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