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Another manifestation of the curse is that the sampling density is proportional to $N^{1/p}$, where $p$ is the dimension of the input space and $N$ is the sample size. Thus, if $n = 100$ represents a dense sample for a single input problem, then $n=100^{10}$ is the sample size required for the same sampling density with $10$ inputs.`

This is from Elements of Statistical Learning... Shouldn't $n=N^p$, isn't formula wrong?

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    $\begingroup$ No, the formula is correct. For a fixed number of points, the sampling density decreases as the dimension increases. $\endgroup$ – Matthew Drury Aug 5 '17 at 15:19
  • $\begingroup$ Oh, I was thinking about it in wrong way. So because distribution density decreases we need need more(100^10)inputs for it to be proportional to one dimensional estimation? $\endgroup$ – econ Aug 5 '17 at 15:25
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Don't know anything about this topic but, if the sample density is 100 and it is proportional to $N^{1/p}$, then you have: $$ 100 = \alpha N^{1/p} $$ which implies $$ \left(\dfrac{100}{\alpha}\right)^p = N $$ So if the constant of proportionality is 1, the required sample size is $100^{10}$.

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