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Suppose $Y_i$ are iid random variables and $\frac{ \max_{i \leq n} Y_i - a_n}{b_n} \rightarrow X$ weakly, where $X$ is nonconstant and $a_n, b_n > 0$. Let $f$ be continuous of compact support. I wish to show that $ \sum_{i=1}^n f(\frac{Y_i-a_n}{b_n}) \rightarrow \int f dN$ weakly where $N$ is a Poisson point process.

I've managed to show using independence that $nP(\frac{Y_n - a_n}{b_n} \geq \alpha) \rightarrow -\log(F_X(\alpha))$ where $F_X$ is the cdf of $X$. But I'm unsure of how to apply the Poisson convergence theorem, and bring in the $f$. I am sure that whatever $N$ turns out to be will be obtained by the Poisson convergence theorem.

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  • $\begingroup$ I think maybe you mean $n \log P(\frac{Y_n-a_n}{b_n} \leq \alpha) \to \log F_X(\alpha)$. $\endgroup$ – Shalop Feb 7 '18 at 7:38
  • $\begingroup$ Actually I think I had forgotten a negative sign in front of the log. I've put it in. $\endgroup$ – user394438 Feb 7 '18 at 7:53
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    $\begingroup$ I've never heard "Poisson convergence theorem." Anyways, I did some asymptotic expansions in the simple case where $f=1_{(-\infty, \alpha]}$. Based on characteristic function computations, I believe that the intensity measure $\mu$ of the Poisson Point process should be given by $$\mu(\alpha,\beta]= \log F_X(\beta) - \log F_X(\alpha),\;\;\;\;\;\;\; \alpha<\beta \in \Bbb R.$$ i.e., the associated increasing function is $\log F_X$. I could be wrong though. $\endgroup$ – Shalop Feb 7 '18 at 8:05
  • $\begingroup$ Yes I agree with that. $\endgroup$ – user394438 Feb 7 '18 at 8:21
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We give a direct proof.

As in the comment above, we define the intensity measure as $\mu(a,b] := \log F_X(b)-\log F_X(a)$. To prove the claim, it suffices to prove convergence of the associated Laplace functionals, i.e., for all $f \in C_c^{\infty}(\Bbb R)$ one has

$$E\bigg[\exp\bigg({f\big(\frac{Y_i-a_n}{b_n}\big)\bigg)}\bigg]^n \;\;\;\stackrel{ n \to \infty}{\longrightarrow} \;\;\;\exp \bigg( \int_{\Bbb R} (e^{f(x)}-1)\mu(dx) \bigg).$$ If $g$ is a $C^1$ function supported on some interval $[a,b]$, then as a consequence of Fubini's theorem, we have $E[g(X)] = \int_a^b g'(x) P(X \geq x) dx$. Applying to $g:=e^f-1$, we find that $E[e^{f(X)}] = 1+\int_a^bf'(x)e^{f(x)}P(X \geq x)dx.$ So we find that \begin{align*} E\bigg[\exp\bigg({f\big(\frac{Y_i-a_n}{b_n}\big)\bigg)}\bigg] = 1+\int_a^b f'(x)e^{f(x)} \big( 1- F_Y(a_n x+b_n)) \big)dy \end{align*} where $F_Y$ denotes the cdf of the $Y_i$. On the other hand, the given convergence in distribution of $\max Y_i$ implies that $F_Y(a_nx+b_n)$ behaves asymptotically like $F_X(x)^{1/n}$, which in turn behaves like $1+\frac{1}{n}\log F_X(x)+O(\frac{1}{n^2})$. Consequently, we find that \begin{align} \lim_{n \to \infty} E\bigg[\exp\bigg({f\big(\frac{Y_i-a_n}{b_n}\big)\bigg)}\bigg]^n &= \lim_{n \to \infty} \bigg[ 1-\frac{1}{n} \int_a^b f'(x)e^{f(x)} \log F_X(x)dx \bigg]^n \\ &= \exp \bigg[ -\int_a^bf'(x) e^{f(x)} \log F_X(x) dx \bigg] \\ &= \exp \bigg[ \int_{\Bbb R} (e^{f(x)}-1) \mu(dx) \bigg] \end{align} where we again used fubini's theorem in the final line.

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