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kkbis · 2020年02月26日

问一道题：NO.PZ2020011101000051

  1. FRM I
  2. Quantitative

问题如下：

Suppose you are interested in approximating the expected value of an option. Based on an initial sample of 100 replications, you estimate that the fair value of the option is USD 47 using the mean of these 100 replications. You also note that the standard deviation of these 100 replications is USD 12.30. How many simulations would you need to run in order to obtain a 95% confidence interval that is less than 1% of the fair value of the option? How many would you need to run to get within 0.1%?

选项：

解释：

The standard deviation is USD 12.30, and a 95% confidence interval is [μ^−1.96∗12.30/n,μ^+1.96∗12.30/n][\widehat\mu - 1.96 * 12.30/\sqrt n, \widehat\mu + 1.96 * 12.30/\sqrt n] and so the width is 2∗1.96∗12.30n2 * 1.96 * 12.30\sqrt n . If we want this value to be 1% of USD 47.00, then 2∗1.96∗12.30n=0.47⇒n=2∗1.96∗12.30/0.47=102.52 * 1.96 * 12.30\sqrt n= 0.47\Rightarrow\sqrt n= 2 * 1.96 * 12.30 /0.47 = 102.5 (so 103). Using .1%, we would need 1,025.8 (replace 0.47 with 0.047) and so 1,026 replication

请问为什么1%对应的是0.47？0.1%对应的是0.047呢？根号n=103之后，n为什么就等于1026了呢？

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3 个答案

品职答疑小助手雍 · 2021年01月05日

是需要平方，谢谢指出，跟后台反馈修改啦~

Sunny · 2021年01月02日

题目问的是number of stimulation，是不是应该最终求的是n，而不是根号n？答案中结果不需要平方一下吗

品职答疑小助手雍 · 2020年02月27日

同学你好，根据标准误的求法，只有标准差才需要除以根号n，均值不需要。

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NO.PZ2020011101000051 问题如下 Suppose you are interestein approximating the expectevalue of option. Baseon initisample of 100 replications, you estimate ththe fair value of the option is US47 using the meof these 100 replications. You also note ththe stanrviation of these 100 replications is US12.30. How many simulations woulyou neeto run in orr to obtain a 95% confinintervthis less th1% of the fair value of the option? How many woulyou neeto run to get within 0.1%? The stanrviation is US12.30, ana 95% confinintervis [μ^−1.96∗12.30/n,μ^+1.96∗12.30/n][\wihat\mu - 1.96 * 12.30/\sqrt n, \wihat\mu + 1.96 * 12.30/\sqrt n][μ​−1.96∗12.30/n​,μ​+1.96∗12.30/n​] anso the wih is 2∗1.96∗12.30/n2 * 1.96 * 12.30/\sqrt n2∗1.96∗12.30/n​ . If we want this value to 1% of US47.00, then 2∗1.96∗12.30/n2*1.96 * 12.30/\sqrt n2∗1.96∗12.30/n​=0.47 ⇒n=2∗1.96∗12.30/0.47=102.5\Rightarrow\sqrt n= 2 * 1.96 * 12.30 /0.47 = 102.5⇒n​=2∗1.96∗12.30/0.47=102.5 (so 103), n=1032=10609Using 0.1%, we woulnee1,025.8 (repla0.47 with 0.047) anso 1,026 replication, so n=10262 =1052676 0.47怎么来的呢，1%对应的置信度也不是0.47呀？

2024-09-15 13:02 1 · 回答

NO.PZ2020011101000051问题如下 Suppose you are interestein approximating the expectevalue of option. Baseon initisample of 100 replications, you estimate ththe fair value of the option is US47 using the meof these 100 replications. You also note ththe stanrviation of these 100 replications is US12.30. How many simulations woulyou neeto run in orr to obtain a 95% confinintervthis less th1% of the fair value of the option? How many woulyou neeto run to get within 0.1%? The stanrviation is US12.30, ana 95% confinintervis [μ^−1.96∗12.30/n,μ^+1.96∗12.30/n][\wihat\mu - 1.96 * 12.30/\sqrt n, \wihat\mu + 1.96 * 12.30/\sqrt n][μ​−1.96∗12.30/n​,μ​+1.96∗12.30/n​] anso the wih is 2∗1.96∗12.30/n2 * 1.96 * 12.30/\sqrt n2∗1.96∗12.30/n​ . If we want this value to 1% of US47.00, then 2∗1.96∗12.30/n2*1.96 * 12.30/\sqrt n2∗1.96∗12.30/n​=0.47 ⇒n=2∗1.96∗12.30/0.47=102.5\Rightarrow\sqrt n= 2 * 1.96 * 12.30 /0.47 = 102.5⇒n​=2∗1.96∗12.30/0.47=102.5 (so 103), n=1032=10609Using 0.1%, we woulnee1,025.8 (repla0.47 with 0.047) anso 1,026 replication, so n=10262 =1052676 根号n是102.5求n不是应该先平方再取整吗？先取整再平方误差很大

2023-05-21 11:25 1 · 回答

NO.PZ2020011101000051问题如下 Suppose you are interestein approximating the expectevalue of option. Baseon initisample of 100 replications, you estimate ththe fair value of the option is US47 using the meof these 100 replications. You also note ththe stanrviation of these 100 replications is US12.30. How many simulations woulyou neeto run in orr to obtain a 95% confinintervthis less th1% of the fair value of the option? How many woulyou neeto run to get within 0.1%? The stanrviation is US12.30, ana 95% confinintervis [μ^−1.96∗12.30/n,μ^+1.96∗12.30/n][\wihat\mu - 1.96 * 12.30/\sqrt n, \wihat\mu + 1.96 * 12.30/\sqrt n][μ​−1.96∗12.30/n​,μ​+1.96∗12.30/n​] anso the wih is 2∗1.96∗12.30/n2 * 1.96 * 12.30/\sqrt n2∗1.96∗12.30/n​ . If we want this value to 1% of US47.00, then 2∗1.96∗12.30/n2*1.96 * 12.30/\sqrt n2∗1.96∗12.30/n​=0.47 ⇒n=2∗1.96∗12.30/0.47=102.5\Rightarrow\sqrt n= 2 * 1.96 * 12.30 /0.47 = 102.5⇒n​=2∗1.96∗12.30/0.47=102.5 (so 103), n=1032=10609Using 0.1%, we woulnee1,025.8 (repla0.47 with 0.047) anso 1,026 replication, so n=10262 =1052676 想问一下老师能根据经验总结一下需要记哪些关键zhi

2023-05-08 10:39 1 · 回答

NO.PZ2020011101000051问题如下 Suppose you are interestein approximating the expectevalue of option. Baseon initisample of 100 replications, you estimate ththe fair value of the option is US47 using the meof these 100 replications. You also note ththe stanrviation of these 100 replications is US12.30. How many simulations woulyou neeto run in orr to obtain a 95% confinintervthis less th1% of the fair value of the option? How many woulyou neeto run to get within 0.1%? The stanrviation is US12.30, ana 95% confinintervis [μ^−1.96∗12.30/n,μ^+1.96∗12.30/n][\wihat\mu - 1.96 * 12.30/\sqrt n, \wihat\mu + 1.96 * 12.30/\sqrt n][μ​−1.96∗12.30/n​,μ​+1.96∗12.30/n​] anso the wih is 2∗1.96∗12.30/n2 * 1.96 * 12.30/\sqrt n2∗1.96∗12.30/n​ . If we want this value to 1% of US47.00, then 2∗1.96∗12.30/n2*1.96 * 12.30/\sqrt n2∗1.96∗12.30/n​=0.47 ⇒n=2∗1.96∗12.30/0.47=102.5\Rightarrow\sqrt n= 2 * 1.96 * 12.30 /0.47 = 102.5⇒n​=2∗1.96∗12.30/0.47=102.5 (so 103), n=1032=10609Using 0.1%, we woulnee1,025.8 (repla0.47 with 0.047) anso 1,026 replication, so n=10262 =1052676 问题原理不明没法应用

2023-02-02 17:28 1 · 回答

NO.PZ2020011101000051 请问为什么1%对应的是0.47？0.1%对应的是0.047呢？

2021-05-02 19:35 1 · 回答