How to minimize the minimum mean square error of this difference












0












$begingroup$


I am trying to minimize the mean square error. More precisely, I am trying to minimize the following optimization problem



$$arg min _{bf{w_1},bf{w_2}}mathbb{E} ,,[|{bf s} - {bf Wy}|^2 ]$$
$${bf W} = begin{bmatrix}
{bf w_1} &{bf 0 } \
{bf 0 } & {bf w_2 }
end{bmatrix}$$



where ${bf W}$ is a $2times N$ matrix and where ${bf w_i}$ is a $1times N/2$ for $iin[1:2]$ and ${bf 0}$ is $1times N/2$ vector while ${bf s}$ is 2 $times $ 1 and ${bf y}$ is $N times 1$ vectors.



Any hints or ideas on finding the minimizing vectors of this problem?



It is also given that



$${bf y = A F s + z}$$ where ${bf A}$ is $Ntimes N$ matrix while



$${bf F} = begin{bmatrix}
{bf f_1} &{bf 0} \
{bf 0} & {bf f_2}
end{bmatrix}$$



where ${bf f_i}$ are $N/2times 1$ vector and ${bf Z}$ is $Ntimes 1$ vector



I am thinking of starting to start the solution as following
$$argmin _{bf{w_1},bf{w_2}}mathbb{E} ,,[[{bf s} - {bf Wy}]^H[{bf s} - {bf Wy}]]$$



But I assume given the specific block diagonal structure of matrix ${bf W}$ and ${bf F}$ it should be easier to solve ...



Thanks






optimization mean-square-error







share|cite|improve this question











$endgroup$












  • $begingroup$
    There must be something wrong with the dimensions. $mathbf W$ is a $2times N$-matrix and $mathbf y$ is a $2times 1$-vector, Then $mathbf{ Wy}$ is not possible.
    $endgroup$
    – callculus
    Oct 22 '15 at 0:17












  • $begingroup$
    sorry had typo thanks
    $endgroup$
    – Tyrone
    Oct 22 '15 at 0:19










  • $begingroup$
    I hope I´m not wrong, but it seems very similar to linear regression. Here it is $(s-Wy)'(s-Wy)=(s'-y'W')(s-Wy)=s's-s'Wy-y'W's-y'W'Wy$ But at linear regression it is optimized w.r.t $y$, not $W$. But I think it wouldn´t be bad to have a look on this subject.
    $endgroup$
    – callculus
    Oct 22 '15 at 0:43












  • $begingroup$
    thnk you for the reference
    $endgroup$
    – Tyrone
    Oct 22 '15 at 1:41
















0












$begingroup$


I am trying to minimize the mean square error. More precisely, I am trying to minimize the following optimization problem



$$arg min _{bf{w_1},bf{w_2}}mathbb{E} ,,[|{bf s} - {bf Wy}|^2 ]$$
$${bf W} = begin{bmatrix}
{bf w_1} &{bf 0 } \
{bf 0 } & {bf w_2 }
end{bmatrix}$$



where ${bf W}$ is a $2times N$ matrix and where ${bf w_i}$ is a $1times N/2$ for $iin[1:2]$ and ${bf 0}$ is $1times N/2$ vector while ${bf s}$ is 2 $times $ 1 and ${bf y}$ is $N times 1$ vectors.



Any hints or ideas on finding the minimizing vectors of this problem?



It is also given that



$${bf y = A F s + z}$$ where ${bf A}$ is $Ntimes N$ matrix while



$${bf F} = begin{bmatrix}
{bf f_1} &{bf 0} \
{bf 0} & {bf f_2}
end{bmatrix}$$



where ${bf f_i}$ are $N/2times 1$ vector and ${bf Z}$ is $Ntimes 1$ vector



I am thinking of starting to start the solution as following
$$argmin _{bf{w_1},bf{w_2}}mathbb{E} ,,[[{bf s} - {bf Wy}]^H[{bf s} - {bf Wy}]]$$



But I assume given the specific block diagonal structure of matrix ${bf W}$ and ${bf F}$ it should be easier to solve ...



Thanks






optimization mean-square-error







share|cite|improve this question











$endgroup$












  • $begingroup$
    There must be something wrong with the dimensions. $mathbf W$ is a $2times N$-matrix and $mathbf y$ is a $2times 1$-vector, Then $mathbf{ Wy}$ is not possible.
    $endgroup$
    – callculus
    Oct 22 '15 at 0:17












  • $begingroup$
    sorry had typo thanks
    $endgroup$
    – Tyrone
    Oct 22 '15 at 0:19










  • $begingroup$
    I hope I´m not wrong, but it seems very similar to linear regression. Here it is $(s-Wy)'(s-Wy)=(s'-y'W')(s-Wy)=s's-s'Wy-y'W's-y'W'Wy$ But at linear regression it is optimized w.r.t $y$, not $W$. But I think it wouldn´t be bad to have a look on this subject.
    $endgroup$
    – callculus
    Oct 22 '15 at 0:43












  • $begingroup$
    thnk you for the reference
    $endgroup$
    – Tyrone
    Oct 22 '15 at 1:41














0












0








0





$begingroup$


I am trying to minimize the mean square error. More precisely, I am trying to minimize the following optimization problem



$$arg min _{bf{w_1},bf{w_2}}mathbb{E} ,,[|{bf s} - {bf Wy}|^2 ]$$
$${bf W} = begin{bmatrix}
{bf w_1} &{bf 0 } \
{bf 0 } & {bf w_2 }
end{bmatrix}$$



where ${bf W}$ is a $2times N$ matrix and where ${bf w_i}$ is a $1times N/2$ for $iin[1:2]$ and ${bf 0}$ is $1times N/2$ vector while ${bf s}$ is 2 $times $ 1 and ${bf y}$ is $N times 1$ vectors.



Any hints or ideas on finding the minimizing vectors of this problem?



It is also given that



$${bf y = A F s + z}$$ where ${bf A}$ is $Ntimes N$ matrix while



$${bf F} = begin{bmatrix}
{bf f_1} &{bf 0} \
{bf 0} & {bf f_2}
end{bmatrix}$$



where ${bf f_i}$ are $N/2times 1$ vector and ${bf Z}$ is $Ntimes 1$ vector



I am thinking of starting to start the solution as following
$$argmin _{bf{w_1},bf{w_2}}mathbb{E} ,,[[{bf s} - {bf Wy}]^H[{bf s} - {bf Wy}]]$$



But I assume given the specific block diagonal structure of matrix ${bf W}$ and ${bf F}$ it should be easier to solve ...



Thanks






optimization mean-square-error







share|cite|improve this question











$endgroup$




I am trying to minimize the mean square error. More precisely, I am trying to minimize the following optimization problem



$$arg min _{bf{w_1},bf{w_2}}mathbb{E} ,,[|{bf s} - {bf Wy}|^2 ]$$
$${bf W} = begin{bmatrix}
{bf w_1} &{bf 0 } \
{bf 0 } & {bf w_2 }
end{bmatrix}$$



where ${bf W}$ is a $2times N$ matrix and where ${bf w_i}$ is a $1times N/2$ for $iin[1:2]$ and ${bf 0}$ is $1times N/2$ vector while ${bf s}$ is 2 $times $ 1 and ${bf y}$ is $N times 1$ vectors.



Any hints or ideas on finding the minimizing vectors of this problem?



It is also given that



$${bf y = A F s + z}$$ where ${bf A}$ is $Ntimes N$ matrix while



$${bf F} = begin{bmatrix}
{bf f_1} &{bf 0} \
{bf 0} & {bf f_2}
end{bmatrix}$$



where ${bf f_i}$ are $N/2times 1$ vector and ${bf Z}$ is $Ntimes 1$ vector



I am thinking of starting to start the solution as following
$$argmin _{bf{w_1},bf{w_2}}mathbb{E} ,,[[{bf s} - {bf Wy}]^H[{bf s} - {bf Wy}]]$$



But I assume given the specific block diagonal structure of matrix ${bf W}$ and ${bf F}$ it should be easier to solve ...



Thanks






optimization mean-square-error




optimization mean-square-error






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 22 '15 at 3:00







Tyrone

















asked Oct 21 '15 at 23:46









TyroneTyrone

244418




244418












  • $begingroup$
    There must be something wrong with the dimensions. $mathbf W$ is a $2times N$-matrix and $mathbf y$ is a $2times 1$-vector, Then $mathbf{ Wy}$ is not possible.
    $endgroup$
    – callculus
    Oct 22 '15 at 0:17












  • $begingroup$
    sorry had typo thanks
    $endgroup$
    – Tyrone
    Oct 22 '15 at 0:19










  • $begingroup$
    I hope I´m not wrong, but it seems very similar to linear regression. Here it is $(s-Wy)'(s-Wy)=(s'-y'W')(s-Wy)=s's-s'Wy-y'W's-y'W'Wy$ But at linear regression it is optimized w.r.t $y$, not $W$. But I think it wouldn´t be bad to have a look on this subject.
    $endgroup$
    – callculus
    Oct 22 '15 at 0:43












  • $begingroup$
    thnk you for the reference
    $endgroup$
    – Tyrone
    Oct 22 '15 at 1:41


















  • $begingroup$
    There must be something wrong with the dimensions. $mathbf W$ is a $2times N$-matrix and $mathbf y$ is a $2times 1$-vector, Then $mathbf{ Wy}$ is not possible.
    $endgroup$
    – callculus
    Oct 22 '15 at 0:17












  • $begingroup$
    sorry had typo thanks
    $endgroup$
    – Tyrone
    Oct 22 '15 at 0:19










  • $begingroup$
    I hope I´m not wrong, but it seems very similar to linear regression. Here it is $(s-Wy)'(s-Wy)=(s'-y'W')(s-Wy)=s's-s'Wy-y'W's-y'W'Wy$ But at linear regression it is optimized w.r.t $y$, not $W$. But I think it wouldn´t be bad to have a look on this subject.
    $endgroup$
    – callculus
    Oct 22 '15 at 0:43












  • $begingroup$
    thnk you for the reference
    $endgroup$
    – Tyrone
    Oct 22 '15 at 1:41
















$begingroup$
There must be something wrong with the dimensions. $mathbf W$ is a $2times N$-matrix and $mathbf y$ is a $2times 1$-vector, Then $mathbf{ Wy}$ is not possible.
$endgroup$
– callculus
Oct 22 '15 at 0:17






$begingroup$
There must be something wrong with the dimensions. $mathbf W$ is a $2times N$-matrix and $mathbf y$ is a $2times 1$-vector, Then $mathbf{ Wy}$ is not possible.
$endgroup$
– callculus
Oct 22 '15 at 0:17














$begingroup$
sorry had typo thanks
$endgroup$
– Tyrone
Oct 22 '15 at 0:19




$begingroup$
sorry had typo thanks
$endgroup$
– Tyrone
Oct 22 '15 at 0:19












$begingroup$
I hope I´m not wrong, but it seems very similar to linear regression. Here it is $(s-Wy)'(s-Wy)=(s'-y'W')(s-Wy)=s's-s'Wy-y'W's-y'W'Wy$ But at linear regression it is optimized w.r.t $y$, not $W$. But I think it wouldn´t be bad to have a look on this subject.
$endgroup$
– callculus
Oct 22 '15 at 0:43






$begingroup$
I hope I´m not wrong, but it seems very similar to linear regression. Here it is $(s-Wy)'(s-Wy)=(s'-y'W')(s-Wy)=s's-s'Wy-y'W's-y'W'Wy$ But at linear regression it is optimized w.r.t $y$, not $W$. But I think it wouldn´t be bad to have a look on this subject.
$endgroup$
– callculus
Oct 22 '15 at 0:43














$begingroup$
thnk you for the reference
$endgroup$
– Tyrone
Oct 22 '15 at 1:41




$begingroup$
thnk you for the reference
$endgroup$
– Tyrone
Oct 22 '15 at 1:41










1 Answer
1






active

oldest

votes


















0












$begingroup$

I assume you wish to find the non random matrix $W$ and hence its elements are not boldface in my solution. $mathbf s$ and $mathbf y$ are random vectors, hence they are boldface. First let me write the matrix ${W}
= begin{bmatrix}
{w_{1}^*} &{ 0 } \
{ 0 } & { w_{2}^* }
end{bmatrix}
$

where $*$ denotes the hermitian operation. Now all lower case vectors are column vectors.

This problem can be restated as

$arg min _{{w_1},{w_2}}mathbb{E} ,,left[leftVertbegin{bmatrix}
bf s_1^* \
bf s_{2}^*
end{bmatrix} - begin{bmatrix}
{bf y_{1}^*} &{bf 0 } \
{bf 0 } & {bf y_{2}^* }
end{bmatrix}begin{bmatrix}
w_1 \
w_{2}
end{bmatrix}rightVert^2right] $.

After this, the problem decouples to solving for $w_1$ and $w_2$. For eg. for $w_1$, we need to minimize $J =mathbb E[ ||mathbf s_1^* - mathbf y_1^* w_1||^2]$.

Just expand the inside argument and differentiate w.r.t. $w_1^*$ and put the gradient to $0$. We get the solution for $w_1$ as

$mathbb E [mathbf y_1 mathbf y_1^*] w_1 = mathbb E [mathbf y_1^* mathbf s_1^*]$.

Now, assuming you can find the correlation matrix of $mathbf y_1$ and it is invertible, and the cross correlation between $mathbf y_1$ and $mathbf s_1$, you can find $w_1$. Similarly, you can solve for $w_2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    i am not sure... wht how did you decompose vector y?
    $endgroup$
    – Tyrone
    Oct 22 '15 at 1:49










  • $begingroup$
    The Nx1 vector $mathbf y$ is split into two vectors $mathbf y_1$ and $mathbf y_2$ each N/2 x1, stacked on top of each other. Hope that clears the confusion.
    $endgroup$
    – shaktiman
    Oct 22 '15 at 3:31













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

I assume you wish to find the non random matrix $W$ and hence its elements are not boldface in my solution. $mathbf s$ and $mathbf y$ are random vectors, hence they are boldface. First let me write the matrix ${W}
= begin{bmatrix}
{w_{1}^*} &{ 0 } \
{ 0 } & { w_{2}^* }
end{bmatrix}
$

where $*$ denotes the hermitian operation. Now all lower case vectors are column vectors.

This problem can be restated as

$arg min _{{w_1},{w_2}}mathbb{E} ,,left[leftVertbegin{bmatrix}
bf s_1^* \
bf s_{2}^*
end{bmatrix} - begin{bmatrix}
{bf y_{1}^*} &{bf 0 } \
{bf 0 } & {bf y_{2}^* }
end{bmatrix}begin{bmatrix}
w_1 \
w_{2}
end{bmatrix}rightVert^2right] $.

After this, the problem decouples to solving for $w_1$ and $w_2$. For eg. for $w_1$, we need to minimize $J =mathbb E[ ||mathbf s_1^* - mathbf y_1^* w_1||^2]$.

Just expand the inside argument and differentiate w.r.t. $w_1^*$ and put the gradient to $0$. We get the solution for $w_1$ as

$mathbb E [mathbf y_1 mathbf y_1^*] w_1 = mathbb E [mathbf y_1^* mathbf s_1^*]$.

Now, assuming you can find the correlation matrix of $mathbf y_1$ and it is invertible, and the cross correlation between $mathbf y_1$ and $mathbf s_1$, you can find $w_1$. Similarly, you can solve for $w_2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    i am not sure... wht how did you decompose vector y?
    $endgroup$
    – Tyrone
    Oct 22 '15 at 1:49










  • $begingroup$
    The Nx1 vector $mathbf y$ is split into two vectors $mathbf y_1$ and $mathbf y_2$ each N/2 x1, stacked on top of each other. Hope that clears the confusion.
    $endgroup$
    – shaktiman
    Oct 22 '15 at 3:31


















0












$begingroup$

I assume you wish to find the non random matrix $W$ and hence its elements are not boldface in my solution. $mathbf s$ and $mathbf y$ are random vectors, hence they are boldface. First let me write the matrix ${W}
= begin{bmatrix}
{w_{1}^*} &{ 0 } \
{ 0 } & { w_{2}^* }
end{bmatrix}
$

where $*$ denotes the hermitian operation. Now all lower case vectors are column vectors.

This problem can be restated as

$arg min _{{w_1},{w_2}}mathbb{E} ,,left[leftVertbegin{bmatrix}
bf s_1^* \
bf s_{2}^*
end{bmatrix} - begin{bmatrix}
{bf y_{1}^*} &{bf 0 } \
{bf 0 } & {bf y_{2}^* }
end{bmatrix}begin{bmatrix}
w_1 \
w_{2}
end{bmatrix}rightVert^2right] $.

After this, the problem decouples to solving for $w_1$ and $w_2$. For eg. for $w_1$, we need to minimize $J =mathbb E[ ||mathbf s_1^* - mathbf y_1^* w_1||^2]$.

Just expand the inside argument and differentiate w.r.t. $w_1^*$ and put the gradient to $0$. We get the solution for $w_1$ as

$mathbb E [mathbf y_1 mathbf y_1^*] w_1 = mathbb E [mathbf y_1^* mathbf s_1^*]$.

Now, assuming you can find the correlation matrix of $mathbf y_1$ and it is invertible, and the cross correlation between $mathbf y_1$ and $mathbf s_1$, you can find $w_1$. Similarly, you can solve for $w_2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    i am not sure... wht how did you decompose vector y?
    $endgroup$
    – Tyrone
    Oct 22 '15 at 1:49










  • $begingroup$
    The Nx1 vector $mathbf y$ is split into two vectors $mathbf y_1$ and $mathbf y_2$ each N/2 x1, stacked on top of each other. Hope that clears the confusion.
    $endgroup$
    – shaktiman
    Oct 22 '15 at 3:31
















0












0








0





$begingroup$

I assume you wish to find the non random matrix $W$ and hence its elements are not boldface in my solution. $mathbf s$ and $mathbf y$ are random vectors, hence they are boldface. First let me write the matrix ${W}
= begin{bmatrix}
{w_{1}^*} &{ 0 } \
{ 0 } & { w_{2}^* }
end{bmatrix}
$

where $*$ denotes the hermitian operation. Now all lower case vectors are column vectors.

This problem can be restated as

$arg min _{{w_1},{w_2}}mathbb{E} ,,left[leftVertbegin{bmatrix}
bf s_1^* \
bf s_{2}^*
end{bmatrix} - begin{bmatrix}
{bf y_{1}^*} &{bf 0 } \
{bf 0 } & {bf y_{2}^* }
end{bmatrix}begin{bmatrix}
w_1 \
w_{2}
end{bmatrix}rightVert^2right] $.

After this, the problem decouples to solving for $w_1$ and $w_2$. For eg. for $w_1$, we need to minimize $J =mathbb E[ ||mathbf s_1^* - mathbf y_1^* w_1||^2]$.

Just expand the inside argument and differentiate w.r.t. $w_1^*$ and put the gradient to $0$. We get the solution for $w_1$ as

$mathbb E [mathbf y_1 mathbf y_1^*] w_1 = mathbb E [mathbf y_1^* mathbf s_1^*]$.

Now, assuming you can find the correlation matrix of $mathbf y_1$ and it is invertible, and the cross correlation between $mathbf y_1$ and $mathbf s_1$, you can find $w_1$. Similarly, you can solve for $w_2$.






share|cite|improve this answer









$endgroup$



I assume you wish to find the non random matrix $W$ and hence its elements are not boldface in my solution. $mathbf s$ and $mathbf y$ are random vectors, hence they are boldface. First let me write the matrix ${W}
= begin{bmatrix}
{w_{1}^*} &{ 0 } \
{ 0 } & { w_{2}^* }
end{bmatrix}
$

where $*$ denotes the hermitian operation. Now all lower case vectors are column vectors.

This problem can be restated as

$arg min _{{w_1},{w_2}}mathbb{E} ,,left[leftVertbegin{bmatrix}
bf s_1^* \
bf s_{2}^*
end{bmatrix} - begin{bmatrix}
{bf y_{1}^*} &{bf 0 } \
{bf 0 } & {bf y_{2}^* }
end{bmatrix}begin{bmatrix}
w_1 \
w_{2}
end{bmatrix}rightVert^2right] $.

After this, the problem decouples to solving for $w_1$ and $w_2$. For eg. for $w_1$, we need to minimize $J =mathbb E[ ||mathbf s_1^* - mathbf y_1^* w_1||^2]$.

Just expand the inside argument and differentiate w.r.t. $w_1^*$ and put the gradient to $0$. We get the solution for $w_1$ as

$mathbb E [mathbf y_1 mathbf y_1^*] w_1 = mathbb E [mathbf y_1^* mathbf s_1^*]$.

Now, assuming you can find the correlation matrix of $mathbf y_1$ and it is invertible, and the cross correlation between $mathbf y_1$ and $mathbf s_1$, you can find $w_1$. Similarly, you can solve for $w_2$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Oct 22 '15 at 0:29









shaktimanshaktiman

628




628












  • $begingroup$
    i am not sure... wht how did you decompose vector y?
    $endgroup$
    – Tyrone
    Oct 22 '15 at 1:49










  • $begingroup$
    The Nx1 vector $mathbf y$ is split into two vectors $mathbf y_1$ and $mathbf y_2$ each N/2 x1, stacked on top of each other. Hope that clears the confusion.
    $endgroup$
    – shaktiman
    Oct 22 '15 at 3:31




















  • $begingroup$
    i am not sure... wht how did you decompose vector y?
    $endgroup$
    – Tyrone
    Oct 22 '15 at 1:49










  • $begingroup$
    The Nx1 vector $mathbf y$ is split into two vectors $mathbf y_1$ and $mathbf y_2$ each N/2 x1, stacked on top of each other. Hope that clears the confusion.
    $endgroup$
    – shaktiman
    Oct 22 '15 at 3:31


















$begingroup$
i am not sure... wht how did you decompose vector y?
$endgroup$
– Tyrone
Oct 22 '15 at 1:49




$begingroup$
i am not sure... wht how did you decompose vector y?
$endgroup$
– Tyrone
Oct 22 '15 at 1:49












$begingroup$
The Nx1 vector $mathbf y$ is split into two vectors $mathbf y_1$ and $mathbf y_2$ each N/2 x1, stacked on top of each other. Hope that clears the confusion.
$endgroup$
– shaktiman
Oct 22 '15 at 3:31






$begingroup$
The Nx1 vector $mathbf y$ is split into two vectors $mathbf y_1$ and $mathbf y_2$ each N/2 x1, stacked on top of each other. Hope that clears the confusion.
$endgroup$
– shaktiman
Oct 22 '15 at 3:31




















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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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