Let $Z_i,Z_2,\ldots$ be IID Random Variables with density $f$. Suppose that $P(Z_i>0)=1$ and that $\lambda=\lim_{x \to 0+} f(x)>0$. How can I show that $X_n=n \times \min\{Z_i\}$ has a limiting exponential distribution with mean$1/\lambda$?
I know that the CDF of the first order Statistic is for $t>0$
$$F_{min \{ Z_i \}} (t)= 1- \left[ 1-F(t) \right]^{n}$$
Convergence in distribution requires that $F_n (t) \to F(t)$ for some distribution function $F$ but I cannot quite take the limit of the above, with $t$ replaced by $x/n$, as the distribution is not known. I guess I have to insert the second piece of information somewhere but I do not recognise where.
Could you please give me a hint or two?
Thank you.
Best Answer
(The answer has been reworked to respond to OP's and whuber's comments).
The complementary cdf of $X$ is
$$G_n(x) = \left[1-F_Z\left(x/n\right)\right]^{n}$$
To prove that asymptotically $X$ follows an exponential distribution, we need to show that $$\lim_{n\rightarrow \infty}G_n(x)= e^{-\lambda x}$$
Consider
$$F_Z\left(x/n\right) = \int_0^{x/n}f(t)dt $$
By the properties of the integral, we have
$$\int_0^{x/n}f(t)dt = \frac 1n\int_0^{x}f(t/n)dt$$
Define $$h_n(w) = \left(1+\frac {w}{n}\right)^{n}, \qquad \lim_{n\rightarrow \infty}h_n(w) = e^w=h(w), \;\; w \in \mathbb R$$
and
$$g_n(x) = -\int_0^{x}f(t/n)dt,\;\;\; -\lim_{n\rightarrow \infty}g_n(x) = -\int_0^{x}f(0)dt = -\lambda x = g(x), \;\;x \in \mathbb R_+$$
(To respond to a question by the OP, we can take the limit inside the integral. First note that $n\geq 1$, and we do not send $x$ to infinity. So the argument of $f$ does not explode. So even if it were the case that $f(\infty) \rightarrow \infty$, we do not need to consider this case here. Then, since also $f(0)$ is finite by assumption, $f$ is bounded and dominated convergence holds).
With these definitions we can write
$$G_n(x) = h_n(g_n(x))$$
and the question is
$$ \lim_{n\rightarrow \infty}h_n(g_n(x)) =?\;\; h(g(x)) = e^{-\lambda x},\;\;x \in \mathbb R_+$$
The limit of a composition of function-sequences does not in general equal the composition of their limits (which is what whuber has essentially pointed out in his comment). But this equality will hold if
$(i)$ $h_n$ converges uniformly to $h$ (it does-convergence to $e^w$ is uniform)
$(ii)$ the limit of $h_n$ is a continuous function (it is)
$(iii)$ the functions $g_n(x)$ map $\mathbb R_+$ to $\mathbb R$ (namely, they map their domain into the set where $h_n$ converges -they do).
So the above equality holds and we have proven what we needed to prove.