# Showing convergence to a (non-degenerate) distribution [closed]

The random variables $$X_1$$, ..., $$X_n$$ are i.i.d with density

$$$$f(x) = \begin{cases} \frac{24}{x^4}, & \text{if}\ x\geq 2 \\ 0, & \text{if x < 2} \end{cases}$$$$. Define $$M_n$$ to be $$\max(x_1, ..., x_n)$$. Find the constant c s.t $$M_n/n^c$$ converges to a non-degenerate distribution and obtain the CDF for this limiting distribution.

What I have done so far:

Find the CDF of f(x) which gives me $$F(x) = \begin{cases}1-8x^{-3}, x\geq 2, \\ 0 \text{ otherwise} \end{cases}$$.

Find the CDF of the max order statistic ie. $$M_n$$, which is $$$$P(M_n \leq x) = P(X_1 < x, ..., X_n < x) = \prod_{i=1}^{n} F_x(x) = \begin{cases} (1-\frac{8}{x^3})^{n}, x \geq 2 \\ 0 \text{ otherwise} \end{cases}$$$$

By definition of convergence in distribution, $$$$\lim_{n->\infty} F_{X_n}(t) = F(t)$$$$

Since our goal is to find a converging distribution in $$M_n/n^c$$, let $$Y_n = M_n/n^c$$. $$$$P(Y_n < y) = P(\frac{M_n}{n^c} < y) = P(M_n < n^cy) = (1-\frac{8y^{-3}}{n^{3c}})^{n}$$$$. If we take the limit of $$(1-\frac{8y^{-3}}{n^{3c}})^{n}$$ at c = 1/3, we will have $$e^{-8y^{-3}}$$ by the identity of $$\lim_{n-> \infty} (1-\lambda/n)^n= e^{-\lambda}$$. So it converges to a non degenerate distribution (with it being in the form of the survivor function of an exponential distribution?)

Any directions or comments will be greatly appreciated! (this was a past exam question and I am taking it out for self-study purposes).

• Questions where the unique best answer is "yes" or "no" are not considered suitable here. Could you specify what kind of guidance you are looking for?
– whuber
Feb 20 at 15:19