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ZF Everyday · 2021年05月02日

0.47

NO.PZ2020011101000051

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^1.9612.30/n,μ^+1.9612.30/n][\widehat\mu - 1.96 * 12.30/\sqrt n, \widehat\mu + 1.96 * 12.30/\sqrt n] and so the width is 21.9612.30/n2 * 1.96 * 12.30/\sqrt n .

If we want this value to be 1% of USD 47.00, then 21.9612.30/n2*1.96 * 12.30/\sqrt n=0.47 n=21.9612.30/0.47=102.5\Rightarrow\sqrt n= 2 * 1.96 * 12.30 /0.47 = 102.5 (so 103), n=1032=10609

Using 0.1%, we would need 1,025.8 (replace 0.47 with 0.047) and so 1,026 replication, so n=10262 =1052676

请问为什么1%对应的是0.47?0.1%对应的是0.047呢?

1 个答案
已采纳答案

品职答疑小助手雍 · 2021年05月02日

嗨,努力学习的PZer你好:


因为均值是47,需要interval是在均值加减1%范围内的,那就对应0.47啦

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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

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2023-02-02 17:28 1 · 回答