$\textbf{The Problem:}$ Suppose that $X_1,X_2,\dots$ are iid random variables with PDF $$f(x)=\begin{cases}x^{-2}&\text{if }x\geq1\\0&\text{otherwise.} \end{cases}$$ Let $M_n=\max\{X_1,\dots,X_n\}$. Show that $M_n/n$ converges in distribution, and identify the CDF of the limiting distribution.

$\textbf{My Thoughts:}$ The independence of the random variables implies that $$\begin{align*}\mathsf P(M_n\leq nx)&=\mathsf P(X_1\leq nx)^n\\&=\left(\int_{1}^{nx}\frac{1}{y^2}dy\right)^{n}\\&=\left(1-\frac{1}{nx}\right)^{n} \end{align*}$$ for all $x$ such that $nx>1$. The above converges to $\large e^{-x^{-1}}$ as $n\to\infty$. Therefore I conclude that the CDF of the limiting distribution is given by $$\mathsf F(x)=\begin{cases}e^{-x^{-1}}&\text{if }x>0\\0&\text{otherwise.} \end{cases}$$ Therere the PDf is given by $$\mathsf f_1(x)=\begin{cases}\displaystyle\frac{e^{-x^{-1}}}{x^2}&\text{if }x>0\\0&\text{otherwise.} \end{cases}$$ Then I compute that $$\int_{-\infty}^{\infty}\mathsf f_1(x)dx=1.$$ The proposed CDF gives an appropriate PDF, however I think I made a mistake in taking $x>0$ and hence in the overall calculation.

Is my reasoning above correct?

Thank you for your time and any feedback provided is much appreciated.

  • $\begingroup$ the PDF does not integrate to 1 $\endgroup$ – Cettt Apr 26 '19 at 11:33
  • $\begingroup$ @Cettt I added the PDF I get, and I checked the integrationa and it yields 1. But I most likely have a mistake on my PDF then, please correct me I am wrong in my reasoning. $\endgroup$ – Stackman Apr 26 '19 at 11:58

Your computation is fine. If you feel uncomfortable for introducing the condition $x > 0$, it may help to utilize the indicator notation $\mathbf{1}(\cdots)$, which is defined as

$$ \mathbf{1}(\cdots) = \begin{cases} 1, &\text{if $\cdots$ holds}; \\ 0, &\text{if $\cdots$ does not hold}. \end{cases} $$

Using this, we have $f(x) = x^{-2}\mathbf{1}(x \geq 1)$, and so,

$$ \mathsf{P}(X \leq x) = \int_{-\infty}^{x} f(t) \, \mathrm{d}t = \int_{-\infty}^{x} t^{-2}\mathbf{1}(t \geq 1) \, \mathrm{d}t = (1 - x^{-1}) \mathbf{1}(x \geq 1). $$

Then it follows that

$$ \mathsf{P}(M_n/n \leq x) = \mathsf{P}(X_1 \leq nx)^n = \left(1 - \frac{1}{nx}\right)^n \mathbf{1}(x \geq 1/n) \xrightarrow[n\to\infty]{} e^{-1/x} \mathbf{1}(x > 0). $$

This naturally produces the range of $x$ for which the limiting CDF coincides $e^{-1/x}$.

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