Mixing
Proximal Operator of Weighted $ {L}_{2} $ Norm
$begingroup$
Define the weighted L2 norm $||x||_{2,w}=sqrt{sum_{i=1}^{n}w_ix_i^2}$. Find the formula for $operatorname{prox}_{lambda r}(y)$, where $lambda >0$.
By definition we have $$operatorname{prox}_{lambda r}(y)=argmin_x left(sqrt{sum_{i=1}^{n}w_ix_i^2}+frac{1}{2lambda}||x-y||^2_2right)$$
But I have no idea how to proceed from here. Any idea?
optimization convex-analysis convex-optimization
edited Feb 2 at 19:59
Royi
3,66512454
asked May 3 '17 at 7:01
user112358user112358
900518
$endgroup$
add a comment |
$begingroup$
Define the weighted L2 norm $||x||_{2,w}=sqrt{sum_{i=1}^{n}w_ix_i^2}$. Find the formula for $operatorname{prox}_{lambda r}(y)$, where $lambda >0$.
By definition we have $$operatorname{prox}_{lambda r}(y)=argmin_x left(sqrt{sum_{i=1}^{n}w_ix_i^2}+frac{1}{2lambda}||x-y||^2_2right)$$
But I have no idea how to proceed from here. Any idea?
optimization convex-analysis convex-optimization
edited Feb 2 at 19:59
Royi
3,66512454
asked May 3 '17 at 7:01
user112358user112358
900518
$endgroup$
add a comment |
$begingroup$
Define the weighted L2 norm $||x||_{2,w}=sqrt{sum_{i=1}^{n}w_ix_i^2}$. Find the formula for $operatorname{prox}_{lambda r}(y)$, where $lambda >0$.
By definition we have $$operatorname{prox}_{lambda r}(y)=argmin_x left(sqrt{sum_{i=1}^{n}w_ix_i^2}+frac{1}{2lambda}||x-y||^2_2right)$$
But I have no idea how to proceed from here. Any idea?
optimization convex-analysis convex-optimization
edited Feb 2 at 19:59
Royi
3,66512454
asked May 3 '17 at 7:01
user112358user112358
900518
$endgroup$
Define the weighted L2 norm $||x||_{2,w}=sqrt{sum_{i=1}^{n}w_ix_i^2}$. Find the formula for $operatorname{prox}_{lambda r}(y)$, where $lambda >0$.
By definition we have $$operatorname{prox}_{lambda r}(y)=argmin_x left(sqrt{sum_{i=1}^{n}w_ix_i^2}+frac{1}{2lambda}||x-y||^2_2right)$$
But I have no idea how to proceed from here. Any idea?
optimization convex-analysis convex-optimization
optimization convex-analysis convex-optimization
edited Feb 2 at 19:59
Royi
3,66512454
asked May 3 '17 at 7:01
user112358user112358
900518
edited Feb 2 at 19:59
Royi
3,66512454
asked May 3 '17 at 7:01
user112358user112358
900518
edited Feb 2 at 19:59
Royi
3,66512454
edited Feb 2 at 19:59
Royi
3,66512454
edited Feb 2 at 19:59
Royi
3,66512454
3,66512454
asked May 3 '17 at 7:01
user112358user112358
900518
asked May 3 '17 at 7:01
user112358user112358
900518
asked May 3 '17 at 7:01
user112358user112358
900518
900518
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It's the usual thing in such situations: set the gradient with respect to $x$ to zero and solve for $x$.
In this case, the $i$-th coordinate of the gradient is
$$ frac{w_ix_i}{sqrt{W}} + frac{x_i-y_i}{lambda}$$
where $W=sum_{i=1}^nw_ix_i^2$ is the weighted squared 2-norm. Setting the above to zero, we get
$$ x_i = frac{y_i}{lambda w_i/sqrt{W}+1}.$$
So these expressions for $i=1,ldots,n$ define the minimizer $x$.
answered May 3 '17 at 7:11
amakelovamakelov
2,617619
$endgroup$
$begingroup$
Maybe to clarify the first term can be obtained by applying the chain rule on $f(g(x)), cases{f(x) = sqrt{x}\ g(x) = sum_{i=1}^{n}w_i{x_i}^2}$.
$endgroup$
– mathreadler
May 3 '17 at 10:15
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's the usual thing in such situations: set the gradient with respect to $x$ to zero and solve for $x$.
In this case, the $i$-th coordinate of the gradient is
$$ frac{w_ix_i}{sqrt{W}} + frac{x_i-y_i}{lambda}$$
where $W=sum_{i=1}^nw_ix_i^2$ is the weighted squared 2-norm. Setting the above to zero, we get
$$ x_i = frac{y_i}{lambda w_i/sqrt{W}+1}.$$
So these expressions for $i=1,ldots,n$ define the minimizer $x$.
answered May 3 '17 at 7:11
amakelovamakelov
2,617619
$endgroup$
$begingroup$
Maybe to clarify the first term can be obtained by applying the chain rule on $f(g(x)), cases{f(x) = sqrt{x}\ g(x) = sum_{i=1}^{n}w_i{x_i}^2}$.
$endgroup$
– mathreadler
May 3 '17 at 10:15
add a comment |
$begingroup$
It's the usual thing in such situations: set the gradient with respect to $x$ to zero and solve for $x$.
In this case, the $i$-th coordinate of the gradient is
$$ frac{w_ix_i}{sqrt{W}} + frac{x_i-y_i}{lambda}$$
where $W=sum_{i=1}^nw_ix_i^2$ is the weighted squared 2-norm. Setting the above to zero, we get
$$ x_i = frac{y_i}{lambda w_i/sqrt{W}+1}.$$
So these expressions for $i=1,ldots,n$ define the minimizer $x$.
answered May 3 '17 at 7:11
amakelovamakelov
2,617619
$endgroup$
$begingroup$
Maybe to clarify the first term can be obtained by applying the chain rule on $f(g(x)), cases{f(x) = sqrt{x}\ g(x) = sum_{i=1}^{n}w_i{x_i}^2}$.
$endgroup$
– mathreadler
May 3 '17 at 10:15
add a comment |
$begingroup$
It's the usual thing in such situations: set the gradient with respect to $x$ to zero and solve for $x$.
In this case, the $i$-th coordinate of the gradient is
$$ frac{w_ix_i}{sqrt{W}} + frac{x_i-y_i}{lambda}$$
where $W=sum_{i=1}^nw_ix_i^2$ is the weighted squared 2-norm. Setting the above to zero, we get
$$ x_i = frac{y_i}{lambda w_i/sqrt{W}+1}.$$
So these expressions for $i=1,ldots,n$ define the minimizer $x$.
answered May 3 '17 at 7:11
amakelovamakelov
2,617619
$endgroup$
It's the usual thing in such situations: set the gradient with respect to $x$ to zero and solve for $x$.
In this case, the $i$-th coordinate of the gradient is
$$ frac{w_ix_i}{sqrt{W}} + frac{x_i-y_i}{lambda}$$
where $W=sum_{i=1}^nw_ix_i^2$ is the weighted squared 2-norm. Setting the above to zero, we get
$$ x_i = frac{y_i}{lambda w_i/sqrt{W}+1}.$$
So these expressions for $i=1,ldots,n$ define the minimizer $x$.
answered May 3 '17 at 7:11
amakelovamakelov
2,617619
edited May 3 '17 at 7:34
answered May 3 '17 at 7:11
amakelovamakelov
2,617619
answered May 3 '17 at 7:11
amakelovamakelov
2,617619
answered May 3 '17 at 7:11
amakelovamakelov
2,617619
2,617619
$begingroup$
Maybe to clarify the first term can be obtained by applying the chain rule on $f(g(x)), cases{f(x) = sqrt{x}\ g(x) = sum_{i=1}^{n}w_i{x_i}^2}$.
$endgroup$
– mathreadler
May 3 '17 at 10:15
add a comment |
$begingroup$
Maybe to clarify the first term can be obtained by applying the chain rule on $f(g(x)), cases{f(x) = sqrt{x}\ g(x) = sum_{i=1}^{n}w_i{x_i}^2}$.
$endgroup$
– mathreadler
May 3 '17 at 10:15
$begingroup$
Maybe to clarify the first term can be obtained by applying the chain rule on $f(g(x)), cases{f(x) = sqrt{x}\ g(x) = sum_{i=1}^{n}w_i{x_i}^2}$.
$endgroup$
– mathreadler
May 3 '17 at 10:15
$begingroup$
Maybe to clarify the first term can be obtained by applying the chain rule on $f(g(x)), cases{f(x) = sqrt{x}\ g(x) = sum_{i=1}^{n}w_i{x_i}^2}$.
$endgroup$
– mathreadler
May 3 '17 at 10:15
add a comment |
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