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

\begin{equation} f(x) = \begin{cases} \frac{24}{x^4}, & \text{if}\ x\geq 2 \\ 0, & \text{if x < 2} \end{cases} \end{equation}. 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 \begin{equation} 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}\end{equation}

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

Since our goal is to find a converging distribution in $M_n/n^c$, let $Y_n = M_n/n^c$. \begin{equation}P(Y_n < y) = P(\frac{M_n}{n^c} < y) = P(M_n < n^cy) = (1-\frac{8y^{-3}}{n^{3c}})^{n} \end{equation}. 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).

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  • $\begingroup$ 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? $\endgroup$
    – whuber
    Feb 20 at 15:19