proving converse of equality involving distribution of minimum observation

Question

Suppose constants $v_n$ are such that: $\lim_{n \to \infty} nF(v_n) =d \in [0,\infty]$ where F is the Cumulative distribution function of $X_i \sim $ i.i.d. random variables. Then the question is to show that for $m_n = \min\{X_1,\dots.,X_n \}$ $$ \lim_{n \to \infty} P(m_n > v_n) =e^{-d}. $$ Now this follows by the following argument: $$ \lim_{n \to \infty} P(m_n > v_n) = \lim_{n \to \infty} (1-F_x(v_n))^n $$ $$ = \lim_{n \to \infty} \left(1-\frac{nF_x(v_n)}{n}\right)^n = e^{-d} $$ I want to know whether the converse is true, that is, whether knowing $\lim_{n \to \infty} P(m_n > v_n) =e^{-d} \implies \lim_{n \to \infty} nF(v_n) =d \in [0,\infty]$ and how I could prove it.

Answer 1

You could prove using the contraposition.

Since $\limsup_n nF(v_n)>d+ \epsilon$ implies that by your argument that

$$\liminf_{n \to \infty} P(m_n > v_n) \leq e^{-(d+\epsilon)}< e^{-d} $$

So $\limsup_n nF(v_n) \leq d$

In the same manner $\liminf_{n \to \infty} nF(v_n)< d -\epsilon$ we have that

$$\limsup_{n \to \infty} P(m_n > v_n) \geq e^{-(d-\epsilon)}> e^{-d} $$

So $\liminf_n nF(v_n) \geq d$ and therefore $\lim_{n \to \infty} nF(v_n) =d \in [0,\infty]$