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Concavity of $ln(x^alpha - y)$ function

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1

$begingroup$

I would like to prove that $f(x,y) = ln(x^alpha - y)$, with $alpha in (0,1)$ is strictly concave function. I can prove that $g(x,y) = ln(x - y)$ is strictly concancave. That is, let $z_1 = (x_1,y_1)$, $z_2 = (x_2,y_2)$, $lambda in (0,1)$ and $tildelambda = 1 - lambda$. Then, if $x_1 neq x_2$

beginequation
g(lambda z_1 +tildelambdaz_2)> lambda g(z_1) +tildelambdag(z_2)
endequation

With this fact, how I can shwo that $f$ is strictly concave?

calculus real-analysis

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asked Jun 15 '18 at 23:26

orrilloorrillo

315111

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1

$begingroup$

I would like to prove that $f(x,y) = ln(x^alpha - y)$, with $alpha in (0,1)$ is strictly concave function. I can prove that $g(x,y) = ln(x - y)$ is strictly concancave. That is, let $z_1 = (x_1,y_1)$, $z_2 = (x_2,y_2)$, $lambda in (0,1)$ and $tildelambda = 1 - lambda$. Then, if $x_1 neq x_2$

beginequation
g(lambda z_1 +tildelambdaz_2)> lambda g(z_1) +tildelambdag(z_2)
endequation

With this fact, how I can shwo that $f$ is strictly concave?

calculus real-analysis

share|cite|improve this question

asked Jun 15 '18 at 23:26

orrilloorrillo

315111

$endgroup$

add a comment | 

$begingroup$

I would like to prove that $f(x,y) = ln(x^alpha - y)$, with $alpha in (0,1)$ is strictly concave function. I can prove that $g(x,y) = ln(x - y)$ is strictly concancave. That is, let $z_1 = (x_1,y_1)$, $z_2 = (x_2,y_2)$, $lambda in (0,1)$ and $tildelambda = 1 - lambda$. Then, if $x_1 neq x_2$

beginequation
g(lambda z_1 +tildelambdaz_2)> lambda g(z_1) +tildelambdag(z_2)
endequation

With this fact, how I can shwo that $f$ is strictly concave?

calculus real-analysis

share|cite|improve this question

asked Jun 15 '18 at 23:26

orrilloorrillo

315111

$endgroup$

share|cite|improve this question

asked Jun 15 '18 at 23:26

orrilloorrillo

315111

share|cite|improve this question

asked Jun 15 '18 at 23:26

orrilloorrillo

315111

asked Jun 15 '18 at 23:26

orrilloorrillo

315111

asked Jun 15 '18 at 23:26

orrilloorrillo

315111

1 Answer
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0

$begingroup$

Since $x^alpha$ and $-y$ are concave, then so is $x^alpha-y$.
Since concavity implies log-concavity, $f(x,y)$ is concave.

I think the same should follow from strict concavity. The sum of a strictly concave function and a concave function is strictly concave.

share|cite|improve this answer

answered Mar 11 at 20:06

Thomas AhleThomas Ahle

1,4541321

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0

$begingroup$

Since $x^alpha$ and $-y$ are concave, then so is $x^alpha-y$.
Since concavity implies log-concavity, $f(x,y)$ is concave.

I think the same should follow from strict concavity. The sum of a strictly concave function and a concave function is strictly concave.

share|cite|improve this answer

answered Mar 11 at 20:06

Thomas AhleThomas Ahle

1,4541321

$endgroup$

add a comment | 

0

$begingroup$

Since $x^alpha$ and $-y$ are concave, then so is $x^alpha-y$.
Since concavity implies log-concavity, $f(x,y)$ is concave.

I think the same should follow from strict concavity. The sum of a strictly concave function and a concave function is strictly concave.

share|cite|improve this answer

answered Mar 11 at 20:06

Thomas AhleThomas Ahle

1,4541321

$endgroup$

add a comment | 

$begingroup$

Since $x^alpha$ and $-y$ are concave, then so is $x^alpha-y$.
Since concavity implies log-concavity, $f(x,y)$ is concave.

I think the same should follow from strict concavity. The sum of a strictly concave function and a concave function is strictly concave.

share|cite|improve this answer

answered Mar 11 at 20:06

Thomas AhleThomas Ahle

1,4541321

$endgroup$

share|cite|improve this answer

answered Mar 11 at 20:06

Thomas AhleThomas Ahle

1,4541321

answered Mar 11 at 20:06

Thomas AhleThomas Ahle

1,4541321

answered Mar 11 at 20:06

Thomas AhleThomas Ahle

1,4541321