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陶朱 · 2024年05月08日

为什么这里是1.96直接乘样本标准差,为什么不是1、96乘样本标准差除以根号下N

NO.PZ2024030508000066

问题如下:

An analyst at a hedge fund is constructing a 95% confidence interval for the 2-month point forecast of the price of a standard option contract on 100 shares of a stock, measured in EUR, using the linear time trend model utilized by the fund. The analyst refers to the model estimated using monthly option prices (OP) over the last 5 years, which is denoted by the following equation:

𝑂𝑃𝑇 + = 318.6 + 12.5(𝑇 + ) + 𝜀𝑇 +

where T (current month) is equal to 0, h is the horizon of the forecast, and 𝜀𝑇 + is Gaussian white noise. The estimate of the residual standard deviation, σ, is 4.62. What is the correct 95% confidence interval for the price of the option contract?

选项:

A.[EUR 322.04, EUR 340.15]

B.[EUR 334.54, EUR 352.65]

C.

[EUR 336.02, EUR 351.18]

D.[EUR 338.98, EUR 348.22]

解释:

Explanation: B is correct. The forecast confidence interval depends on the variance of the forecast error. When the error is Gaussian white noise N(0, σ2), then the constructed confidence interval for the future value follows a normal distribution. The 95% confidence interval for the forecast of the option contract price is given by:

𝐸𝑇 [𝑂𝑃𝑇 + ] ± 1.96𝜎

Note that the time T expectation or forecast of OPT + h is given by:

𝐸𝑇 [𝑂𝑃𝑇 + ] = 318.6 + 12,5(𝑇 + )

Therefore,

𝐸0 [𝑂𝑃0 + 2] = 318.6 + 12.5(0 + 2)

𝐸[𝑂𝑃2] = 318.6 + 12.5(2)

𝐸[𝑂𝑃2] = 343.6

The 95% confidence interval for the forecast is then constructed as:

𝐸[𝑂𝑃2] ± 1.96𝜎 = 343.6 ± 1.96 (4.62) = [334.54 , 352.65].

A is incorrect. This is just 𝐸[𝑂𝑃1 ] ± 1.96𝜎.

C is incorrect. This is the 90% confidence interval, 𝐸[𝑂𝑃2] ± 1.64𝜎.

D is incorrect. This is just 𝐸[𝑂𝑃2] ± 𝜎.

Learning Objective: Calculate the estimated trend value and form an interval forecast for a time series.

Reference: Global Association of Risk Professionals. Quantitative Analysis. New York, NY: Pearson, 2023, Chapter 11, Non-Stationary Time Series [QA-11].

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1 个答案

品职答疑小助手雍 · 2024年05月09日

同学你好,因为这4.62本身就是残差的标准差了,直接拿来计算回归结果的范围就行。

本题不是用中心极限定理求均值的标准差。

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