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临江仙 · 2020年03月17日

问一道题:NO.PZ2020021205000060

问题如下:

A stock has an expected return of 15% and a volatility of 20%. The current price of the stock is USD 50. Estimate a 99% confidence interval for the price at the end of one day.

选项:

解释:

Here, we are dealing with a short time period, and so it is reasonable to assume that the return is normally distributed. The return has a mean of 15% X (1 /252) = 0.0595%, and a standard deviation of 20% X 1/252\sqrt{1/252} = 1.2599%. The 99% confidence interval for the percentage return is between:

0.0595 - 1.2599 X N1N^{-1}(0.995) = -3.186%

and

0.0595 + 1.2599 X N1N^{-1}(0.995)= +3.305%

The confidence interval for the stock price is therefore between 50 X 0.96814 = 48.4 and

50 X 1.03305 = 51.7.

一般题目中,什么时间算短期,什么时间算长期。

个人感觉:短期就利用μ服从正态分布来计算;

长期就利用lnSt来计算。

但题目中什么算短期,什么算长期,这个不知道,求指导

1 个答案
已采纳答案

品职答疑小助手雍 · 2020年03月18日

同学你好,我的印象感觉就是按天的都是短期,到月这种数据跨度大一些的就可以算长期了。一般题目的长短跨度会给的比较明显

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NO.PZ2020021205000060 问题如下 A stohexpectereturn of 15% ana volatility of 20%. The current priof the stois US50. Estimate a 99% confinintervfor the prithe enof one y. Here, we are aling with a short time perio anso it is reasonable to assume ththe return is normally stribute The return ha meof 15% X (1 /252) = 0.0595%, ana stanrviation of 20% X 1/252\sqrt{1/252}1/252​ = 1.2599%. The 99% confinintervfor the percentage return is between:0.0595 - 1.2599 X N−1N^{-1}N−1(0.995) = -3.186%an.0595 + 1.2599 X N−1N^{-1}N−1(0.995)= +3.305%The confinintervfor the stopriis therefore between 50 X 0.96814 = 48.4 an0 X 1.03305 = 51.7. The confinintervfor the stopriis therefore between 50 X 0.96814 = 48.4 an0 X 1.03305 = 51.7.老师,上面这个一步想确认一下,是分别用50*exp(-3.186%) 、50*exp(+3.305%)吗?

2024-10-17 14:18 1 · 回答

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