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Suppose $Y$ is an exponential random variable with mean $\theta$. Assume $X_1, \dots, X_n$ are independent uniform $(0, \theta)$ random variables, $\theta > 0$ a constant. Let $X_{(n)}$ be the largest order statistic.

Show that $$n[X_{(n)}-\theta]\overset{d}{\to}(-Y)$$ as $n \to \infty$.

Denote $Z = n[X_{(n)}-\theta]$. Since $\theta \geq X_{(n)}$, it follows that the support of $Z$ is $(-\infty, 0)$. Let $z < 0$. Then $$\mathbb{P}(Z \leq z) = \mathbb{P}(n[X_{(n)}-\theta]\leq z) = \mathbb{P}\left(X_{(n)} \leq \dfrac{z}{n}+\theta\right)\text{.}$$ For $x \in (0, \theta)$, $$\mathbb{P}(X_{(n)} \leq x) = [\mathbb{P}(X_1 \leq x)]^{n}=\left(\dfrac{x}{\theta}\right)^{n}\text{.}$$ Hence, as long as $z \in (-n\theta, n(1-\theta))$, $$\mathbb{P}\left(X_{(n)} \leq \dfrac{z}{n}+\theta\right) = \left(\dfrac{\frac{z}{n}+\theta}{\theta}\right)^n = \left(1+\dfrac{z/\theta}{n}\right)^n$$ which, as $n \to \infty$, goes to $e^{z/\theta}$.

Since $Y$ is exponential with mean $\theta$, set $W = -Y$. For $w < 0$, $$\mathbb{P}(W \leq w) = \mathbb{P}(-Y \leq w) = \mathbb{P}(Y \geq w) = e^{-w/\theta}\text{.}$$ Why is it that these two don't match? I.e., $$\lim_{n \to \infty}\mathbb{P}(Z \leq z) \neq \mathbb{P}(W \leq z)\text{?}$$

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You made a sign flip error, note that $$ \def\P{\mathbf P} \P(W \le w) = \P(-Y \le w) = \P(Y \ge {\color{red}-w}) = e^{w/\theta} $$ Hence - as it should be - $$ \lim_{n\to \infty} \P(Z \le z) = e^{z/\theta} = ˙\P(-Y \le z) $$ Therefore $Z_n = n(X^{(n)} - \theta) \to -Y$ in distribution.