Distribution of sup of uniform random variables

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.

• 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. – wolfies Dec 4 '18 at 6:39

$$P\{\rho \leq a\}=(P\{|\lambda_1| \leq a\})^{n} =a^{n}$$for $$0 \leq a \leq 1$$. The density function is $$n a^{n-1}$$ for $$0.
• 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. – Jiexiong687691 Dec 4 '18 at 5:39
• @Jiexiong687691 It is given that $\lambda_i$ are i.i.d. So the events $\{\lambda_i \leq a\}$ are independent. – Kabo Murphy Dec 4 '18 at 5:41
• Can $\{\lambda_{i}\leq a\}$ are independent imply that $\{|\lambda_{i}| \leq a\}$ are independent? – Jiexiong687691 Dec 4 '18 at 5:50
• 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. – Kabo Murphy Dec 4 '18 at 5:53
• 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 $n\to \infty$? – Jiexiong687691 Dec 4 '18 at 6:25