Take the 2-minute tour ×
Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It's 100% free, no registration required.

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.

share|improve this question
1  
One approach is to apply the Fisher-Tippett-Gnedenko theorem to $-Z_i$. –  whuber Jun 9 at 17:05
    
@whuber Thank you, I didn't even know that such thing existed. I wonder though whether a simpler way exists as my knowledge does not stretch that far yet. –  JohnK Jun 9 at 17:08
1  
I believe a study of that theorem will pay off here, if only to suggest that what you are trying to prove might not be true--or at least requires some strong additional assumptions about $F$. –  whuber Jun 9 at 17:14
1  
Thank you for the reference: in reading it I see I overlooked the assumption that the limiting value of the density at zero is strictly positive: that's a crucial assumption. Intuitively, it tells you that the distribution of the minimum of a very large number of independent variables will be controlled by the value of $f$ near $0$. So one way to appreciate this problem would be to replace $F$ by, say, a uniform distribution, which would have to be U$(0,1/\lambda)$. You can compute the distribution of $X_n$ exactly in this case: what is it and what is its limit as $n\to\infty$? –  whuber Jun 9 at 19:13
1  
Thank you @AlecosPapadopoulos. You can certainly post an answer, if you like so I can mark it as the correct one. My sole objection is that the limit we are taking in this case is the limit of a pdf and I do not think we are supposed to do that,in fact I have seen many counterexamples of this practice. Are we not supposed to examine CDF's only? –  JohnK Jun 9 at 20:15
show 4 more comments

2 Answers 2

up vote 3 down vote accepted

(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 functions 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 $(i)$ the limit of $h_n$ is a continuous function (it is), $(ii)$ the functions $g_n(x)$ map $\mathbb R_+$ to $\mathbb R$ (they do), and $(iii)$ $h_n$ converges uniformly to $h$. This too holds (convergence to $e^w$ is uniform). So the above equality holds and we have proven what we needed to prove.

share|improve this answer
    
Beautiful, thank you. Are we in the clear about moving the limit inside the integral? –  JohnK Jun 9 at 21:17
1  
There is a subtle issue that seems not to be justified in this demonstration: when taking the limits, the same $n$ is involved in the limit of the integrals and the limit of the powers (the antepenultimate and penultimate lines, respectively): without specific analysis, you cannot take those limits separately. (It was the consideration of this point that originally caused me to present a slightly longer answer.) –  whuber Jun 9 at 21:27
    
@whuber whenever you have some time, I would appreciate if you could have a look at my re-worked answer - I tried to respond to your comment. –  Alecos Papadopoulos Jun 10 at 0:22
    
You have definitely answered my question, thank you. –  JohnK Jun 10 at 7:54
1  
I believe you can seriously consider amazon.com/… –  Alecos Papadopoulos Jun 11 at 22:27
show 1 more comment

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.

share|improve this answer
    
Referring to the question's comment thread, when $n$ is very large so that $t/n$ is very small (but positive), we can characterize $\lambda\frac{t}{n} - F\left(\frac{t}{n}\right)$ as the discrepancy between $F$ and a Uniform$(\lambda)$ distribution near $0$. Thus this argument is simply a semi-rigorous restatement of the intuition expressed in those comments. (For even more rigor one would write a full epsilon-delta limiting argument, but it should now be obvious how that proof would go.) –  whuber Jun 9 at 20:58
    
Thank you very much both for your answer and for your comments. –  JohnK Jun 9 at 21:15
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.