Limiting distribution of the first order statistic of a general distribution

Question

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.

Answer 1

(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.

Answer 2

To prove convergence in distribution we need to show that the complementary distribution of $X_n$, written $G_n$ where $G_n(x)=\Pr(X_n\gt x)$, gets close to an exponential function for $n$ sufficiently large. To this end, let $t\gt 0$ be an arbitrary point at which to evaluate $G_n(t)$. Note that the independence of the $Z_i$ implies

$$G_n(t) = \left(1 - F\left(\frac{t}{n}\right)\right)^n = \left(1 - \lambda\frac{t}{n} + \left[\lambda\frac{t}{n} - F\left(\frac{t}{n}\right)\right]\right)^n.$$

The term in square brackets is the problem--if it weren't there the limit would obviously be exponential--so we will use the only information available to us to estimate it and hope that it's very small for large $n$. The existence of the limit

$$\lambda = {\lim}_{x\to 0^{+}} f\left(x\right)$$

implies

$$\left|\lambda\frac{t}{n} - F\left(\frac{t}{n}\right)\right| = \left|\int_0^{t/n} (\lambda - f(u)) du\right| \le \frac{t}{n}\sup_{0\le u\le t/n}\left(|\lambda - f(u)|\right) = \frac{t}{n}\varepsilon(n)$$

for some function $\varepsilon$ that approaches $0$ for large arguments. Substitute this into the foregoing and assume $n$ is so large that $F\left(\frac{t}{n}\right)\lt 1$, so that we may take logarithms, and use the Taylor series of the logarithm near $1$ to estimate

$$\eqalign{ \log(G(t))=n\log\left(1 - F\left(\frac{t}{n}\right)\right) &= n\log\left(1 - \lambda\frac{t}{n} + \left[\lambda\frac{t}{n} - F\left(\frac{t}{n}\right)\right]\right) \\ &= n\log\left(1 - \left(\lambda-\varepsilon(n)\right)\frac{t}{n}\right) \\ &= -\left(\lambda-\varepsilon(n)\right)t + \left[(\lambda - \varepsilon(n))t\right]^2O\left(\frac{1}{n}\right). }$$

Clearly (applying theorems about the limits of products and sums of continuous functions) this has a limit as $n\to \infty$ and it equals $-\lambda t$, showing that $G(t)=\exp(\log(G(t))$ has the limiting value $\exp(-\lambda t)$, QED.