# Confusion with Convergence in Distribution of Maximum of iid Random Variables

$$\textbf{The Problem:}$$ 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.

• the PDF does not integrate to 1 – Cettt Apr 26 '19 at 11:33
• @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. – 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 $$e^{-1/x}$$.