determining the limiting distribution of a consistent estimator

Question

Setting:

$X_i$ ~ i.i.d $Uniform(\theta,2\theta)$, and say we have shown that $X_{(1)}$, the first ordered statistic, is a consistent estimator for $\theta$. Now we want to show that

$$n \left( \frac{X_{(1)}}{\theta} - 1 \right) \longrightarrow_d G_1$$

Where $G_1$ is some non degenerate distribution. I have seen that

$$P\left( n(\frac{X_{(1)}}{\theta} - 1) > t \right) = (1-\frac{t}{n})^n \longrightarrow e^{-t}$$

and understand how to show this. The claim is that $e^{-t}$ is the tail distribution of a random variable having the exponential distribution with scale paramenter = 1.

This implies:

$$n \left( \frac{X_{(1)}}{\theta} - 1 \right) \longrightarrow_d Exp(0,1)$$

Knowing that convergence in probability implies convergence in distribution provides a little insight as to why this method works/is a good idea, but I still feel that I do not understand it completely.

For a fixed t, aren't we showing that it converges in probability to a constant? (Maybe an application of Slutsky then gets us to convergence to an exponential distribution, but I do not quite see how).

Moreover, why does the tail distribution imply that the entire variable behaves as an exponential?

Many thanks for your insights!

Answer 1

You are confusing a number of things. What you have shown is that the complementary cumulative distribution function [CDF] of the suitably scaled first order order statistic converges pointwise (in $t$) to the complementary CDF of the exponential distribution, which is the definition of convergence in distribution.