# Probability of limit of discrete uniform r.v

Let $$\mathrm{X}_{\mathrm{1}}=1$$ and $$\mathrm{X}_{\mathrm{i}}, 1<\mathrm{i}\leq\mathrm{N}$$, be $$\mathrm{N}$$ independent and uniformly distributed random variables over the set $$\{1 / \mathrm{i}, 2 / \mathrm{i}, \ldots,(\mathrm{i}-1) / \mathrm{i}\} .$$ As $$\mathrm{N}$$ tends towards infinity,what is the probability that $$\mathrm{N} \cdot \min \left(\mathrm{X}_{1}, \mathrm{X}_{2}, \ldots, \mathrm{X}_{N}\right)$$ is greater than 2018?

Tentative solution: \begin{align*}P\left(N\min \left(X_{1}, \ldots, X_{N}\right) > 2018\right) &= P\left(\min \left(X_{1}, \ldots, X_{N}\right)>\frac{2018}{N}\right) \\ &= P\left(X_{1}>\frac{2018}{N}, \ldots, X_{N}>\frac{2018}{N}\right) \\ &= P\left(X_{1}>\frac{2018}{N}\right) \ldots P\left(X_{N}>\frac{2018}{N}\right) \end{align*}

I am not sure on how to deal with $$\lim_{N\to\infty} P(X_{1}>\frac{2018}{N}) \ldots P\left(X_{N}>\frac{2018}{N}\right)$$

• Evaluate one of the terms? Sep 1, 2021 at 8:05
• If $N> 2018$ then $P\left(X_{N}>\frac{2018}{N}\right) <1$ Sep 1, 2021 at 22:33
• Yes I understand, I think this version it's how the problem was supposed to be. Sep 2, 2021 at 8:25
• Cross-posted at math.stackexchange.com/questions/4237551/….
– JimB
Sep 3, 2021 at 5:06
• Answered at math.stackexchange.com/questions/4237551/… Nov 15, 2021 at 14:42