Concavity of $ln(x^alpha - y)$ functionTwo dimensional distributionHow do I show that $ d((x_1, x_2), (y_1, y_2)) = |y_1 - x_1| + |y_2 - x_2|$ is a metric?Prove statement related to dot productStrict Log-ConcavityStokes theorem for CuboidShow that $(M, xi)$ is a complete metric space.Prove and draw this metricFind an expression for a sequence of numbers ordered within an intervalProving $f(x,y,z)=big(fracxa+x+y+z, fracya+x+y+z, fracza+x+y+z big)$ is injectiveProve the triangle inequality in R^2
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Concavity of $ln(x^alpha - y)$ function
Two dimensional distributionHow do I show that $ d((x_1, x_2), (y_1, y_2)) = |y_1 - x_1| + |y_2 - x_2|$ is a metric?Prove statement related to dot productStrict Log-ConcavityStokes theorem for CuboidShow that $(M, xi)$ is a complete metric space.Prove and draw this metricFind an expression for a sequence of numbers ordered within an intervalProving $f(x,y,z)=big(fracxa+x+y+z, fracya+x+y+z, fracza+x+y+z big)$ is injectiveProve the triangle inequality in R^2
$begingroup$
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
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
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
asked Jun 15 '18 at 23:26
orrilloorrillo
315111
$endgroup$
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
calculus real-analysis
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
asked Jun 15 '18 at 23:26
orrilloorrillo
315111
asked Jun 15 '18 at 23:26
orrilloorrillo
315111
315111
add a comment |
add a comment |
1 Answer
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$begingroup$
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.
answered Mar 11 at 20:06
Thomas AhleThomas Ahle
1,4541321
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add a comment |
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$begingroup$
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.
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.
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.
answered Mar 11 at 20:06
Thomas AhleThomas Ahle
1,4541321
$endgroup$
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.
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
answered Mar 11 at 20:06
Thomas AhleThomas Ahle
1,4541321
1,4541321
add a comment |
add a comment |
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