Maximizing a weighted sum of logarithms
up vote
1
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I'm trying to show that for:
$$p^* equiv max_{x} sum^{k=n}_{k=1} a_k ln x_k
$$
Where:
$
x in mathbb{R}^n \
x geq 0 \
sum_{k=1}^{k=n} x_k=c \
c > 0 \
forall a_k, a_k > 0 \
a equiv sum_{k=1}^{k=n} a_k
$
That:
$$
p^* = aln{frac{c}{a}}+sum^{k=n}_{k=1} a_k ln a_k
$$
I'm really not certain where even to begin with this derivation.
inequality optimization summation logarithms convex-optimization
edited Nov 22 at 17:57
Batominovski
33.5k33292
asked Nov 14 at 20:59
mmam
811826
add a comment |
up vote
1
down vote
favorite
I'm trying to show that for:
$$p^* equiv max_{x} sum^{k=n}_{k=1} a_k ln x_k
$$
Where:
$
x in mathbb{R}^n \
x geq 0 \
sum_{k=1}^{k=n} x_k=c \
c > 0 \
forall a_k, a_k > 0 \
a equiv sum_{k=1}^{k=n} a_k
$
That:
$$
p^* = aln{frac{c}{a}}+sum^{k=n}_{k=1} a_k ln a_k
$$
I'm really not certain where even to begin with this derivation.
inequality optimization summation logarithms convex-optimization
edited Nov 22 at 17:57
Batominovski
33.5k33292
asked Nov 14 at 20:59
mmam
811826
Try the KKT approach.
– LinAlg
Nov 14 at 21:36
@LinAlg what's a good resource for KKT? I'm not super familiar.
– mmam
Nov 14 at 22:08
en.wikipedia.org/wiki/…
– LinAlg
Nov 14 at 23:02
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to show that for:
$$p^* equiv max_{x} sum^{k=n}_{k=1} a_k ln x_k
$$
Where:
$
x in mathbb{R}^n \
x geq 0 \
sum_{k=1}^{k=n} x_k=c \
c > 0 \
forall a_k, a_k > 0 \
a equiv sum_{k=1}^{k=n} a_k
$
That:
$$
p^* = aln{frac{c}{a}}+sum^{k=n}_{k=1} a_k ln a_k
$$
I'm really not certain where even to begin with this derivation.
inequality optimization summation logarithms convex-optimization
edited Nov 22 at 17:57
Batominovski
33.5k33292
asked Nov 14 at 20:59
mmam
811826
I'm trying to show that for:
$$p^* equiv max_{x} sum^{k=n}_{k=1} a_k ln x_k
$$
Where:
$
x in mathbb{R}^n \
x geq 0 \
sum_{k=1}^{k=n} x_k=c \
c > 0 \
forall a_k, a_k > 0 \
a equiv sum_{k=1}^{k=n} a_k
$
That:
$$
p^* = aln{frac{c}{a}}+sum^{k=n}_{k=1} a_k ln a_k
$$
I'm really not certain where even to begin with this derivation.
inequality optimization summation logarithms convex-optimization
inequality optimization summation logarithms convex-optimization
edited Nov 22 at 17:57
Batominovski
33.5k33292
asked Nov 14 at 20:59
mmam
811826
edited Nov 22 at 17:57
Batominovski
33.5k33292
asked Nov 14 at 20:59
mmam
811826
edited Nov 22 at 17:57
Batominovski
33.5k33292
edited Nov 22 at 17:57
Batominovski
33.5k33292
edited Nov 22 at 17:57
Batominovski
33.5k33292
33.5k33292
asked Nov 14 at 20:59
mmam
811826
asked Nov 14 at 20:59
mmam
811826
asked Nov 14 at 20:59
mmam
811826
811826
Try the KKT approach.
– LinAlg
Nov 14 at 21:36
@LinAlg what's a good resource for KKT? I'm not super familiar.
– mmam
Nov 14 at 22:08
en.wikipedia.org/wiki/…
– LinAlg
Nov 14 at 23:02
add a comment |
Try the KKT approach.
– LinAlg
Nov 14 at 21:36
@LinAlg what's a good resource for KKT? I'm not super familiar.
– mmam
Nov 14 at 22:08
en.wikipedia.org/wiki/…
– LinAlg
Nov 14 at 23:02
Try the KKT approach.
– LinAlg
Nov 14 at 21:36
Try the KKT approach.
– LinAlg
Nov 14 at 21:36
@LinAlg what's a good resource for KKT? I'm not super familiar.
– mmam
Nov 14 at 22:08
@LinAlg what's a good resource for KKT? I'm not super familiar.
– mmam
Nov 14 at 22:08
en.wikipedia.org/wiki/…
– LinAlg
Nov 14 at 23:02
en.wikipedia.org/wiki/…
– LinAlg
Nov 14 at 23:02
add a comment |
2 Answers
2
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I shall come back with a Lagrangian approach tomorrow. However, I now present a solution using the Weighted AM-GM Inequality.
We have that
$$sum_{k=1}^n,frac{a_k}{a},left(frac{a,x_k}{a_k}right)=sum_{k=1}^n,x_k=c,,$$
with
$$sum_{k=1}^n,frac{a_i}{a}=frac{sumlimits_{k=1}^n,a_k}{a}=frac{a}{a}=1,.$$
By the Weighted AM-GM Inequality, we have
$$c=sum_{k=1}^n,frac{a_k}{a},left(frac{a,x_k}{a_k}right)geq prod_{k=1}^n,left(frac{a,x_k}{a_k}right)^{frac{a_k}{a}},.$$
Take logarithm on both sides of the inequality above, we get
$$ln(c)geq sum_{k=1}^n,frac{a_k}{a},lnleft(frac{a,x_k}{a_k}right)=frac{1}{a},sum_{k=1}^n,Big(a_k,lnleft(x_kright)-a_k,lnleft(a_kright)+a_k,ln(a)Big),.$$
Ergo,
$$sum_{k=1}^n,a_k,lnleft(x_kright)leq a,ln(c)-left(sum_{k=1}^n,a_kright),ln(a)+sum_{k=1}^n,a_k,lnleft(a_kright)=a,lnleft(frac{c}{a}right)+sum_{k=1}^n,a_k,lnleft(a_kright),.$$
Note that inequality above becomes an equality if and only if $$frac{x_1}{a_1}=frac{x_2}{a_2}=ldots=frac{x_n}{a_n}=frac{c}{a},,$$
or equivalently,
$$left(x_1,x_2,ldots,x_nright)=left(frac{c,a_1}{a},frac{c,a_2}{a},ldots,frac{c,a_n}{a}right),.$$
This shows that $$p^*= a,lnleft(frac{c}{a}right)+sum_{k=1}^n,a_k,lnleft(a_kright),.$$
answered Nov 22 at 18:09
Batominovski
33.5k33292
add a comment |
up vote
0
down vote
With the Lagrange multiplier method you want to maximize $p(x)=sum a_k ln(x_k) +lambda (c-sum x_k)$.
Differentiating w.r.t. the $x$'s and setting equal to zero gives $frac{a_k}{x_k}=lambda$.
Imposing the constraint $c=sum x_k=frac{sum a_k}{lambda}=frac{a}{lambda}$. It follows that $lambda=a/c$.
So $x_k=frac{a_k }{ lambda}=frac{c a_k}{a}$.
So $p^*=sum a_k ln(x_k)=sum a_k ln(frac{c a_k}{a})=sum a_k (ln(frac{c}{a}) + ln(a_k))=aln(frac{c}{a})+sum a_k ln(a_k)$
answered Nov 22 at 18:37
user121049
1,352164
add a comment |
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2 Answers
2
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2 Answers
2
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oldest
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active
oldest
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active
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up vote
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I shall come back with a Lagrangian approach tomorrow. However, I now present a solution using the Weighted AM-GM Inequality.
We have that
$$sum_{k=1}^n,frac{a_k}{a},left(frac{a,x_k}{a_k}right)=sum_{k=1}^n,x_k=c,,$$
with
$$sum_{k=1}^n,frac{a_i}{a}=frac{sumlimits_{k=1}^n,a_k}{a}=frac{a}{a}=1,.$$
By the Weighted AM-GM Inequality, we have
$$c=sum_{k=1}^n,frac{a_k}{a},left(frac{a,x_k}{a_k}right)geq prod_{k=1}^n,left(frac{a,x_k}{a_k}right)^{frac{a_k}{a}},.$$
Take logarithm on both sides of the inequality above, we get
$$ln(c)geq sum_{k=1}^n,frac{a_k}{a},lnleft(frac{a,x_k}{a_k}right)=frac{1}{a},sum_{k=1}^n,Big(a_k,lnleft(x_kright)-a_k,lnleft(a_kright)+a_k,ln(a)Big),.$$
Ergo,
$$sum_{k=1}^n,a_k,lnleft(x_kright)leq a,ln(c)-left(sum_{k=1}^n,a_kright),ln(a)+sum_{k=1}^n,a_k,lnleft(a_kright)=a,lnleft(frac{c}{a}right)+sum_{k=1}^n,a_k,lnleft(a_kright),.$$
Note that inequality above becomes an equality if and only if $$frac{x_1}{a_1}=frac{x_2}{a_2}=ldots=frac{x_n}{a_n}=frac{c}{a},,$$
or equivalently,
$$left(x_1,x_2,ldots,x_nright)=left(frac{c,a_1}{a},frac{c,a_2}{a},ldots,frac{c,a_n}{a}right),.$$
This shows that $$p^*= a,lnleft(frac{c}{a}right)+sum_{k=1}^n,a_k,lnleft(a_kright),.$$
answered Nov 22 at 18:09
Batominovski
33.5k33292
add a comment |
up vote
2
down vote
I shall come back with a Lagrangian approach tomorrow. However, I now present a solution using the Weighted AM-GM Inequality.
We have that
$$sum_{k=1}^n,frac{a_k}{a},left(frac{a,x_k}{a_k}right)=sum_{k=1}^n,x_k=c,,$$
with
$$sum_{k=1}^n,frac{a_i}{a}=frac{sumlimits_{k=1}^n,a_k}{a}=frac{a}{a}=1,.$$
By the Weighted AM-GM Inequality, we have
$$c=sum_{k=1}^n,frac{a_k}{a},left(frac{a,x_k}{a_k}right)geq prod_{k=1}^n,left(frac{a,x_k}{a_k}right)^{frac{a_k}{a}},.$$
Take logarithm on both sides of the inequality above, we get
$$ln(c)geq sum_{k=1}^n,frac{a_k}{a},lnleft(frac{a,x_k}{a_k}right)=frac{1}{a},sum_{k=1}^n,Big(a_k,lnleft(x_kright)-a_k,lnleft(a_kright)+a_k,ln(a)Big),.$$
Ergo,
$$sum_{k=1}^n,a_k,lnleft(x_kright)leq a,ln(c)-left(sum_{k=1}^n,a_kright),ln(a)+sum_{k=1}^n,a_k,lnleft(a_kright)=a,lnleft(frac{c}{a}right)+sum_{k=1}^n,a_k,lnleft(a_kright),.$$
Note that inequality above becomes an equality if and only if $$frac{x_1}{a_1}=frac{x_2}{a_2}=ldots=frac{x_n}{a_n}=frac{c}{a},,$$
or equivalently,
$$left(x_1,x_2,ldots,x_nright)=left(frac{c,a_1}{a},frac{c,a_2}{a},ldots,frac{c,a_n}{a}right),.$$
This shows that $$p^*= a,lnleft(frac{c}{a}right)+sum_{k=1}^n,a_k,lnleft(a_kright),.$$
answered Nov 22 at 18:09
Batominovski
33.5k33292
add a comment |
up vote
2
down vote
up vote
2
down vote
I shall come back with a Lagrangian approach tomorrow. However, I now present a solution using the Weighted AM-GM Inequality.
We have that
$$sum_{k=1}^n,frac{a_k}{a},left(frac{a,x_k}{a_k}right)=sum_{k=1}^n,x_k=c,,$$
with
$$sum_{k=1}^n,frac{a_i}{a}=frac{sumlimits_{k=1}^n,a_k}{a}=frac{a}{a}=1,.$$
By the Weighted AM-GM Inequality, we have
$$c=sum_{k=1}^n,frac{a_k}{a},left(frac{a,x_k}{a_k}right)geq prod_{k=1}^n,left(frac{a,x_k}{a_k}right)^{frac{a_k}{a}},.$$
Take logarithm on both sides of the inequality above, we get
$$ln(c)geq sum_{k=1}^n,frac{a_k}{a},lnleft(frac{a,x_k}{a_k}right)=frac{1}{a},sum_{k=1}^n,Big(a_k,lnleft(x_kright)-a_k,lnleft(a_kright)+a_k,ln(a)Big),.$$
Ergo,
$$sum_{k=1}^n,a_k,lnleft(x_kright)leq a,ln(c)-left(sum_{k=1}^n,a_kright),ln(a)+sum_{k=1}^n,a_k,lnleft(a_kright)=a,lnleft(frac{c}{a}right)+sum_{k=1}^n,a_k,lnleft(a_kright),.$$
Note that inequality above becomes an equality if and only if $$frac{x_1}{a_1}=frac{x_2}{a_2}=ldots=frac{x_n}{a_n}=frac{c}{a},,$$
or equivalently,
$$left(x_1,x_2,ldots,x_nright)=left(frac{c,a_1}{a},frac{c,a_2}{a},ldots,frac{c,a_n}{a}right),.$$
This shows that $$p^*= a,lnleft(frac{c}{a}right)+sum_{k=1}^n,a_k,lnleft(a_kright),.$$
answered Nov 22 at 18:09
Batominovski
33.5k33292
I shall come back with a Lagrangian approach tomorrow. However, I now present a solution using the Weighted AM-GM Inequality.
We have that
$$sum_{k=1}^n,frac{a_k}{a},left(frac{a,x_k}{a_k}right)=sum_{k=1}^n,x_k=c,,$$
with
$$sum_{k=1}^n,frac{a_i}{a}=frac{sumlimits_{k=1}^n,a_k}{a}=frac{a}{a}=1,.$$
By the Weighted AM-GM Inequality, we have
$$c=sum_{k=1}^n,frac{a_k}{a},left(frac{a,x_k}{a_k}right)geq prod_{k=1}^n,left(frac{a,x_k}{a_k}right)^{frac{a_k}{a}},.$$
Take logarithm on both sides of the inequality above, we get
$$ln(c)geq sum_{k=1}^n,frac{a_k}{a},lnleft(frac{a,x_k}{a_k}right)=frac{1}{a},sum_{k=1}^n,Big(a_k,lnleft(x_kright)-a_k,lnleft(a_kright)+a_k,ln(a)Big),.$$
Ergo,
$$sum_{k=1}^n,a_k,lnleft(x_kright)leq a,ln(c)-left(sum_{k=1}^n,a_kright),ln(a)+sum_{k=1}^n,a_k,lnleft(a_kright)=a,lnleft(frac{c}{a}right)+sum_{k=1}^n,a_k,lnleft(a_kright),.$$
Note that inequality above becomes an equality if and only if $$frac{x_1}{a_1}=frac{x_2}{a_2}=ldots=frac{x_n}{a_n}=frac{c}{a},,$$
or equivalently,
$$left(x_1,x_2,ldots,x_nright)=left(frac{c,a_1}{a},frac{c,a_2}{a},ldots,frac{c,a_n}{a}right),.$$
This shows that $$p^*= a,lnleft(frac{c}{a}right)+sum_{k=1}^n,a_k,lnleft(a_kright),.$$
answered Nov 22 at 18:09
Batominovski
33.5k33292
answered Nov 22 at 18:09
Batominovski
33.5k33292
answered Nov 22 at 18:09
Batominovski
33.5k33292
answered Nov 22 at 18:09
Batominovski
33.5k33292
33.5k33292
add a comment |
add a comment |
up vote
0
down vote
With the Lagrange multiplier method you want to maximize $p(x)=sum a_k ln(x_k) +lambda (c-sum x_k)$.
Differentiating w.r.t. the $x$'s and setting equal to zero gives $frac{a_k}{x_k}=lambda$.
Imposing the constraint $c=sum x_k=frac{sum a_k}{lambda}=frac{a}{lambda}$. It follows that $lambda=a/c$.
So $x_k=frac{a_k }{ lambda}=frac{c a_k}{a}$.
So $p^*=sum a_k ln(x_k)=sum a_k ln(frac{c a_k}{a})=sum a_k (ln(frac{c}{a}) + ln(a_k))=aln(frac{c}{a})+sum a_k ln(a_k)$
answered Nov 22 at 18:37
user121049
1,352164
add a comment |
up vote
0
down vote
With the Lagrange multiplier method you want to maximize $p(x)=sum a_k ln(x_k) +lambda (c-sum x_k)$.
Differentiating w.r.t. the $x$'s and setting equal to zero gives $frac{a_k}{x_k}=lambda$.
Imposing the constraint $c=sum x_k=frac{sum a_k}{lambda}=frac{a}{lambda}$. It follows that $lambda=a/c$.
So $x_k=frac{a_k }{ lambda}=frac{c a_k}{a}$.
So $p^*=sum a_k ln(x_k)=sum a_k ln(frac{c a_k}{a})=sum a_k (ln(frac{c}{a}) + ln(a_k))=aln(frac{c}{a})+sum a_k ln(a_k)$
answered Nov 22 at 18:37
user121049
1,352164
add a comment |
up vote
0
down vote
up vote
0
down vote
With the Lagrange multiplier method you want to maximize $p(x)=sum a_k ln(x_k) +lambda (c-sum x_k)$.
Differentiating w.r.t. the $x$'s and setting equal to zero gives $frac{a_k}{x_k}=lambda$.
Imposing the constraint $c=sum x_k=frac{sum a_k}{lambda}=frac{a}{lambda}$. It follows that $lambda=a/c$.
So $x_k=frac{a_k }{ lambda}=frac{c a_k}{a}$.
So $p^*=sum a_k ln(x_k)=sum a_k ln(frac{c a_k}{a})=sum a_k (ln(frac{c}{a}) + ln(a_k))=aln(frac{c}{a})+sum a_k ln(a_k)$
answered Nov 22 at 18:37
user121049
1,352164
With the Lagrange multiplier method you want to maximize $p(x)=sum a_k ln(x_k) +lambda (c-sum x_k)$.
Differentiating w.r.t. the $x$'s and setting equal to zero gives $frac{a_k}{x_k}=lambda$.
Imposing the constraint $c=sum x_k=frac{sum a_k}{lambda}=frac{a}{lambda}$. It follows that $lambda=a/c$.
So $x_k=frac{a_k }{ lambda}=frac{c a_k}{a}$.
So $p^*=sum a_k ln(x_k)=sum a_k ln(frac{c a_k}{a})=sum a_k (ln(frac{c}{a}) + ln(a_k))=aln(frac{c}{a})+sum a_k ln(a_k)$
answered Nov 22 at 18:37
user121049
1,352164
answered Nov 22 at 18:37
user121049
1,352164
answered Nov 22 at 18:37
user121049
1,352164
answered Nov 22 at 18:37
user121049
1,352164
1,352164
add a comment |
add a comment |
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Try the KKT approach.
– LinAlg
Nov 14 at 21:36
@LinAlg what's a good resource for KKT? I'm not super familiar.
– mmam
Nov 14 at 22:08
en.wikipedia.org/wiki/…
– LinAlg
Nov 14 at 23:02