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梦梦 · 2024年05月27日

老师,方差是怎么求的呢?

NO.PZ2020010303000011

问题如下:

If the return on a stock, R, is normally distributed with a daily mean of 8%/252 and a daily variance of (20%)2/252(20\%)^2/252, find the values where

a. Pr(R < r) = .001

b. Pr(R < r) = .01

c. Pr(R < r) = .05

解释:

a. The mean is 0.031% per day and the variance is 1.58 per day (so that the standard deviation is 1.26% per day). To find these values, we transform the variable to be standard normal, so that

Pr(R<r)=.001=Pr(Z<rμσ)=.001 Pr(R < r) = .001 = Pr(Z<\frac{r-\mu}{\sigma})= .001

The value for the standard normal is -3.09

(NORM.S.INV(0.001) in Excel) so that 3.09σ+μ=3.86%-3.09 * \sigma + \mu = -3.86\%.

b. The same idea an be used here where z = -2.33 so that Pr(Z < z) = .01. Transforming this value, r=2.33σ+μ=2.89%r = -2.33 * \sigma + \mu = -2.89\%.

c. Here the value of z is -1.645 so that r=1.645σ+μ=2.04%r = -1.645 * \sigma + \mu = -2.04\%.

These are all common VaR quan-tiles and suggest that there is a 5% chance that the return would be less than -2.04% on any given day, a 1% change that it would be less than -2.89%, and a one in 1,000 chance that the return would be less than -3.86%, if returns were normally distributed.


老师好,方差是用20%^2/252吗?这样算,得出的是方差是0.0158%,而不是1.58%。另外,20%^2大于252,能直接用20%^2除以252吗?

2 个答案

李坏_品职助教 · 2024年05月27日

嗨,努力学习的PZer你好:


有可能是格式显示错误了,我刚才用电脑看到的题库是这样的:


那个>肯定不对,现在题库里是对的,是/

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梦梦 · 2024年05月31日

好的,谢谢

李坏_品职助教 · 2024年05月27日

嗨,爱思考的PZer你好:


题目告诉我们daily variance的条件:

0.2^2 / 252 = 0.0158%, 所以标准差 = 根号0.0158% = 1.26%.

答案此处的variance是写错了。


你说的20%^2大于252,我不明白是什么意思?20%的平方 = 4%啊,而且这个和大于小于没关系,这个就是题目的条件。


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梦梦 · 2024年05月27日

是不是笔误?“问一道题:”下面您仔细看,有0.2^2大于252,我看半天也没明白什么意思

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NO.PZ2020010303000011 问题如下 If the return on a stock, R, is normally stributewith a ily meof 8%/252 ana ily varianof (20%)2/252(20\%)^2/252(20%)2/252, finthe values where Pr(R r) = .001Pr(R r) = .01Pr(R r) = .05 The meis 0.031% per y anthe varianis 1.58 per y (so ththe stanrviation is 1.26% per y). To finthese values, we transform the variable to stanrnormal, so thatPr(R r)=.001=Pr(Z r−μσ)=.001 Pr(R r) = .001 = Pr(Z \frac{r-\mu}{\sigma})= .001 Pr(R r)=.001=Pr(Z σr−μ​)=.001The value for the stanrnormis -3.09(NORM.S.INV(0.001) in Excel) so th−3.09∗σ+μ=−3.86%-3.09 * \sigma + \mu = -3.86\%−3.09∗σ+μ=−3.86%.The same ia usehere where z = -2.33 so thPr(Z z) = .01. Transforming this value, r=−2.33∗σ+μ=−2.89%r = -2.33 * \sigma + \mu = -2.89\%r=−2.33∗σ+μ=−2.89%.Here the value of z is -1.645 so thr=−1.645∗σ+μ=−2.04%r = -1.645 * \sigma + \mu = -2.04\%r=−1.645∗σ+μ=−2.04%.These are all common Vquan-tiles ansuggest ththere is a 5% chanththe return woulless th-2.04% on any given y, a 1% change thit woulless th-2.89%, ana one in 1,000 chanththe return woulless th-3.86%, if returns were normally stribute 请问这是已知概率,反求z的取值吗?查表方式可以计算?

2024-01-15 00:20 4 · 回答

NO.PZ2020010303000011 问题如下 If the return on a stock, R, is normally stributewith a ily meof 8%/252 ana ily varianof (20%)2/252(20\%)^2/252(20%)2/252, finthe values where Pr(R r) = .001Pr(R r) = .01Pr(R r) = .05 The meis 0.031% per y anthe varianis 1.58 per y (so ththe stanrviation is 1.26% per y). To finthese values, we transform the variable to stanrnormal, so thatPr(R r)=.001=Pr(Z r−μσ)=.001 Pr(R r) = .001 = Pr(Z \frac{r-\mu}{\sigma})= .001 Pr(R r)=.001=Pr(Z σr−μ​)=.001The value for the stanrnormis -3.09(NORM.S.INV(0.001) in Excel) so th−3.09∗σ+μ=−3.86%-3.09 * \sigma + \mu = -3.86\%−3.09∗σ+μ=−3.86%.The same ia usehere where z = -2.32 so thPr(Z z) = .01. Transforming this value, r=−2.32∗σ+μ=−2.89%r = -2.32 * \sigma + \mu = -2.89\%r=−2.32∗σ+μ=−2.89%.Here the value of z is -1.645 so thr=−1.645∗σ+μ=−2.04%r = -1.645 * \sigma + \mu = -2.04\%r=−1.645∗σ+μ=−2.04%.These are all common Vquan-tiles ansuggest ththere is a 5% chanththe return woulless th-2.04% on any given y, a 1% change thit woulless th-2.89%, ana one in 1,000 chanththe return woulless th-3.86%, if returns were normally stribute

2022-05-17 21:52 2 · 回答

老师,这道题你了,但是我没有理解,请详细下这套题的分析过程。

2020-02-22 18:13 1 · 回答

a ily varianof (20%)²>252 怎么推出“the varianis 1.58 per y (so ththe stanrviation is 1.26% per y)”?

2020-02-15 20:01 1 · 回答