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小爽加油呀 · 2022年12月25日

这个题的思路可以不计算吗

NO.PZ2018122701000033

问题如下:

Basel II requires a backtest of a bank’s internal value at risk (VaR) model (IMA). Assume the bank’s ten-day 99% VaR is $1 million (minimum of 99% is hard-wired per Basel). The null hypothesis is: the VaR model is accurate. Out of 1,000 observations, 25 exceptions are observed (we saw the actual loss exceed the VaR 25 out of 1000 observations).  (Binomial CDF)

选项:

A.

We will probably call the VaR model good (accurate) but we risk a Type I error.

B.

We will probably call the VaR model good (accurate) but we risk a Type II error.

C.

We will probably call the model bad (inaccurate) but we risk a Type I error.

D.

We will probably call the model bad (inaccurate) but we risk a Type II error.

解释:

C is correct.

考点 : Backtesting VaR

解析 :H0 : the VaR model is accurate. Hα: the VaR model is inaccurate.

Z=xpTp(1p)T=251%×10001%×(11%)×1000=4.77Z=\frac{x-pT}{\sqrt{p(1-p)T}}=\frac{25-1\%\times1000}{\sqrt{1\%\times(1-1\%)\times1000}}=4.77

As 4.77 is larger than 2.58, we reject the null hypothesis. Therefore, the model is bad model, and this implies a risk of type I error.

老师,可以直接想成 如果正常的话,拒绝应该是1000*1%=10个,但是实际上是25个,多拒绝了不该拒绝的,所以是弃真,type1,可以吗

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已采纳答案

品职答疑小助手雍 · 2022年12月26日

同学你好,不可以。

这里要用到假设检验,不能只比较均值(只把25拿来和10比较本题中自然差距挺大的,那我要是给11个exception让你做判断呢,你的方法和答案方法的结论就是相反的了)。

出不出现excption是服从二项分布的,根据中心极限定理1000的样本量可以认为算是服从np,np*(1-p)的正态分布了。

也就是1000*1%,1000*1%*99%的正态分布。

后面就是题目中用的常规假设检验的步骤了。

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NO.PZ2018122701000033问题如下 Basel II requires a backtest of a bank’s internal value risk (VaR) mol (IMA). Assume the bank’s ten-y 99% Vis $1 million (minimum of 99% is harwireper Basel). The null hypothesis is: the Vmol is accurate. Out of 1,000 observations, 25 exceptions are observed (we sthe actuloss exceethe V25 out of 1000 observations).  (BinomiC) We will probably call the Vmol good (accurate) but we risk a Type I error. We will probably call the Vmol good (accurate) but we risk a Type II error. We will probably call the mol bad (inaccurate) but we risk a Type I error. We will probably call the mol bad (inaccurate) but we risk a Type II error. C is correct. 考点 Backtesting V解析 H0 : the Vmol is accurate. Hα: the Vmol is inaccurate.Z=x−pTp(1−p)T=25−1%×10001%×(1−1%)×1000=4.77Z=\frac{x-pT}{\sqrt{p(1-p)T}}=\frac{25-1\%\times1000}{\sqrt{1\%\times(1-1\%)\times1000}}=4.77Z=p(1−p)T​x−pT​=1%×(1−1%)×1000​25−1%×1000​=4.774.77 is larger th2.58, we rejethe null hypothesis. Therefore, the mol is bmol, anthis implies a risk of type I error. 如果是bmol就存在 type I error,goomol就存在type II error是这样吗?

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