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[This question is a refinement of my other question here for a specific $b_{n,k}$]

Consider

  • a mapping $u$ on $\mathbb{N} \mapsto (0,\infty)$ with $k \mapsto u(k)\equiv u_k$, bounded in absolute value by $\bar{u}<\infty$.

  • a mapping $a$ on $\mathbb{N} \mapsto \mathbb{R}$ with $n\mapsto a(n)\equiv a_n$.

  • a mapping $b$ on $\mathbb{N} \mapsto \mathbb{R}$ with $n\mapsto b(n)\equiv b_n$.

$\forall t \in \mathbb{R}$ $\forall n \in \mathbb{N}$, let $$ c_n(t)\equiv \sum_{k=1}^{2n} \log(G(a_n*(t+u_k)+b_n)) $$ and $$ \tilde{c}_n(t) \equiv \sum_{k=1}^{2n} (1-G(a_n*(t+u_k)+b_n)) $$ Assume also that

  • $G:\mathbb{R}\rightarrow [0,1]$

  • $G$ continuous and strictly monotone increasing on $\mathbb{R}$

  • $\lim_{x\rightarrow \infty}G(x)=1$

  • $\lim_{x\rightarrow -\infty}G(x)=0$

  • ADDED: $G$ is such that $$ \exists \text{ }A:\mathbb{R}\rightarrow (0,\infty) \text{ s.t. } \lim_{s\rightarrow \infty }\frac{1-G(s+v A(s))}{1-G(s)} = e^{-v}\text{ }\forall v \in \mathbb{R} $$ Moreover, $A(s)\equiv \frac{1-G(s)}{g(s)}$ where $G(s)\equiv \int_{-\infty}^s g(e) de$ $\forall s \in \mathbb{R}$.

  • $b_n\equiv G^{-1}(1-\frac{1}{n})$, so that $\lim_{n\rightarrow \infty}b_{n}=\infty$ (which implies $\lim_{n\rightarrow \infty}G(b_{n})=1$)

  • $a_n\equiv A(b_n)$.

  • Under these conditions it can be shown that $\lim_{n\rightarrow \infty} G(a_n*(t+u_k)+b_n)=1$ $\forall k$ $\forall t$.

I want to show that $\forall t \in \mathbb{R}$ $$ \lim_{n\rightarrow \infty}(c_n(t)+\tilde{c}_n(t))=0 $$ or, equivalently, $$ c_n(t)=-\tilde{c}_n(t)+o(1) $$ and your help would be extremely appreciated

I have attempted to start the proof (incomplete!)

CASE 1: $t$ such that $t+\bar{u}\leq 0$ $\Rightarrow$ $t+u_k\leq 0$. In this case $$ 0<G(a_n*(t+u_k)+b_n)\leq G(a_n*(t+\bar{u})+b_n)\leq G(b_n)=1-\frac{1}{n} \hspace{1cm} \text{$\forall n, k$} $$ $$ \Downarrow $$ $$ 0<G(a_n*(t+u_k)+b_n)\leq 1-\frac{1}{n} \hspace{1cm} \text{$\forall n, k$} $$ We know that for $x\in (0,1)$ $1-x+\log(x)\leq -(1-x)^2/2$. Applying this rule to $1-\frac{1}{n}$, we get $$ 1-(1-\frac{1}{n})+\log(1-\frac{1}{n})\leq -\frac{1}{2n^2} \hspace{1cm} \text{$\forall n$} $$ We can show that if $0<G(a_n*(t+u_k)+b_n)\leq 1-\frac{1}{n}$, then $$1-G(a_n*(t+u_k)+b_n)+\log(G(a_n*(t+u_k)+b_n))\leq 1-(1-\frac{1}{n})+\log(1-\frac{1}{n}) \hspace{1cm} \text{$\forall n, k$}$$

Hence, $$ 1-G(a_n*(t+u_k)+b_n)+\log(G(a_n*(t+u_k)+b_n))\leq -\frac{1}{2n^2} \hspace{1cm} \text{$\forall n, k$} $$ $$ \Downarrow $$ $$ \sum_{k=1}^{2n} (1-G(a_n*(t+u_k)+b_n)+\log(G(a_n*(t+u_k)+b_n)))\leq -\frac{1}{n} \hspace{1cm} \text{$\forall n$} $$ How do I conclude that the limit is zero?

CASE 2: $t$ such that $t+\bar{u}\geq 0$. This is much more complicated because we may have $t+u_k\geq 0$ for some $k$ and $t+u_k\leq 0$ for some other. I divide into two subcases.

SUBCASE 2.1: $\max_{k} \{t+u_k\}\leq 0$. In this case $$ 0<G(a_n*(t+u_k)+b_n)\leq G(b_n)=1-\frac{1}{n}\leq G(a_n*(t+\bar{u})+b_n) \hspace{1cm} \text{$\forall n, k$} $$ and we go back to case 1.

SUBCASE 2.2: $\max_{k} \{t+u_k\}\geq 0$. In this case $$ 1-\frac{1}{n}=G(b_n)\leq G(a_n*(t+u_k)+b_n)\leq G(a_n*(t+\bar{u})+b_n) \hspace{1cm} \text{$\forall n$, for at least one $k$} $$ $$ 0<G(a_n*(t+u_k)+b_n)\leq G(b_n)=1-\frac{1}{n}\leq G(a_n*(t+\bar{u})+b_n) \hspace{1cm} \text{$\forall n$, for the remaining $k$} $$ I don't know how to proceed

Star
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  • Please see my attempted incomplete proof, thank you – Star Apr 24 '18 at 15:11
  • I have also corrected an horrible typo (I was using for two different objects the notation $a_n$), apologies – Star Apr 24 '18 at 15:57
  • The statement on $\displaystyle\lim_{n\to\infty}a_n$ is wrong. Taking, say, $g(s)=\displaystyle\frac{1}{\pi(1+s^2)}$, you get $a_n=\displaystyle\frac{\pi}{n}\csc^2\frac{\pi}{n}\to\infty$ when $n\to\infty$. This "hit or miss" doesn't break the rest - I'm still looking at it. – metamorphy Apr 25 '18 at 17:40
  • Thanks. Notice that the integral of $g$ should give me a continuous and strictly monotone increasing on $\mathbb{R}$ function $G$. Indeed, we can think of $G$ as a cumulative distribution function and $g$ as the probability density function. – Star Apr 25 '18 at 18:50
  • This example of $g$ still "breaks" the statement if $t+\bar{u}<-\pi$ (in this case $a_n(t+\bar{u})+b_n\to-\infty$ when $n\to\infty$). As for the rest, I fear that simply bounding $u_k$ is not enough (for a definite resolution). – metamorphy Apr 25 '18 at 19:02
  • There is something ensuring that $G(a_n(t+u_k)+b_n)=G(\lim_{n\rightarrow \infty}a_n(t+u_k)+b_n) =G(\infty)=1 $ $\forall t,u_k$. Let me think more where I have left that assumption. – Star Apr 25 '18 at 19:18
  • I have added an assumption (this exercise comes from Extreme Value Theory and that assumptions ensures that $G$ is in the domain of attraction of a Gumbel). – Star Apr 25 '18 at 20:03
  • Could you please give an example of $G$ (or maybe $g$) satisfying these new conditions? – metamorphy Apr 26 '18 at 18:15
  • For example, $G$ is the cdf of a standard normal distribution and $g$ is the pdf of a standard normal distribution. – Star May 02 '18 at 09:31
  • OK, got it - under these conditions everything is easy; see my answer. – metamorphy May 03 '18 at 19:25

1 Answers1

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One can even claim that, for any $v\in\mathbb{R}$, $c_n(t)+\bar{c}_n(t) \to 0$ with $n\to\infty$ uniformly for $t \geqslant v$.

For $x\in(0,1)$, $L(x)=1-x+\log x$ is monotonically increasing, so $t+u_k \geqslant v$ implies $$ 0 \geqslant c_n(t)+\bar{c}_n(t)=\sum_{k=1}^{2n}L(G(a_n(t+u_k)+b_n)) \geqslant 2nL(G(a_n v+b_n)).$$ But the "ADDED" condition implies $\displaystyle\lim_{n\to\infty}n(1-G(a_n v+b_n))=e^{-v}$ (take $s=b_n$), so $$ \lim_{n\to\infty}nL(G(a_n v+b_n))=\lim_{n\to\infty}n(1-G(a_n v+b_n))\frac{L(G(a_n v+b_n))}{1-G(a_n v+b_n)}$$ exists and is equal to $0$ because $\displaystyle\lim_{x\to 1}\frac{L(x)}{1-x}=0$.

metamorphy
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  • @methamorphy if you have time, look at my other question https://math.stackexchange.com/questions/2751845/local-uniform-convergence-of-the-terms-of-a-sum-is-sufficient-for-convergence-of – Star May 04 '18 at 11:47