Let $Z_1, Z_2, ...$ be independent and identically distributed random variables with some density $f$. Suppose that $P(Z_i > 0) = 1$, and that

$$ \lambda = \lim_{x\to 0} f(x) > 0$$

Let $X_n = nZ_{(1)}$, where $Z_{(1)}$ denotes the minimum of all of the $Z_i$ variables.

Show that $X_n$ converges in distribution to $Y$, where $Y$ has an exponential distribution with mean $\frac{1}{\lambda}$.

I have no idea how to start this. Please help.


2 Answers 2


Sketch of argument: For $x > 0$,

$$P(nZ_{(1)} > x) = P\left(Z_{(1)} > \frac xn\right) = \prod_{k=1}^nP\left(Z_k > \frac xn\right) = \left(1-F_Z\left(\frac xn\right)\right)^n.$$ But for small $a>0$, $(1-a)^n \approx \exp(-na)$ (Taylor series match for first two terms) and so we have that $$P(nZ_{(1)} > x) \approx \exp\left(-nF_Z\left(\frac xn\right)\right).$$ Now argue that $F_Z\left(\frac xn\right)$ is approximately equal to the limiting value $\lambda$ of the density $f_Z$ as the argument approaches $0$ times the length $\frac xn$ of the short interval $\left[0,\frac xn\right]$ over which we must integrate the pdf to find the CDF value at $\frac xn$ to arrive at the conclusion that $$\lim_{n\to \infty} P(nZ_{(1)} > x) = \exp(-\lambda x),$$ and so the limiting distribution of $nZ_{(1)}$ is exponential with parameter $\lambda$. Putting in all the epsilons and deltas to make this a proof is left as an exercise for the OP.



This is definition of converging in disribution.

$\lim _{n\to \infty }F_{n}(x)=F(x)$

So need to find the CDF of $X_n$.

Because $X_n =nZ_{(1)}$, So need to find CDF of $Z_{(1)}$. You can search the distribution of minimum ... on internet.

Then go back step by step and use the conditions given in the question, you will finish.

  • $\begingroup$ Yes, that makes a lot of sense. Thanks, @user15855. $\endgroup$
    – MSE
    Dec 2, 2018 at 13:58

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