# Prove convergence in distribution for n times the minimum of an unknown positive distribution

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 $$\frac{1}{\lambda}$$.

Hints:

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.

• Yes, that makes a lot of sense. Thanks, @user15855. – MSE Dec 2 '18 at 13:58

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.