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I'm trying to understand algorithm L from reservoir sampling. One of the steps mentions that given n samples independently drawn from a uniform distribution U(0,1), we can simulate drawing n samples and getting the max with just drawing 1 sample can raising it to the power of 1/n.

That is max{random()_1, ...random()_n} = random()^(1/n)

Does anyone know how we get this result? Thanks

Joel Lim
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Mar 12 '23 at 09:32

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Define $z \sim D_z :=\max(r_1,\dots, r_n)$ for $r_i \sim_{\text{iid}} U(0,1)$ is $P(z \le s) = s^n$, (Proof), and reference its distribution by $D_z$.

Further, we can define $a \sim D_a:=U(0,1)^{1/n}$ and $t=a^{n}$, we have $P(a \le s) = P(t\le s^{n}) =s^{n}$.

Therefore, $P(z \le s)=s^n=P(a \le s)$, implying $D_a=D_z$ as suggested.

Snared
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