, X^2(T − 1)]. Setting this equal to the chi-square value on the right-hand side of Equation (3.6) gives the lower bound for the variance of q2.5% × s2/(T − 1).

不懂為何 q2.5% = q[2.5%, X^2(T − 1)] 帶到 Equation (3.6) 變成 q2.5% × s2/(T − 1)
作者: jazlynn1031    时间: 2015-5-8 20:01

跪求解...
作者: cashking    时间: 2015-6-2 10:35

a. q2.5%=198.1 怎麼算出來的?
可以查Chi square分布表,选择自由度为239、左尾累积概率为0.025,查出对应Chi square数值大体上为198.1。或者使用excel函数,输入=CHISQ.INV(0.025,239), 等于198.07
b.不知道你这个问题的背景,只能猜测一下:因为自由度为N的chi square变量相当于N个标准正态变量的和,所以sqrt(198.1/239) × 3.24%的构建方法实际上假定了3.24%这个数值所代表的变量应当服从一个正态分布。




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