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Olinaaaaa · 2024年04月24日

我其他都没问题 就不明白u是可以抵消吗

  1. FRM I
  2. Quantitative

NO.PZ2020010304000050

问题如下：

If you are given a 99% confidence interval for the mean return on the Nasdaq 100 of [2.32%, 12.78%], what is the sample mean and standard error? If this confidence interval is based on 37 years of data, assumed to be iid, what is the sample standard deviation?

解释：

The mean is the midpoint of a symmetric confidence interval (the usual type), and so is 7.55%.

The 99% CI is constructed as [μ∧−c∗σ∧,μ∧+c∗σ∧]\left[\overset\wedge\mu-c*\overset\wedge\sigma, \overset\wedge\mu+c*\overset\wedge\sigma\right] and so c∗σ∧c*\overset\wedge\sigma = 12.78% - 7.55% = 5.23%. The critical value for a 99% CI corresponds to the point where there is 0.5% in each tail, or 2.57, and so σ∧\overset\wedge\sigma =5.23% /2.57=2.03% needs to be surveyed.

If this confidence interval is based on 37 years of data, assumed to be iid, the sample standard deviation is 2.03%×37=12.34%2.03\%\times\sqrt{37}=12.34\%

我的问题在直接用2.58倍的deviation算吗 为什么都没有用到u

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1 个答案

李坏_品职助教 · 2024年04月24日

嗨，从没放弃的小努力你好：


μ在计算过程中抵消了。

方程1： μ + 2.58 * standard error = 12.78%，

方程2：μ - 2.58*standard error = 2.32%，

俩方程上下相减，μ就没有了。2.58*2*standard error = 10.46%，所以standard error = 2.03%


standard error求出来是2.03%，而standard error = standard deviation / 根号n，

所以2.03% = standard deviation / 根号37，所以standard deviation = 2.03% *根号37 = 12.34%

----------------------------------------------
虽然现在很辛苦，但努力过的感觉真的很好，加油！

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NO.PZ2020010304000050 问题如下 If you are given a 99% confinintervfor the mereturn on the Nasq 100 of [2.32%, 12.78%], whis the sample meanstanrerror? If this confinintervis baseon 37 years of tassumeto ii whis the sample stanrviation? The meis the mioint of a symmetric confininterv(the usutype), anso is 7.55%.The 99% is constructe[μ∧−c∗σ∧,μ∧+c∗σ∧]\left[\overset\wee\mu-c*\overset\wee\sigm\overset\wee\mu+c*\overset\wee\sigma\right][μ∧​−c∗σ∧,μ∧​+c∗σ∧] anso c∗σ∧c*\overset\wee\sigmac∗σ∧ = 12.78% - 7.55% = 5.23%. The criticvalue for a 99% correspon to the point where there is 0.5% in eatail, or 2.57, anso σ∧\overset\wee\sigmaσ∧ =5.23% /2.57=2.03% nee to surveye If this confinintervis baseon 37 years of tassumeto ii the sample stanrviation is 2.03%×37=12.34%2.03\%\times\sqrt{37}=12.34\%2.03%×37​=12.34% mioint = me这一步还懂， σ ∧ 为什么等于一半的 置信区间。 还有一个解答里面5.23不是减法，是用除法算的，更不理解了，求解答

2024-04-20 11:58 2 · 回答

NO.PZ2020010304000050 问题如下 If you are given a 99% confinintervfor the mereturn on the Nasq 100 of [2.32%, 12.78%], whis the sample meanstanrerror? If this confinintervis baseon 37 years of tassumeto ii whis the sample stanrviation? The meis the mioint of a symmetric confininterv(the usutype), anso is 7.55%.The 99% is constructe[μ∧−c∗σ∧,μ∧+c∗σ∧]\left[\overset\wee\mu-c*\overset\wee\sigm\overset\wee\mu+c*\overset\wee\sigma\right][μ∧​−c∗σ∧,μ∧​+c∗σ∧] anso c∗σ∧c*\overset\wee\sigmac∗σ∧ = 12.78% - 7.55% = 5.23%. The criticvalue for a 99% correspon to the point where there is 0.5% in eatail, or 2.57, anso σ∧\overset\wee\sigmaσ∧ =5.23% /2.57=2.03% nee to surveye If this confinintervis baseon 37 years of tassumeto ii the sample stanrviation is 2.03%×37=12.34%2.03\%\times\sqrt{37}=12.34\%2.03%×37​=12.34% correspon to the point where there is 0.5% in eatail, or 2.57。样本置信区间求总体均值，首先根据对称性求得样本均值，由于总体方差和样本容量未知，所以用t分布，由于置信区间是99%，所以α=0.5。这样理解对吗？但是之后怎么求2.57呢？

2024-02-03 16:31 1 · 回答

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2023-03-26 04:42 1 · 回答

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2022-02-13 09:49 1 · 回答