# Asymptotic Distribution of Minimum Uniform Random Variables

I've been working on this problem for a while, and I've made some progress, but I'm still stuck on some parts. I was hoping to get some assistance with this!

Let $$M_n = \min(X_1, ..., X_n)$$ where $$X_i$$ are i.i.d Unif$$(a,b)$$.

First part is to find the distribution of $$M_n$$. This was quite simple, and I ended up getting the CDF as:

$$F_M(m) = 0$$ for $$m ≤ a \\$$

$$F_M(m) = 1 - \left[ 1 - \dfrac{m-a}{b-a}\right]^n$$ for $$m \in (a,b] \\$$

$$F_M(m) = 1$$ for $$x > b$$

Now that the CDF has been found, the next step is to find the asymptotic distribution of $$P_n = n(M_n - a)$$. This is the part I'm stuck on. I thought about working through and finding the distribution of $$P_n$$ by using $$Pr(P_n ≤ t)$$. But when I do this, I end up with the CDF of $$P_n$$ as $$n \rightarrow \infty$$ being 0 for all values of $$t$$, which I know is incorrect since the next part of the question relies on calculating confidence intervals. So, the asymptotic distribution for $$P_n$$ is needed, and shouldn't be 0.

Finally, once I get that asymptotic distribution of $$P_n$$, how can I go about calculating a confidence interval, given specific values say, $$b=3$$, $$n=10$$, and $$M_n=1$$.

• It may help to begin with a standard uniform first and see how that proceeds (since there's fewer constants hanging around to distract you from the required steps, for one thing). One approach is to make use of the behavior of $(1-\frac{x}{t})^t$ as $t$ grows – Glen_b -Reinstate Monica Sep 14 '19 at 3:59
• A confidence interval for which parameter? $a$? – jbowman Sep 15 '19 at 15:39

First, note that you can simplify your problem by setting $$a=0$$ without loss of generality, which gets rid of $$a$$ as a parameter.

Let $$z = nm$$, then $$m = z/n$$ and, substituting,

$$F_Z(z) = 1 -\left[1-{z \over nb}\right]^n$$

for $$0 < z < nb$$.

Remembering our limits, specifically, the one relating $$(1-x/n)^n$$ to $$e^{-x}$$, we can write:

$$\lim_{n \to \infty}F_Z(z) = 1 - e^{-z/b}$$

which is the c.d.f. of an Exponential variate with mean $$b$$.

Some straightforward substitution gets us back to your problem:

$$\lim_{n \to \infty}F_{P_n}(p) = 1 - e^{-p/(b-a)}$$

• Thank you! As an addition, I've found the CDF to simply be $Exp(b-a)$. Do you have any insight on how to do the second part of the question? Calculating confidence intervals? I'm confused because my distribution will be $Exp(10-a)$ which is not fully known, so doing the confidence interval won't work? Or do I need to estimate $a$ using the fact that $Y_n = 6.5$ and $Y_n \xrightarrow{p} a$? – theDerivative Sep 14 '19 at 19:15
• Confidence intervals for which parameters? You are assuming you know $b$, so that leaves only $a$, but your question ... "do I need to estimate $a$..." leaves me uncertain about what you are actually trying to do. – jbowman Sep 16 '19 at 17:00