Show $f(x,y,z) = frac{log(y/x)}{log(z/x)}$ is quasi concave for $x ge y > 0$, $x ge z > 0$.Convexity of...
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Show $f(x,y,z) = frac{log(y/x)}{log(z/x)}$ is quasi concave for $x ge y > 0$, $x ge z > 0$.
Convexity of a SetHow do I show that $ d((x_1, x_2), (y_1, y_2)) = |y_1 - x_1| + |y_2 - x_2|$ is a metric?How to show $f(x,y,z)=(xy, yz,xz,x^2-y^2)$ is injective?A proof of property of log-concaveShowing that the Triangle Inequality holds for the $L_infty$ norm as a metric.Proving ${(x_1, x_2) in mathbb{R}^2_{+} mid x_1 x_2 geq 1}$ is convexStokes theorem for CuboidEstimating Parameters of Log-Concave ModelQuasiconcavity of the product functionConcavity of $ln(x^alpha - y)$ function
$begingroup$
I'd like to show that
$$
frac{logfrac{y_1+y_2}{x_1+x_2}}{logfrac{z_1+z_2}{x_1+x_2}}
ge
minleft{
frac{log(y_1/x_1)}{log(z_1/x_1)},
frac{log(y_2/x_2)}{log(z_2/x_2)}
right}.
$$
This would follow if $f(x,y,z) =
frac{log(y/x)}{log(z/x)}$ was quasi concave.
One way to show this is if $f(x,y,z)gealpha$ defines a convex set for all $alphage 0$.
Plotting for various values of $alpha$, this is clearly the case.
From the first order condition, it suffices if
$f(vec y) le f(vec x) implies nabla f(vec x)^T(vec y-vec x) ge 0.$
Let $x'ge y',z'$ be such that $f(x',y',z')le f(x,y,z)$ then we must show
$$z log left(frac{x}{z}right) (y x'-x y')
ge y log left(frac{x}{y}right) (z x'-x z'),$$
which unfortunately doesn't seem any easier.
Alternatively, since $log x$ is quasi-linear, perhaps it suffices to show that $(x-y)/(x-z)$ is quasi-concave? I don't know a theorem to this effect, however.
Do you see a nice argument I might use to simplify this problem? Perhaps just tackling the original inequality directly?
real-analysis convex-analysis convex-optimization
asked Mar 11 at 11:21
Thomas AhleThomas Ahle
1,4641322
$endgroup$
add a comment |
$begingroup$
I'd like to show that
$$
frac{logfrac{y_1+y_2}{x_1+x_2}}{logfrac{z_1+z_2}{x_1+x_2}}
ge
minleft{
frac{log(y_1/x_1)}{log(z_1/x_1)},
frac{log(y_2/x_2)}{log(z_2/x_2)}
right}.
$$
This would follow if $f(x,y,z) =
frac{log(y/x)}{log(z/x)}$ was quasi concave.
One way to show this is if $f(x,y,z)gealpha$ defines a convex set for all $alphage 0$.
Plotting for various values of $alpha$, this is clearly the case.
From the first order condition, it suffices if
$f(vec y) le f(vec x) implies nabla f(vec x)^T(vec y-vec x) ge 0.$
Let $x'ge y',z'$ be such that $f(x',y',z')le f(x,y,z)$ then we must show
$$z log left(frac{x}{z}right) (y x'-x y')
ge y log left(frac{x}{y}right) (z x'-x z'),$$
which unfortunately doesn't seem any easier.
Alternatively, since $log x$ is quasi-linear, perhaps it suffices to show that $(x-y)/(x-z)$ is quasi-concave? I don't know a theorem to this effect, however.
Do you see a nice argument I might use to simplify this problem? Perhaps just tackling the original inequality directly?
real-analysis convex-analysis convex-optimization
asked Mar 11 at 11:21
Thomas AhleThomas Ahle
1,4641322
$endgroup$
$begingroup$
Could you clearly define the domain? Can $x/y/z$ all be negative?
$endgroup$
– LinAlg
Mar 11 at 19:05
$begingroup$
@LinAlg Right! They are all positive. They are also at most 1, though that doesn't seem to make a difference.
$endgroup$
– Thomas Ahle
Mar 11 at 19:50
1
$begingroup$
Ok, just to make sure, $log(z/x)$ can be both negative and positive, right? Or did you mean $x geq y > 0$ and $x > z > 0$?
$endgroup$
– LinAlg
Mar 11 at 19:53
$begingroup$
@LinAlg I didn't even think of that obvious ambiguity! It is always negative, or if we write it as $log(x/y)/log(x/z)$.
$endgroup$
– Thomas Ahle
Mar 11 at 19:56
add a comment |
$begingroup$
I'd like to show that
$$
frac{logfrac{y_1+y_2}{x_1+x_2}}{logfrac{z_1+z_2}{x_1+x_2}}
ge
minleft{
frac{log(y_1/x_1)}{log(z_1/x_1)},
frac{log(y_2/x_2)}{log(z_2/x_2)}
right}.
$$
This would follow if $f(x,y,z) =
frac{log(y/x)}{log(z/x)}$ was quasi concave.
One way to show this is if $f(x,y,z)gealpha$ defines a convex set for all $alphage 0$.
Plotting for various values of $alpha$, this is clearly the case.
From the first order condition, it suffices if
$f(vec y) le f(vec x) implies nabla f(vec x)^T(vec y-vec x) ge 0.$
Let $x'ge y',z'$ be such that $f(x',y',z')le f(x,y,z)$ then we must show
$$z log left(frac{x}{z}right) (y x'-x y')
ge y log left(frac{x}{y}right) (z x'-x z'),$$
which unfortunately doesn't seem any easier.
Alternatively, since $log x$ is quasi-linear, perhaps it suffices to show that $(x-y)/(x-z)$ is quasi-concave? I don't know a theorem to this effect, however.
Do you see a nice argument I might use to simplify this problem? Perhaps just tackling the original inequality directly?
real-analysis convex-analysis convex-optimization
asked Mar 11 at 11:21
Thomas AhleThomas Ahle
1,4641322
$endgroup$
I'd like to show that
$$
frac{logfrac{y_1+y_2}{x_1+x_2}}{logfrac{z_1+z_2}{x_1+x_2}}
ge
minleft{
frac{log(y_1/x_1)}{log(z_1/x_1)},
frac{log(y_2/x_2)}{log(z_2/x_2)}
right}.
$$
This would follow if $f(x,y,z) =
frac{log(y/x)}{log(z/x)}$ was quasi concave.
One way to show this is if $f(x,y,z)gealpha$ defines a convex set for all $alphage 0$.
Plotting for various values of $alpha$, this is clearly the case.
From the first order condition, it suffices if
$f(vec y) le f(vec x) implies nabla f(vec x)^T(vec y-vec x) ge 0.$
Let $x'ge y',z'$ be such that $f(x',y',z')le f(x,y,z)$ then we must show
$$z log left(frac{x}{z}right) (y x'-x y')
ge y log left(frac{x}{y}right) (z x'-x z'),$$
which unfortunately doesn't seem any easier.
Alternatively, since $log x$ is quasi-linear, perhaps it suffices to show that $(x-y)/(x-z)$ is quasi-concave? I don't know a theorem to this effect, however.
Do you see a nice argument I might use to simplify this problem? Perhaps just tackling the original inequality directly?
real-analysis convex-analysis convex-optimization
real-analysis convex-analysis convex-optimization
asked Mar 11 at 11:21
Thomas AhleThomas Ahle
1,4641322
asked Mar 11 at 11:21
Thomas AhleThomas Ahle
1,4641322
edited Mar 11 at 19:56
Thomas Ahle
asked Mar 11 at 11:21
Thomas AhleThomas Ahle
1,4641322
asked Mar 11 at 11:21
Thomas AhleThomas Ahle
1,4641322
asked Mar 11 at 11:21
Thomas AhleThomas Ahle
1,4641322
1,4641322
$begingroup$
Could you clearly define the domain? Can $x/y/z$ all be negative?
$endgroup$
– LinAlg
Mar 11 at 19:05
$begingroup$
@LinAlg Right! They are all positive. They are also at most 1, though that doesn't seem to make a difference.
$endgroup$
– Thomas Ahle
Mar 11 at 19:50
1
$begingroup$
Ok, just to make sure, $log(z/x)$ can be both negative and positive, right? Or did you mean $x geq y > 0$ and $x > z > 0$?
$endgroup$
– LinAlg
Mar 11 at 19:53
$begingroup$
@LinAlg I didn't even think of that obvious ambiguity! It is always negative, or if we write it as $log(x/y)/log(x/z)$.
$endgroup$
– Thomas Ahle
Mar 11 at 19:56
add a comment |
$begingroup$
Could you clearly define the domain? Can $x/y/z$ all be negative?
$endgroup$
– LinAlg
Mar 11 at 19:05
$begingroup$
@LinAlg Right! They are all positive. They are also at most 1, though that doesn't seem to make a difference.
$endgroup$
– Thomas Ahle
Mar 11 at 19:50
1
$begingroup$
Ok, just to make sure, $log(z/x)$ can be both negative and positive, right? Or did you mean $x geq y > 0$ and $x > z > 0$?
$endgroup$
– LinAlg
Mar 11 at 19:53
$begingroup$
@LinAlg I didn't even think of that obvious ambiguity! It is always negative, or if we write it as $log(x/y)/log(x/z)$.
$endgroup$
– Thomas Ahle
Mar 11 at 19:56
$begingroup$
Could you clearly define the domain? Can $x/y/z$ all be negative?
$endgroup$
– LinAlg
Mar 11 at 19:05
$begingroup$
Could you clearly define the domain? Can $x/y/z$ all be negative?
$endgroup$
– LinAlg
Mar 11 at 19:05
$begingroup$
@LinAlg Right! They are all positive. They are also at most 1, though that doesn't seem to make a difference.
$endgroup$
– Thomas Ahle
Mar 11 at 19:50
$begingroup$
@LinAlg Right! They are all positive. They are also at most 1, though that doesn't seem to make a difference.
$endgroup$
– Thomas Ahle
Mar 11 at 19:50
1
1
$begingroup$
Ok, just to make sure, $log(z/x)$ can be both negative and positive, right? Or did you mean $x geq y > 0$ and $x > z > 0$?
$endgroup$
– LinAlg
Mar 11 at 19:53
$begingroup$
Ok, just to make sure, $log(z/x)$ can be both negative and positive, right? Or did you mean $x geq y > 0$ and $x > z > 0$?
$endgroup$
– LinAlg
Mar 11 at 19:53
$begingroup$
@LinAlg I didn't even think of that obvious ambiguity! It is always negative, or if we write it as $log(x/y)/log(x/z)$.
$endgroup$
– Thomas Ahle
Mar 11 at 19:56
$begingroup$
@LinAlg I didn't even think of that obvious ambiguity! It is always negative, or if we write it as $log(x/y)/log(x/z)$.
$endgroup$
– Thomas Ahle
Mar 11 at 19:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let me give you a possible method. Consider the superlevel set ${(x,y,z) : log(y/x) / log(z/x) geq alpha, x geq y > 0, x > z > 0 }$. Since $log(z/x)$ is negative, the sublevel set is equivalent to the set:
$${(x,y,z) : log(y/x) leq alpha log(z/x), x geq y > 0, x > z > 0 }$$
$$={(x,y,z) : y leq x^{1-alpha}z^{alpha}, x geq y > 0, x > z > 0 }.$$
Consider the Hessian of $f(x,z) = x^{1-alpha}z^{alpha}$. If it is negative semidefinite, the set is convex and you are done. Otherwise, try the same but with $x$ or $z$ on one side.
answered Mar 11 at 20:51
LinAlgLinAlg
10.1k1521
$endgroup$
$begingroup$
The eigenvalues of the Hessian are $0$ and $-(1-alpha) a x^{-alpha-1} z^{alpha-2} left(x^2+z^2right)$, so I guess it is negative semidefinite for all $alphale1$. That suffices for my purposes! Is ${(x,y,z) : y le f(x,y)}$ always convex when $f$ is concave?
$endgroup$
– Thomas Ahle
Mar 11 at 21:16
1
$begingroup$
@ThomasAhle the answer to your last question is yes; the equivalent set in convex optimization is ${x : f(x) leq 0}$ where $f$ is convex.
$endgroup$
– LinAlg
Mar 12 at 0:44
add a comment |
Your Answer
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$begingroup$
Let me give you a possible method. Consider the superlevel set ${(x,y,z) : log(y/x) / log(z/x) geq alpha, x geq y > 0, x > z > 0 }$. Since $log(z/x)$ is negative, the sublevel set is equivalent to the set:
$${(x,y,z) : log(y/x) leq alpha log(z/x), x geq y > 0, x > z > 0 }$$
$$={(x,y,z) : y leq x^{1-alpha}z^{alpha}, x geq y > 0, x > z > 0 }.$$
Consider the Hessian of $f(x,z) = x^{1-alpha}z^{alpha}$. If it is negative semidefinite, the set is convex and you are done. Otherwise, try the same but with $x$ or $z$ on one side.
answered Mar 11 at 20:51
LinAlgLinAlg
10.1k1521
$endgroup$
$begingroup$
The eigenvalues of the Hessian are $0$ and $-(1-alpha) a x^{-alpha-1} z^{alpha-2} left(x^2+z^2right)$, so I guess it is negative semidefinite for all $alphale1$. That suffices for my purposes! Is ${(x,y,z) : y le f(x,y)}$ always convex when $f$ is concave?
$endgroup$
– Thomas Ahle
Mar 11 at 21:16
1
$begingroup$
@ThomasAhle the answer to your last question is yes; the equivalent set in convex optimization is ${x : f(x) leq 0}$ where $f$ is convex.
$endgroup$
– LinAlg
Mar 12 at 0:44
add a comment |
$begingroup$
Let me give you a possible method. Consider the superlevel set ${(x,y,z) : log(y/x) / log(z/x) geq alpha, x geq y > 0, x > z > 0 }$. Since $log(z/x)$ is negative, the sublevel set is equivalent to the set:
$${(x,y,z) : log(y/x) leq alpha log(z/x), x geq y > 0, x > z > 0 }$$
$$={(x,y,z) : y leq x^{1-alpha}z^{alpha}, x geq y > 0, x > z > 0 }.$$
Consider the Hessian of $f(x,z) = x^{1-alpha}z^{alpha}$. If it is negative semidefinite, the set is convex and you are done. Otherwise, try the same but with $x$ or $z$ on one side.
answered Mar 11 at 20:51
LinAlgLinAlg
10.1k1521
$endgroup$
$begingroup$
The eigenvalues of the Hessian are $0$ and $-(1-alpha) a x^{-alpha-1} z^{alpha-2} left(x^2+z^2right)$, so I guess it is negative semidefinite for all $alphale1$. That suffices for my purposes! Is ${(x,y,z) : y le f(x,y)}$ always convex when $f$ is concave?
$endgroup$
– Thomas Ahle
Mar 11 at 21:16
1
$begingroup$
@ThomasAhle the answer to your last question is yes; the equivalent set in convex optimization is ${x : f(x) leq 0}$ where $f$ is convex.
$endgroup$
– LinAlg
Mar 12 at 0:44
add a comment |
$begingroup$
Let me give you a possible method. Consider the superlevel set ${(x,y,z) : log(y/x) / log(z/x) geq alpha, x geq y > 0, x > z > 0 }$. Since $log(z/x)$ is negative, the sublevel set is equivalent to the set:
$${(x,y,z) : log(y/x) leq alpha log(z/x), x geq y > 0, x > z > 0 }$$
$$={(x,y,z) : y leq x^{1-alpha}z^{alpha}, x geq y > 0, x > z > 0 }.$$
Consider the Hessian of $f(x,z) = x^{1-alpha}z^{alpha}$. If it is negative semidefinite, the set is convex and you are done. Otherwise, try the same but with $x$ or $z$ on one side.
answered Mar 11 at 20:51
LinAlgLinAlg
10.1k1521
$endgroup$
Let me give you a possible method. Consider the superlevel set ${(x,y,z) : log(y/x) / log(z/x) geq alpha, x geq y > 0, x > z > 0 }$. Since $log(z/x)$ is negative, the sublevel set is equivalent to the set:
$${(x,y,z) : log(y/x) leq alpha log(z/x), x geq y > 0, x > z > 0 }$$
$$={(x,y,z) : y leq x^{1-alpha}z^{alpha}, x geq y > 0, x > z > 0 }.$$
Consider the Hessian of $f(x,z) = x^{1-alpha}z^{alpha}$. If it is negative semidefinite, the set is convex and you are done. Otherwise, try the same but with $x$ or $z$ on one side.
answered Mar 11 at 20:51
LinAlgLinAlg
10.1k1521
answered Mar 11 at 20:51
LinAlgLinAlg
10.1k1521
answered Mar 11 at 20:51
LinAlgLinAlg
10.1k1521
answered Mar 11 at 20:51
LinAlgLinAlg
10.1k1521
10.1k1521
$begingroup$
The eigenvalues of the Hessian are $0$ and $-(1-alpha) a x^{-alpha-1} z^{alpha-2} left(x^2+z^2right)$, so I guess it is negative semidefinite for all $alphale1$. That suffices for my purposes! Is ${(x,y,z) : y le f(x,y)}$ always convex when $f$ is concave?
$endgroup$
– Thomas Ahle
Mar 11 at 21:16
1
$begingroup$
@ThomasAhle the answer to your last question is yes; the equivalent set in convex optimization is ${x : f(x) leq 0}$ where $f$ is convex.
$endgroup$
– LinAlg
Mar 12 at 0:44
add a comment |
$begingroup$
The eigenvalues of the Hessian are $0$ and $-(1-alpha) a x^{-alpha-1} z^{alpha-2} left(x^2+z^2right)$, so I guess it is negative semidefinite for all $alphale1$. That suffices for my purposes! Is ${(x,y,z) : y le f(x,y)}$ always convex when $f$ is concave?
$endgroup$
– Thomas Ahle
Mar 11 at 21:16
1
$begingroup$
@ThomasAhle the answer to your last question is yes; the equivalent set in convex optimization is ${x : f(x) leq 0}$ where $f$ is convex.
$endgroup$
– LinAlg
Mar 12 at 0:44
$begingroup$
The eigenvalues of the Hessian are $0$ and $-(1-alpha) a x^{-alpha-1} z^{alpha-2} left(x^2+z^2right)$, so I guess it is negative semidefinite for all $alphale1$. That suffices for my purposes! Is ${(x,y,z) : y le f(x,y)}$ always convex when $f$ is concave?
$endgroup$
– Thomas Ahle
Mar 11 at 21:16
$begingroup$
The eigenvalues of the Hessian are $0$ and $-(1-alpha) a x^{-alpha-1} z^{alpha-2} left(x^2+z^2right)$, so I guess it is negative semidefinite for all $alphale1$. That suffices for my purposes! Is ${(x,y,z) : y le f(x,y)}$ always convex when $f$ is concave?
$endgroup$
– Thomas Ahle
Mar 11 at 21:16
1
1
$begingroup$
@ThomasAhle the answer to your last question is yes; the equivalent set in convex optimization is ${x : f(x) leq 0}$ where $f$ is convex.
$endgroup$
– LinAlg
Mar 12 at 0:44
$begingroup$
@ThomasAhle the answer to your last question is yes; the equivalent set in convex optimization is ${x : f(x) leq 0}$ where $f$ is convex.
$endgroup$
– LinAlg
Mar 12 at 0:44
add a comment |
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$begingroup$
Could you clearly define the domain? Can $x/y/z$ all be negative?
$endgroup$
– LinAlg
Mar 11 at 19:05
$begingroup$
@LinAlg Right! They are all positive. They are also at most 1, though that doesn't seem to make a difference.
$endgroup$
– Thomas Ahle
Mar 11 at 19:50
1
$begingroup$
Ok, just to make sure, $log(z/x)$ can be both negative and positive, right? Or did you mean $x geq y > 0$ and $x > z > 0$?
$endgroup$
– LinAlg
Mar 11 at 19:53
$begingroup$
@LinAlg I didn't even think of that obvious ambiguity! It is always negative, or if we write it as $log(x/y)/log(x/z)$.
$endgroup$
– Thomas Ahle
Mar 11 at 19:56