maximizing concave function with parameter
We want to maximize the convex function:
$$vec{q} cdot vec{x} - lambda||vec{x} - vec{1}||_2^2$$ where $lambda$ is some parameter.
I'm looking at a solution that states that the maximum is a piecewise function depending on whether $||q|| leq 2 cdot lambda$ or otherwise.
The value of $x$ that maximizes the function above is claimed to be: $$begin{cases}
(1/2)vec{q}/lambda + vec{1} & ||vec{q}|| leq 2lambda \
vec{q}/||vec{q}|| + vec{1} & ||vec{q}||> 2lambda\
end{cases}$$
I didn't even consider how the norm of $vec{q}$ might relate to $lambda$ when trying to compute this myself. In fact, I arrived at only the solution for when $||vec{q}|| leq 2lambda$ by taking the derivative of $vec{q} cdot vec{x} - lambda||vec{x} - vec{1}||_2^2$ with respect to $vec{x}$ coordinate-wise, setting it equal to $0$, and then solving for $vec{x}$ again coordinate-wise.
Can someone please explain why we should care whether or not $||vec{q}|| > 2lambda$, and how to compute the $x$ that maximizes the above function when $||vec{q}|| > 2lambda$?
Edit: there is a constraint on $vec{x}$ that was mentioned elsewhere in the solution: $vec{x}$ is in the $2-$norm unit ball centered at $1$.
optimization convex-analysis convex-optimization maxima-minima
asked Dec 11 '18 at 5:19
0k33
12010
add a comment |
We want to maximize the convex function:
$$vec{q} cdot vec{x} - lambda||vec{x} - vec{1}||_2^2$$ where $lambda$ is some parameter.
I'm looking at a solution that states that the maximum is a piecewise function depending on whether $||q|| leq 2 cdot lambda$ or otherwise.
The value of $x$ that maximizes the function above is claimed to be: $$begin{cases}
(1/2)vec{q}/lambda + vec{1} & ||vec{q}|| leq 2lambda \
vec{q}/||vec{q}|| + vec{1} & ||vec{q}||> 2lambda\
end{cases}$$
I didn't even consider how the norm of $vec{q}$ might relate to $lambda$ when trying to compute this myself. In fact, I arrived at only the solution for when $||vec{q}|| leq 2lambda$ by taking the derivative of $vec{q} cdot vec{x} - lambda||vec{x} - vec{1}||_2^2$ with respect to $vec{x}$ coordinate-wise, setting it equal to $0$, and then solving for $vec{x}$ again coordinate-wise.
Can someone please explain why we should care whether or not $||vec{q}|| > 2lambda$, and how to compute the $x$ that maximizes the above function when $||vec{q}|| > 2lambda$?
Edit: there is a constraint on $vec{x}$ that was mentioned elsewhere in the solution: $vec{x}$ is in the $2-$norm unit ball centered at $1$.
optimization convex-analysis convex-optimization maxima-minima
asked Dec 11 '18 at 5:19
0k33
12010
add a comment |
We want to maximize the convex function:
$$vec{q} cdot vec{x} - lambda||vec{x} - vec{1}||_2^2$$ where $lambda$ is some parameter.
I'm looking at a solution that states that the maximum is a piecewise function depending on whether $||q|| leq 2 cdot lambda$ or otherwise.
The value of $x$ that maximizes the function above is claimed to be: $$begin{cases}
(1/2)vec{q}/lambda + vec{1} & ||vec{q}|| leq 2lambda \
vec{q}/||vec{q}|| + vec{1} & ||vec{q}||> 2lambda\
end{cases}$$
I didn't even consider how the norm of $vec{q}$ might relate to $lambda$ when trying to compute this myself. In fact, I arrived at only the solution for when $||vec{q}|| leq 2lambda$ by taking the derivative of $vec{q} cdot vec{x} - lambda||vec{x} - vec{1}||_2^2$ with respect to $vec{x}$ coordinate-wise, setting it equal to $0$, and then solving for $vec{x}$ again coordinate-wise.
Can someone please explain why we should care whether or not $||vec{q}|| > 2lambda$, and how to compute the $x$ that maximizes the above function when $||vec{q}|| > 2lambda$?
Edit: there is a constraint on $vec{x}$ that was mentioned elsewhere in the solution: $vec{x}$ is in the $2-$norm unit ball centered at $1$.
optimization convex-analysis convex-optimization maxima-minima
asked Dec 11 '18 at 5:19
0k33
12010
We want to maximize the convex function:
$$vec{q} cdot vec{x} - lambda||vec{x} - vec{1}||_2^2$$ where $lambda$ is some parameter.
I'm looking at a solution that states that the maximum is a piecewise function depending on whether $||q|| leq 2 cdot lambda$ or otherwise.
The value of $x$ that maximizes the function above is claimed to be: $$begin{cases}
(1/2)vec{q}/lambda + vec{1} & ||vec{q}|| leq 2lambda \
vec{q}/||vec{q}|| + vec{1} & ||vec{q}||> 2lambda\
end{cases}$$
I didn't even consider how the norm of $vec{q}$ might relate to $lambda$ when trying to compute this myself. In fact, I arrived at only the solution for when $||vec{q}|| leq 2lambda$ by taking the derivative of $vec{q} cdot vec{x} - lambda||vec{x} - vec{1}||_2^2$ with respect to $vec{x}$ coordinate-wise, setting it equal to $0$, and then solving for $vec{x}$ again coordinate-wise.
Can someone please explain why we should care whether or not $||vec{q}|| > 2lambda$, and how to compute the $x$ that maximizes the above function when $||vec{q}|| > 2lambda$?
Edit: there is a constraint on $vec{x}$ that was mentioned elsewhere in the solution: $vec{x}$ is in the $2-$norm unit ball centered at $1$.
optimization convex-analysis convex-optimization maxima-minima
optimization convex-analysis convex-optimization maxima-minima
asked Dec 11 '18 at 5:19
0k33
12010
asked Dec 11 '18 at 5:19
0k33
12010
edited Dec 11 '18 at 21:54
asked Dec 11 '18 at 5:19
0k33
12010
asked Dec 11 '18 at 5:19
0k33
12010
asked Dec 11 '18 at 5:19
0k33
12010
12010
add a comment |
add a comment |
1 Answer
1
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The function (which is concave, not convex) is always maximized at $x= 1 + frac{q}{2 lambda}$ for $lambda > 0$. Perhaps there is some constraint on $x$. Also, the solution you suggest for the second case is a number, not a vector.
answered Dec 11 '18 at 6:34
mineiro
812
1
you actually need $lambda>0$ as otherwise the function is unbounded (and not concave)
– LinAlg
Dec 11 '18 at 14:25
1
Yes, that’s right. I edit my answer.
– mineiro
Dec 11 '18 at 15:39
@mineiro you are so right. I've edited my question to add the constraint (and I copied down the second part of the piecewise totally wrong, I have no idea where I got it from). Even with the constraint, it's not immediately coming to me!
– 0k33
Dec 11 '18 at 21:56
I figured it out! never mind!
– 0k33
Dec 13 '18 at 2:09
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
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oldest
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active
oldest
votes
The function (which is concave, not convex) is always maximized at $x= 1 + frac{q}{2 lambda}$ for $lambda > 0$. Perhaps there is some constraint on $x$. Also, the solution you suggest for the second case is a number, not a vector.
answered Dec 11 '18 at 6:34
mineiro
812
1
you actually need $lambda>0$ as otherwise the function is unbounded (and not concave)
– LinAlg
Dec 11 '18 at 14:25
1
Yes, that’s right. I edit my answer.
– mineiro
Dec 11 '18 at 15:39
@mineiro you are so right. I've edited my question to add the constraint (and I copied down the second part of the piecewise totally wrong, I have no idea where I got it from). Even with the constraint, it's not immediately coming to me!
– 0k33
Dec 11 '18 at 21:56
I figured it out! never mind!
– 0k33
Dec 13 '18 at 2:09
add a comment |
The function (which is concave, not convex) is always maximized at $x= 1 + frac{q}{2 lambda}$ for $lambda > 0$. Perhaps there is some constraint on $x$. Also, the solution you suggest for the second case is a number, not a vector.
answered Dec 11 '18 at 6:34
mineiro
812
1
you actually need $lambda>0$ as otherwise the function is unbounded (and not concave)
– LinAlg
Dec 11 '18 at 14:25
1
Yes, that’s right. I edit my answer.
– mineiro
Dec 11 '18 at 15:39
@mineiro you are so right. I've edited my question to add the constraint (and I copied down the second part of the piecewise totally wrong, I have no idea where I got it from). Even with the constraint, it's not immediately coming to me!
– 0k33
Dec 11 '18 at 21:56
I figured it out! never mind!
– 0k33
Dec 13 '18 at 2:09
add a comment |
The function (which is concave, not convex) is always maximized at $x= 1 + frac{q}{2 lambda}$ for $lambda > 0$. Perhaps there is some constraint on $x$. Also, the solution you suggest for the second case is a number, not a vector.
answered Dec 11 '18 at 6:34
mineiro
812
The function (which is concave, not convex) is always maximized at $x= 1 + frac{q}{2 lambda}$ for $lambda > 0$. Perhaps there is some constraint on $x$. Also, the solution you suggest for the second case is a number, not a vector.
answered Dec 11 '18 at 6:34
mineiro
812
edited Dec 11 '18 at 15:40
answered Dec 11 '18 at 6:34
mineiro
812
answered Dec 11 '18 at 6:34
mineiro
812
answered Dec 11 '18 at 6:34
mineiro
812
812
1
you actually need $lambda>0$ as otherwise the function is unbounded (and not concave)
– LinAlg
Dec 11 '18 at 14:25
1
Yes, that’s right. I edit my answer.
– mineiro
Dec 11 '18 at 15:39
@mineiro you are so right. I've edited my question to add the constraint (and I copied down the second part of the piecewise totally wrong, I have no idea where I got it from). Even with the constraint, it's not immediately coming to me!
– 0k33
Dec 11 '18 at 21:56
I figured it out! never mind!
– 0k33
Dec 13 '18 at 2:09
add a comment |
1
you actually need $lambda>0$ as otherwise the function is unbounded (and not concave)
– LinAlg
Dec 11 '18 at 14:25
1
Yes, that’s right. I edit my answer.
– mineiro
Dec 11 '18 at 15:39
@mineiro you are so right. I've edited my question to add the constraint (and I copied down the second part of the piecewise totally wrong, I have no idea where I got it from). Even with the constraint, it's not immediately coming to me!
– 0k33
Dec 11 '18 at 21:56
I figured it out! never mind!
– 0k33
Dec 13 '18 at 2:09
1
1
you actually need $lambda>0$ as otherwise the function is unbounded (and not concave)
– LinAlg
Dec 11 '18 at 14:25
you actually need $lambda>0$ as otherwise the function is unbounded (and not concave)
– LinAlg
Dec 11 '18 at 14:25
1
1
Yes, that’s right. I edit my answer.
– mineiro
Dec 11 '18 at 15:39
Yes, that’s right. I edit my answer.
– mineiro
Dec 11 '18 at 15:39
@mineiro you are so right. I've edited my question to add the constraint (and I copied down the second part of the piecewise totally wrong, I have no idea where I got it from). Even with the constraint, it's not immediately coming to me!
– 0k33
Dec 11 '18 at 21:56
@mineiro you are so right. I've edited my question to add the constraint (and I copied down the second part of the piecewise totally wrong, I have no idea where I got it from). Even with the constraint, it's not immediately coming to me!
– 0k33
Dec 11 '18 at 21:56
I figured it out! never mind!
– 0k33
Dec 13 '18 at 2:09
I figured it out! never mind!
– 0k33
Dec 13 '18 at 2:09
add a comment |
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