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Let $(\lambda_i)_{i=1}^n$ be a colloction of iid random variables and $\lambda_i$ is uniform on $[-1,1]$. What is the distribution of $\rho$, where $\rho=\text{sup}_{i}(|\lambda_{i}|)$? And how to calculate $\mathbb {P}(\rho<1)$? I use matlab to plot this distribution, it looks like a flip of the exponential distribution. But I don't know how to get the exact density function.

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  • $\begingroup$ The absolute value is standard Uniform(0,1), so the question is simply to find the pdf of the sample maximum from a standard Uniform rv. $\endgroup$ – wolfies Dec 4 '18 at 6:39
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$P\{\rho \leq a\}=(P\{|\lambda_1| \leq a\})^{n} =a^{n}$for $0 \leq a \leq 1$. The density function is $ n a^{n-1}$ for $0<a<1$.

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  • $\begingroup$ Why $\mathbb {P}(\rho \leq a)=(\mathbb {P}(|\lambda_1| \leq a))^n$? My understanding is that if I order the value of $\lambda_i$'s from smallest to largest, like $\lambda_1, \lambda_2, \ldots, \lambda_n$, then $\mathbb {P}(\rho \leq 1)=\mathbb {P}(|\lambda_1| \leq 1, |\lambda_n| \leq 1)$, but $|\lambda_1| \leq 1$ and $|\lambda_n| \leq 1$ are not independent. $\endgroup$ – Jiexiong687691 Dec 4 '18 at 5:39
  • $\begingroup$ @Jiexiong687691 It is given that $\lambda_i$ are i.i.d. So the events $\{\lambda_i \leq a\}$ are independent. $\endgroup$ – Kabo Murphy Dec 4 '18 at 5:41
  • $\begingroup$ Can $\{\lambda_{i}\leq a\}$ are independent imply that $\{|\lambda_{i}| \leq a\}$ are independent? $\endgroup$ – Jiexiong687691 Dec 4 '18 at 5:50
  • $\begingroup$ Yes. If a collection of random variable is independent then any measurable functions of them are also independent. In fact, in this case $|\lambda_i| \leq a$ iff $-a \leq \lambda_i \leq a$ and these events are independent. $\endgroup$ – Kabo Murphy Dec 4 '18 at 5:53
  • $\begingroup$ Sorry, I have a further question that what happens to the random variable $Y_n=\text{max}(|\lambda_1|, |\lambda_n|)$ where $\lambda_1=\text{min}\lambda_i$ and $\lambda_n=\text{max}\lambda_n$. $\lambda_i$'s are iid random variables on [-1,1] when $n\to \infty$? $\endgroup$ – Jiexiong687691 Dec 4 '18 at 6:25

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