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小伟 · 2020年03月01日

问一道题:NO.PZ2020011101000015

问题如下:

Suppose all residual autocorrelations are 1.5/T1.5/ \sqrt T, where T is the sample size. Would these violate the confidence bands in an ACF plot?

选项:

解释:

No, the confidence bands are ±1.96/T\pm 1.96/ \sqrt T and so these would not. If many autocorrelations are jointly relatively large, then this series is likely autocorrelated. A Ljung-Box test would likely detect the joint autocorrelation.

这道题的讲解还是没看明白

1 个答案

orange品职答疑助手 · 2020年03月01日

同学你好,这题是原版书上的题,也有点偏。这道题考的是残差项的confidence bands。本题没有明确说confidence level是95%。我看了下原版书,原版书这块它都是用的95%。如果真的考起来,并且它题目里没有写出是95%的confidence level,就默认是95%的去做。不过考到的几率很低。这个知识点较偏。

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NO.PZ2020011101000015问题如下 Suppose all resiautocorrelations are 1.5/T1.5/ \sqrt T1.5/T​, where T is the sample size. Woulthese violate the confinban in Aplot? No, the confinban are ±1.96/T\pm 1.96/ \sqrt T±1.96/T​ anso these woulnot. If many autocorrelations are jointly relatively large, then this series is likely autocorrelate A Ljung-Box test woullikely tethe joint autocorrelation.“resiautocorrelations are(1.5/跟号T)”这句话是说e1和e2,e2和e3等等,各个e之间的correlation结果是(1.5/跟号T)?而不是 H0:p1=p2=…pn=0,Ha至少一个p不等于0的假设检验结果是(1.5/跟号T)?

2024-08-13 09:27 4 · 回答

NO.PZ2020011101000015 问题如下 Suppose all resiautocorrelations are 1.5/T1.5/ \sqrt T1.5/T​, where T is the sample size. Woulthese violate the confinban in Aplot? No, the confinban are ±1.96/T\pm 1.96/ \sqrt T±1.96/T​ anso these woulnot. If many autocorrelations are jointly relatively large, then this series is likely autocorrelate A Ljung-Box test woullikely tethe joint autocorrelation. 印象中好像有个公式方差/根号下n,计算样本标准误的吧。这里的1.5/ 根号下T、1.96根号下T,是什么意思?

2024-01-29 08:48 1 · 回答

这道题目能翻译下吗?要考察的点是什么啊?已经看了之前的问题和回答,还是不是很清楚题目在问啥,谢谢。

2020-03-29 10:11 1 · 回答

请问一下confinban的公式在笔记上哪里呢

2020-03-04 21:38 1 · 回答

回答思路不是很懂,这里为什么要用t test呢?因为题干里提到all resiautocorrelation,为什么不用F test呢?

2020-03-01 15:19 1 · 回答