还是没有理清mean exception 和统计量判断。

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. 我可以计算出Z= (x-pxT)/sqt[px(1-p)xT] = (25-1%x1000)/sqt[1x(1-1%)x1000] = 4.77 大于2.58，所以拒绝原假设，因此这是一个bmol。但是从哪判断这是Type I 还是type II risk？ 看了之前的，还是不知道什么意思。

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？

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. 大概能懂题意，我们对银行的一个模型进行backtesing， 然后银行的模型给的VaR是99%，而我们实际测出来的是1000次25个exception。所以我们的结论是这个模型不准，但是因为原假设H0是模型是准确的，而我们的结果拒绝了原假设，所以我们犯了一类错误？有点懵了，那我们既然犯错误了，那对于这个mol不准确的结论成立吗？或者说这个犯一类错误的意义是啥，有点没懂。不知道表达清楚没有。

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. 这个题可不可以另一种解法，回测的标准是99%，所以p＝0.01，代入T＝1000，和z＝2.58，反求x，算出x＝18+。而题目说exceptions是25，则说明这个模型做的不好，25在拒绝域，易犯第一类错误，这样也能得出正确答案

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. 如题