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I have a question that follows from this question. That discussion shows that if $F(x) = 1 - \frac{1}{2}x^2$ for $x \in [-1, 0)$ and $X_i$ are independent random variables with CDF $F$, then $\max_{1 \leq i \leq n} X_i$ converges almost surely to 0 as $n\to \infty$.

The question now is whether there exists $a_n > 0$ and $b_n$ such that $b_n + a_n\max_{1 \leq i \leq n} X_i$ converges in distribution to a random variable that isn't degenerate.

My thinking is that if such a random variable existed, then there would have to be $x_1 \neq x_2 \neq x_3$ such that $F(x_1) \neq F(x_2) \neq F(x_3)$, but I don't think there can be. Unless $a_n$ and $b_n$ were random variables.

Is this on the right track?

(This question seems similar but there's an additional function $h$.)

D.W.
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johnsmith
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1 Answers1

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Take $b_n=0$ and $a_n=\sqrt{n}$ for all $n \in \{1, 2, 3, ...\}$. Define $$Z_n = b_n + a_n\max_{i \in \{1, ..., n\}}\{X_i\} = \sqrt{n}\max_{i \in \{1, ..., n\}}\{X_i\}$$ Fix $x\leq 0$. Then $x/\sqrt{n}\in [-1, 0]$ for all sufficiently large $n$ and we obtain for all sufficiently large $n$: \begin{align} &P[Z_n\leq x]\\ &= P\left[\sqrt{n}\max_{i \in \{1, ..., n\}}\{X_i\}\leq x\right]\\ &=P\left[X_1\leq x/\sqrt{n}, X_2\leq x/\sqrt{n}, ..., X_n\leq x/\sqrt{n}\right]\\ &=\left(1-\frac{(1/2)x^2}{n}\right)^n \end{align} Thus, for all $x\leq 0$ we obtain: $$\lim_{n\rightarrow\infty} P[Z_n\leq x] = e^{-(1/2)x^2}$$ which is a valid continuous CDF.

Michael
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