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Domain of attraction $F(x)=exp(-x-sin(x))$

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$begingroup$

I need to show that $F(x)=exp(-x-sin(x)),x >0$ is in no domain of attraction. That means that there exist no $a_{k} >0, b_{k} in mathbb{R}, k in mathbb{N}$ with $limlimits_{k to infty} frac{U(kx)-b(k)}{a(k)}=D(x)$ where $D(x)=G^{leftarrow}(e^{frac{1}{x}}),$ $G$ is a nondegenerate distribution function and $U=(frac{1}{1-F})^{leftarrow}$.
As a hint i know that the following holds $$ limlimits_{k to infty} U(n_{k}x)-log(n_{k})= U_{1}(x)$$
where $U_{1}$ is the inverse of $exp(x+sin(x))$ and $n_{k}=[exp(2pi k)].$
Any approach?

probability-distributions stochastic-analysis extreme-value-theorem

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asked Mar 13 at 7:34

John DoeJohn Doe

183

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add a comment | 

1

$begingroup$

I need to show that $F(x)=exp(-x-sin(x)),x >0$ is in no domain of attraction. That means that there exist no $a_{k} >0, b_{k} in mathbb{R}, k in mathbb{N}$ with $limlimits_{k to infty} frac{U(kx)-b(k)}{a(k)}=D(x)$ where $D(x)=G^{leftarrow}(e^{frac{1}{x}}),$ $G$ is a nondegenerate distribution function and $U=(frac{1}{1-F})^{leftarrow}$.
As a hint i know that the following holds $$ limlimits_{k to infty} U(n_{k}x)-log(n_{k})= U_{1}(x)$$
where $U_{1}$ is the inverse of $exp(x+sin(x))$ and $n_{k}=[exp(2pi k)].$
Any approach?

probability-distributions stochastic-analysis extreme-value-theorem

share|cite|improve this question

asked Mar 13 at 7:34

John DoeJohn Doe

183

$endgroup$

add a comment | 

$begingroup$

I need to show that $F(x)=exp(-x-sin(x)),x >0$ is in no domain of attraction. That means that there exist no $a_{k} >0, b_{k} in mathbb{R}, k in mathbb{N}$ with $limlimits_{k to infty} frac{U(kx)-b(k)}{a(k)}=D(x)$ where $D(x)=G^{leftarrow}(e^{frac{1}{x}}),$ $G$ is a nondegenerate distribution function and $U=(frac{1}{1-F})^{leftarrow}$.
As a hint i know that the following holds $$ limlimits_{k to infty} U(n_{k}x)-log(n_{k})= U_{1}(x)$$
where $U_{1}$ is the inverse of $exp(x+sin(x))$ and $n_{k}=[exp(2pi k)].$
Any approach?

probability-distributions stochastic-analysis extreme-value-theorem

share|cite|improve this question

asked Mar 13 at 7:34

John DoeJohn Doe

183

$endgroup$

share|cite|improve this question

asked Mar 13 at 7:34

John DoeJohn Doe

183

share|cite|improve this question

asked Mar 13 at 7:34

John DoeJohn Doe

183

asked Mar 13 at 7:34

John DoeJohn Doe

183

asked Mar 13 at 7:34

John DoeJohn Doe

183