完美样本

题目说明样本中离群点所占比例为 $e$，因此样本中不是离群点的比例可以表示为 $1-e$。

由于我们希望以 99% 的置信度至少得到一个完全不含离群点的样本，因此希望满足 $P(\textrm{Every Sample Contains Outlier}) \leq 0.01$。

$$\begin{equation*} P(\textrm{Sample Contains 0 Outliers}) = (1-e)^s \end{equation*} \\\ \\ P(\textrm{Sample Contains at Least 1 Outlier)} = (1 - (1-e)^s) \\\ \\ P(\textrm{All Samples Contain at Least 1 Outlier}) = (1 - (1-e)^s)^N \\\ \\ P(\textrm{All Samples Contain at Least 1 Outlier}) \leq (1-p) \\\ \\ (1 - (1-e)^s)^N \leq (1-p)$$

至此，我们已经得到一个约束条件：所有样本都含有离群点的概率必须小于 1%。进一步解出 $N$，得到所需样本数：

$$\begin{equation*} (1 - (1-e)^s)^N \leq (1-p) \end{equation*} \\\ \\ N \gt \frac{\log(1-p)}{\log(1 - (1-e)^s)} \\\ \\ \boxed{N \gt \frac{\log(1-0.99)}{\log(1 - (1-e)^s)}}$$

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Original Explanation

The problem states that the proportion of points in a sample that are outliers is $e$, and therefore the proportion of points in the sample that are not outliers can be represented as $1-e$.

Since we want to be 99% confident that we get at least one sample that is completely void of outliers, we want to make the $P(\textrm{Every Sample Contains Outlier}) \leq 0.01$

$$\begin{equation*} P(\textrm{Sample Contains 0 Outliers}) = (1-e)^s \end{equation*} \\\ \\ P(\textrm{Sample Contains at Least 1 Outlier)} = (1 - (1-e)^s) \\\ \\ P(\textrm{All Samples Contain at Least 1 Outlier}) = (1 - (1-e)^s)^N \\\ \\ P(\textrm{All Samples Contain at Least 1 Outlier}) \leq (1-p) \\\ \\ (1 - (1-e)^s)^N \leq (1-p)$$

Finally, we've arrived at an equation that restricts the probability that all the samples contain an outlier to less than 1%. Finally, we can solve for $N$ to find the number of samples we would need:

$$\begin{equation*} (1 - (1-e)^s)^N \leq (1-p) \end{equation*} \\\ \\ N \gt \frac{\log(1-p)}{\log(1 - (1-e^s))} \\\ \\ \boxed{N \gt \frac{\log(1-0.99)}{\log(1 - (1-e^s))}}$$