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习题9-5 #67

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simo-an opened this issue Jan 24, 2022 · 0 comments

simo-an commented Jan 24, 2022

习题9-5 举例说明， K近邻方法估计的密度函数不是严格的概率密度函数，其在整个空间上的积分不等于1

解答：

当R足够小时，可将区域R内的概率密度视为一个不变量。

而K近邻方法是设置一种可变宽度的区域，并使得落入每个区域中样本数量为固定的K．这样就不能保证区域R是足够小的，从而导致其在整个空间上的积分不等于1。

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