Convergence in distribution of minimum of IID random variables












2












$begingroup$


I'm stuck on the following problem and could use a hint:




Let $Z_1,ldots,Z_n$ be IID random variables with density $f$. Suppose that $mathbb{P}(Z_i > 0) = 1$ and that $lambda = lim_{x to 0^{+} } f(x) > 0$. Let:



$$X_n = nminlbrace Z_1,ldots,Z_nrbrace$$



Show that $X_n rightsquigarrow Z$ (converges in distribution) where $Z$ has an exponential distribution with mean $frac{1}{lambda}$.




So far I've figured out that:



$$
begin{align*}
mathbb{P}(X_n leq x) &= mathbb{P}(nminlbrace Z_1,ldots,Z_nrbrace leq x) \
&= mathbb{P}(Z_1 leq x/n cup ldots cup Z_n leq x/n) \
&= 1 - mathbb{P}(Z_1 > x/n cap ldots cap Z_n > x/n) \
&= 1 - mathbb{P}(Z_1 > x/n) cdot ldots cdot mathbb{P}(Z_n > x/n) \
&= 1 - mathbb{P}(Z_1 > x/n)^n \
&= 1 - (1-mathbb{P}(Z_1 leq x/n))^n \
&= 1 - (1-F_{Z_1}(x/n))^n
end{align*}
$$



But I'm not sure where to go from here. Although I know the CDF/PDF for $Z$, I don't know it for $Z_1,ldots,Z_n$, and it's not immediately obvious to me how I can leverage the fact that $Z_1,ldots,Z_n$ are guaranteed to be positive, other than if I rewrite $F_{Z_1}(x/n)$ to be an explicit integral of $f$ that the lower bound of the integral can be zero instead of negative infinity.



If I just try to take the limit as $n$ goes to infinity of both sides I can't simplify the RHS expression any further because of the power of $n$. I know that $a_n to a implies (1 + frac{a_n}{n})^n to e^a$ but I haven't managed to successfully apply it -- I assume this somehow helps me collapse the RHS into the CDF for an exponential, because I don't see another way to get $e$.



Finally I don't have any idea how to leverage the definition of lambda. Usually we care about right continuity of the CDF not the density function.






calculus probability convergence exponential-distribution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Consider the Taylor series of $F_{Z_1}$ at x=0. Also, I suspect that you have gotten your X's and Z's mixed up some places.
    $endgroup$
    – Mike Hawk
    Dec 16 '18 at 22:28










  • $begingroup$
    @MikeHawk You were right about mixing up, copy and pasted a latex template and modified it and made a few mistakes. Hopefully fixed now...
    $endgroup$
    – Joseph Garvin
    Dec 16 '18 at 22:42










  • $begingroup$
    @MikeHawk: I don't see how I can use that Taylor series, because I don't know the CDF for $Z_1$, only for $Z$.
    $endgroup$
    – Joseph Garvin
    Jan 8 at 23:06










  • $begingroup$
    But you know $F_{Z_1}'(0)=lambda$, so you just have to show that $(1-lambda x/n+O(1/n^2))^nto e^{-lambda x}$
    $endgroup$
    – Mike Hawk
    Jan 9 at 16:45
















2












$begingroup$


I'm stuck on the following problem and could use a hint:




Let $Z_1,ldots,Z_n$ be IID random variables with density $f$. Suppose that $mathbb{P}(Z_i > 0) = 1$ and that $lambda = lim_{x to 0^{+} } f(x) > 0$. Let:



$$X_n = nminlbrace Z_1,ldots,Z_nrbrace$$



Show that $X_n rightsquigarrow Z$ (converges in distribution) where $Z$ has an exponential distribution with mean $frac{1}{lambda}$.




So far I've figured out that:



$$
begin{align*}
mathbb{P}(X_n leq x) &= mathbb{P}(nminlbrace Z_1,ldots,Z_nrbrace leq x) \
&= mathbb{P}(Z_1 leq x/n cup ldots cup Z_n leq x/n) \
&= 1 - mathbb{P}(Z_1 > x/n cap ldots cap Z_n > x/n) \
&= 1 - mathbb{P}(Z_1 > x/n) cdot ldots cdot mathbb{P}(Z_n > x/n) \
&= 1 - mathbb{P}(Z_1 > x/n)^n \
&= 1 - (1-mathbb{P}(Z_1 leq x/n))^n \
&= 1 - (1-F_{Z_1}(x/n))^n
end{align*}
$$



But I'm not sure where to go from here. Although I know the CDF/PDF for $Z$, I don't know it for $Z_1,ldots,Z_n$, and it's not immediately obvious to me how I can leverage the fact that $Z_1,ldots,Z_n$ are guaranteed to be positive, other than if I rewrite $F_{Z_1}(x/n)$ to be an explicit integral of $f$ that the lower bound of the integral can be zero instead of negative infinity.



If I just try to take the limit as $n$ goes to infinity of both sides I can't simplify the RHS expression any further because of the power of $n$. I know that $a_n to a implies (1 + frac{a_n}{n})^n to e^a$ but I haven't managed to successfully apply it -- I assume this somehow helps me collapse the RHS into the CDF for an exponential, because I don't see another way to get $e$.



Finally I don't have any idea how to leverage the definition of lambda. Usually we care about right continuity of the CDF not the density function.






calculus probability convergence exponential-distribution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Consider the Taylor series of $F_{Z_1}$ at x=0. Also, I suspect that you have gotten your X's and Z's mixed up some places.
    $endgroup$
    – Mike Hawk
    Dec 16 '18 at 22:28










  • $begingroup$
    @MikeHawk You were right about mixing up, copy and pasted a latex template and modified it and made a few mistakes. Hopefully fixed now...
    $endgroup$
    – Joseph Garvin
    Dec 16 '18 at 22:42










  • $begingroup$
    @MikeHawk: I don't see how I can use that Taylor series, because I don't know the CDF for $Z_1$, only for $Z$.
    $endgroup$
    – Joseph Garvin
    Jan 8 at 23:06










  • $begingroup$
    But you know $F_{Z_1}'(0)=lambda$, so you just have to show that $(1-lambda x/n+O(1/n^2))^nto e^{-lambda x}$
    $endgroup$
    – Mike Hawk
    Jan 9 at 16:45














2












2








2





$begingroup$


I'm stuck on the following problem and could use a hint:




Let $Z_1,ldots,Z_n$ be IID random variables with density $f$. Suppose that $mathbb{P}(Z_i > 0) = 1$ and that $lambda = lim_{x to 0^{+} } f(x) > 0$. Let:



$$X_n = nminlbrace Z_1,ldots,Z_nrbrace$$



Show that $X_n rightsquigarrow Z$ (converges in distribution) where $Z$ has an exponential distribution with mean $frac{1}{lambda}$.




So far I've figured out that:



$$
begin{align*}
mathbb{P}(X_n leq x) &= mathbb{P}(nminlbrace Z_1,ldots,Z_nrbrace leq x) \
&= mathbb{P}(Z_1 leq x/n cup ldots cup Z_n leq x/n) \
&= 1 - mathbb{P}(Z_1 > x/n cap ldots cap Z_n > x/n) \
&= 1 - mathbb{P}(Z_1 > x/n) cdot ldots cdot mathbb{P}(Z_n > x/n) \
&= 1 - mathbb{P}(Z_1 > x/n)^n \
&= 1 - (1-mathbb{P}(Z_1 leq x/n))^n \
&= 1 - (1-F_{Z_1}(x/n))^n
end{align*}
$$



But I'm not sure where to go from here. Although I know the CDF/PDF for $Z$, I don't know it for $Z_1,ldots,Z_n$, and it's not immediately obvious to me how I can leverage the fact that $Z_1,ldots,Z_n$ are guaranteed to be positive, other than if I rewrite $F_{Z_1}(x/n)$ to be an explicit integral of $f$ that the lower bound of the integral can be zero instead of negative infinity.



If I just try to take the limit as $n$ goes to infinity of both sides I can't simplify the RHS expression any further because of the power of $n$. I know that $a_n to a implies (1 + frac{a_n}{n})^n to e^a$ but I haven't managed to successfully apply it -- I assume this somehow helps me collapse the RHS into the CDF for an exponential, because I don't see another way to get $e$.



Finally I don't have any idea how to leverage the definition of lambda. Usually we care about right continuity of the CDF not the density function.






calculus probability convergence exponential-distribution







share|cite|improve this question











$endgroup$




I'm stuck on the following problem and could use a hint:




Let $Z_1,ldots,Z_n$ be IID random variables with density $f$. Suppose that $mathbb{P}(Z_i > 0) = 1$ and that $lambda = lim_{x to 0^{+} } f(x) > 0$. Let:



$$X_n = nminlbrace Z_1,ldots,Z_nrbrace$$



Show that $X_n rightsquigarrow Z$ (converges in distribution) where $Z$ has an exponential distribution with mean $frac{1}{lambda}$.




So far I've figured out that:



$$
begin{align*}
mathbb{P}(X_n leq x) &= mathbb{P}(nminlbrace Z_1,ldots,Z_nrbrace leq x) \
&= mathbb{P}(Z_1 leq x/n cup ldots cup Z_n leq x/n) \
&= 1 - mathbb{P}(Z_1 > x/n cap ldots cap Z_n > x/n) \
&= 1 - mathbb{P}(Z_1 > x/n) cdot ldots cdot mathbb{P}(Z_n > x/n) \
&= 1 - mathbb{P}(Z_1 > x/n)^n \
&= 1 - (1-mathbb{P}(Z_1 leq x/n))^n \
&= 1 - (1-F_{Z_1}(x/n))^n
end{align*}
$$



But I'm not sure where to go from here. Although I know the CDF/PDF for $Z$, I don't know it for $Z_1,ldots,Z_n$, and it's not immediately obvious to me how I can leverage the fact that $Z_1,ldots,Z_n$ are guaranteed to be positive, other than if I rewrite $F_{Z_1}(x/n)$ to be an explicit integral of $f$ that the lower bound of the integral can be zero instead of negative infinity.



If I just try to take the limit as $n$ goes to infinity of both sides I can't simplify the RHS expression any further because of the power of $n$. I know that $a_n to a implies (1 + frac{a_n}{n})^n to e^a$ but I haven't managed to successfully apply it -- I assume this somehow helps me collapse the RHS into the CDF for an exponential, because I don't see another way to get $e$.



Finally I don't have any idea how to leverage the definition of lambda. Usually we care about right continuity of the CDF not the density function.






calculus probability convergence exponential-distribution




calculus probability convergence exponential-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 22:40







Joseph Garvin

















asked Dec 16 '18 at 22:22









Joseph GarvinJoseph Garvin

44928




44928












  • $begingroup$
    Consider the Taylor series of $F_{Z_1}$ at x=0. Also, I suspect that you have gotten your X's and Z's mixed up some places.
    $endgroup$
    – Mike Hawk
    Dec 16 '18 at 22:28










  • $begingroup$
    @MikeHawk You were right about mixing up, copy and pasted a latex template and modified it and made a few mistakes. Hopefully fixed now...
    $endgroup$
    – Joseph Garvin
    Dec 16 '18 at 22:42










  • $begingroup$
    @MikeHawk: I don't see how I can use that Taylor series, because I don't know the CDF for $Z_1$, only for $Z$.
    $endgroup$
    – Joseph Garvin
    Jan 8 at 23:06










  • $begingroup$
    But you know $F_{Z_1}'(0)=lambda$, so you just have to show that $(1-lambda x/n+O(1/n^2))^nto e^{-lambda x}$
    $endgroup$
    – Mike Hawk
    Jan 9 at 16:45


















  • $begingroup$
    Consider the Taylor series of $F_{Z_1}$ at x=0. Also, I suspect that you have gotten your X's and Z's mixed up some places.
    $endgroup$
    – Mike Hawk
    Dec 16 '18 at 22:28










  • $begingroup$
    @MikeHawk You were right about mixing up, copy and pasted a latex template and modified it and made a few mistakes. Hopefully fixed now...
    $endgroup$
    – Joseph Garvin
    Dec 16 '18 at 22:42










  • $begingroup$
    @MikeHawk: I don't see how I can use that Taylor series, because I don't know the CDF for $Z_1$, only for $Z$.
    $endgroup$
    – Joseph Garvin
    Jan 8 at 23:06










  • $begingroup$
    But you know $F_{Z_1}'(0)=lambda$, so you just have to show that $(1-lambda x/n+O(1/n^2))^nto e^{-lambda x}$
    $endgroup$
    – Mike Hawk
    Jan 9 at 16:45
















$begingroup$
Consider the Taylor series of $F_{Z_1}$ at x=0. Also, I suspect that you have gotten your X's and Z's mixed up some places.
$endgroup$
– Mike Hawk
Dec 16 '18 at 22:28




$begingroup$
Consider the Taylor series of $F_{Z_1}$ at x=0. Also, I suspect that you have gotten your X's and Z's mixed up some places.
$endgroup$
– Mike Hawk
Dec 16 '18 at 22:28












$begingroup$
@MikeHawk You were right about mixing up, copy and pasted a latex template and modified it and made a few mistakes. Hopefully fixed now...
$endgroup$
– Joseph Garvin
Dec 16 '18 at 22:42




$begingroup$
@MikeHawk You were right about mixing up, copy and pasted a latex template and modified it and made a few mistakes. Hopefully fixed now...
$endgroup$
– Joseph Garvin
Dec 16 '18 at 22:42












$begingroup$
@MikeHawk: I don't see how I can use that Taylor series, because I don't know the CDF for $Z_1$, only for $Z$.
$endgroup$
– Joseph Garvin
Jan 8 at 23:06




$begingroup$
@MikeHawk: I don't see how I can use that Taylor series, because I don't know the CDF for $Z_1$, only for $Z$.
$endgroup$
– Joseph Garvin
Jan 8 at 23:06












$begingroup$
But you know $F_{Z_1}'(0)=lambda$, so you just have to show that $(1-lambda x/n+O(1/n^2))^nto e^{-lambda x}$
$endgroup$
– Mike Hawk
Jan 9 at 16:45




$begingroup$
But you know $F_{Z_1}'(0)=lambda$, so you just have to show that $(1-lambda x/n+O(1/n^2))^nto e^{-lambda x}$
$endgroup$
– Mike Hawk
Jan 9 at 16:45










1 Answer
1






active

oldest

votes


















0












$begingroup$

You need to prove this probability approximates $1-exp -lambda x$, i.e. that $(1-F_{Z_i}(x/n))^napproxexp -lambda x$. But for $ngg x$, $$F_{Z_i}(0)=0implies F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)=frac{lambda x}{n},$$so the claim follows from $exp -t=lim_{ntoinfty}(1-frac{t}{n})^n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    what does the n >> x notation mean?
    $endgroup$
    – Joseph Garvin
    Jan 6 at 18:59






  • 1




    $begingroup$
    @JosephGarvin That $n$ is much greater than $x$.
    $endgroup$
    – J.G.
    Jan 6 at 19:20










  • $begingroup$
    what does the $sim$ notation mean?
    $endgroup$
    – Joseph Garvin
    Jan 6 at 21:54












  • $begingroup$
    @JosephGarvin is asymptotic to.
    $endgroup$
    – J.G.
    Jan 6 at 21:55










  • $begingroup$
    I don't understand how you conclude $ F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)$ ? how're those expressions related?
    $endgroup$
    – Joseph Garvin
    Jan 9 at 0:00











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









0












$begingroup$

You need to prove this probability approximates $1-exp -lambda x$, i.e. that $(1-F_{Z_i}(x/n))^napproxexp -lambda x$. But for $ngg x$, $$F_{Z_i}(0)=0implies F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)=frac{lambda x}{n},$$so the claim follows from $exp -t=lim_{ntoinfty}(1-frac{t}{n})^n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    what does the n >> x notation mean?
    $endgroup$
    – Joseph Garvin
    Jan 6 at 18:59






  • 1




    $begingroup$
    @JosephGarvin That $n$ is much greater than $x$.
    $endgroup$
    – J.G.
    Jan 6 at 19:20










  • $begingroup$
    what does the $sim$ notation mean?
    $endgroup$
    – Joseph Garvin
    Jan 6 at 21:54












  • $begingroup$
    @JosephGarvin is asymptotic to.
    $endgroup$
    – J.G.
    Jan 6 at 21:55










  • $begingroup$
    I don't understand how you conclude $ F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)$ ? how're those expressions related?
    $endgroup$
    – Joseph Garvin
    Jan 9 at 0:00
















0












$begingroup$

You need to prove this probability approximates $1-exp -lambda x$, i.e. that $(1-F_{Z_i}(x/n))^napproxexp -lambda x$. But for $ngg x$, $$F_{Z_i}(0)=0implies F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)=frac{lambda x}{n},$$so the claim follows from $exp -t=lim_{ntoinfty}(1-frac{t}{n})^n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    what does the n >> x notation mean?
    $endgroup$
    – Joseph Garvin
    Jan 6 at 18:59






  • 1




    $begingroup$
    @JosephGarvin That $n$ is much greater than $x$.
    $endgroup$
    – J.G.
    Jan 6 at 19:20










  • $begingroup$
    what does the $sim$ notation mean?
    $endgroup$
    – Joseph Garvin
    Jan 6 at 21:54












  • $begingroup$
    @JosephGarvin is asymptotic to.
    $endgroup$
    – J.G.
    Jan 6 at 21:55










  • $begingroup$
    I don't understand how you conclude $ F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)$ ? how're those expressions related?
    $endgroup$
    – Joseph Garvin
    Jan 9 at 0:00














0












0








0





$begingroup$

You need to prove this probability approximates $1-exp -lambda x$, i.e. that $(1-F_{Z_i}(x/n))^napproxexp -lambda x$. But for $ngg x$, $$F_{Z_i}(0)=0implies F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)=frac{lambda x}{n},$$so the claim follows from $exp -t=lim_{ntoinfty}(1-frac{t}{n})^n$.






share|cite|improve this answer









$endgroup$



You need to prove this probability approximates $1-exp -lambda x$, i.e. that $(1-F_{Z_i}(x/n))^napproxexp -lambda x$. But for $ngg x$, $$F_{Z_i}(0)=0implies F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)=frac{lambda x}{n},$$so the claim follows from $exp -t=lim_{ntoinfty}(1-frac{t}{n})^n$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 16 '18 at 22:50









J.G.J.G.

26.3k22541




26.3k22541












  • $begingroup$
    what does the n >> x notation mean?
    $endgroup$
    – Joseph Garvin
    Jan 6 at 18:59






  • 1




    $begingroup$
    @JosephGarvin That $n$ is much greater than $x$.
    $endgroup$
    – J.G.
    Jan 6 at 19:20










  • $begingroup$
    what does the $sim$ notation mean?
    $endgroup$
    – Joseph Garvin
    Jan 6 at 21:54












  • $begingroup$
    @JosephGarvin is asymptotic to.
    $endgroup$
    – J.G.
    Jan 6 at 21:55










  • $begingroup$
    I don't understand how you conclude $ F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)$ ? how're those expressions related?
    $endgroup$
    – Joseph Garvin
    Jan 9 at 0:00


















  • $begingroup$
    what does the n >> x notation mean?
    $endgroup$
    – Joseph Garvin
    Jan 6 at 18:59






  • 1




    $begingroup$
    @JosephGarvin That $n$ is much greater than $x$.
    $endgroup$
    – J.G.
    Jan 6 at 19:20










  • $begingroup$
    what does the $sim$ notation mean?
    $endgroup$
    – Joseph Garvin
    Jan 6 at 21:54












  • $begingroup$
    @JosephGarvin is asymptotic to.
    $endgroup$
    – J.G.
    Jan 6 at 21:55










  • $begingroup$
    I don't understand how you conclude $ F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)$ ? how're those expressions related?
    $endgroup$
    – Joseph Garvin
    Jan 9 at 0:00
















$begingroup$
what does the n >> x notation mean?
$endgroup$
– Joseph Garvin
Jan 6 at 18:59




$begingroup$
what does the n >> x notation mean?
$endgroup$
– Joseph Garvin
Jan 6 at 18:59




1




1




$begingroup$
@JosephGarvin That $n$ is much greater than $x$.
$endgroup$
– J.G.
Jan 6 at 19:20




$begingroup$
@JosephGarvin That $n$ is much greater than $x$.
$endgroup$
– J.G.
Jan 6 at 19:20












$begingroup$
what does the $sim$ notation mean?
$endgroup$
– Joseph Garvin
Jan 6 at 21:54






$begingroup$
what does the $sim$ notation mean?
$endgroup$
– Joseph Garvin
Jan 6 at 21:54














$begingroup$
@JosephGarvin is asymptotic to.
$endgroup$
– J.G.
Jan 6 at 21:55




$begingroup$
@JosephGarvin is asymptotic to.
$endgroup$
– J.G.
Jan 6 at 21:55












$begingroup$
I don't understand how you conclude $ F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)$ ? how're those expressions related?
$endgroup$
– Joseph Garvin
Jan 9 at 0:00




$begingroup$
I don't understand how you conclude $ F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)$ ? how're those expressions related?
$endgroup$
– Joseph Garvin
Jan 9 at 0:00


















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Ellipse (mathématiques)

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Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
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Quarter-circle Tiles

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up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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Mont Emei

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Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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