Converting a quadratic constraint into a second-order cone constraint
$begingroup$
This is straight from the book: Optimization Methods in Finance.
I'm trying to gain understanding of how the author derived the cone constraints from the the following quadratic constraint:
$$x^TQx + 2p^T x + γ ≤ 0$$
Assuming Matrix $Q$ is positive definite, there exists an invertible matrix, say $R$, satisfying $Q = RR^T$
This allows us to rearrange the equation to:
$$(R^Tx)^T(R^Tx) + 2p^Tx + γ ≤ 0 $$
This is fine and makes sense. However the next step I have had issues understanding:
Define $$ y = (y_1, . . . , y_k)^T = R^T x + R^{−1}p $$
Then we have:
$$ y^Ty = (R^Tx)^T(R^Tx) + 2p^Tx + p^TQ^{-1}p $$
Pictures from the text below show the entire derivation. Any suggestions or clarifications on how the last two lines are derived? Perhaps some linear algebra trick I am not aware of? Thanks!
part1
part2
optimization
asked Nov 25 '18 at 22:23
user3547551user3547551
12
$endgroup$
add a comment |
$begingroup$
This is straight from the book: Optimization Methods in Finance.
I'm trying to gain understanding of how the author derived the cone constraints from the the following quadratic constraint:
$$x^TQx + 2p^T x + γ ≤ 0$$
Assuming Matrix $Q$ is positive definite, there exists an invertible matrix, say $R$, satisfying $Q = RR^T$
This allows us to rearrange the equation to:
$$(R^Tx)^T(R^Tx) + 2p^Tx + γ ≤ 0 $$
This is fine and makes sense. However the next step I have had issues understanding:
Define $$ y = (y_1, . . . , y_k)^T = R^T x + R^{−1}p $$
Then we have:
$$ y^Ty = (R^Tx)^T(R^Tx) + 2p^Tx + p^TQ^{-1}p $$
Pictures from the text below show the entire derivation. Any suggestions or clarifications on how the last two lines are derived? Perhaps some linear algebra trick I am not aware of? Thanks!
part1
part2
optimization
asked Nov 25 '18 at 22:23
user3547551user3547551
12
$endgroup$
4
$begingroup$
No tricks involved. Just the use of basic identities and the fact that $p^Tx=x^Tp$ (since that’s just the dot product of $x$ and $p$).
$endgroup$
– amd
Nov 26 '18 at 0:04
$begingroup$
Thank you, using the dot product rule of x and p did not come to mind!
$endgroup$
– user3547551
Nov 27 '18 at 15:48
add a comment |
$begingroup$
This is straight from the book: Optimization Methods in Finance.
I'm trying to gain understanding of how the author derived the cone constraints from the the following quadratic constraint:
$$x^TQx + 2p^T x + γ ≤ 0$$
Assuming Matrix $Q$ is positive definite, there exists an invertible matrix, say $R$, satisfying $Q = RR^T$
This allows us to rearrange the equation to:
$$(R^Tx)^T(R^Tx) + 2p^Tx + γ ≤ 0 $$
This is fine and makes sense. However the next step I have had issues understanding:
Define $$ y = (y_1, . . . , y_k)^T = R^T x + R^{−1}p $$
Then we have:
$$ y^Ty = (R^Tx)^T(R^Tx) + 2p^Tx + p^TQ^{-1}p $$
Pictures from the text below show the entire derivation. Any suggestions or clarifications on how the last two lines are derived? Perhaps some linear algebra trick I am not aware of? Thanks!
part1
part2
optimization
asked Nov 25 '18 at 22:23
user3547551user3547551
12
$endgroup$
This is straight from the book: Optimization Methods in Finance.
I'm trying to gain understanding of how the author derived the cone constraints from the the following quadratic constraint:
$$x^TQx + 2p^T x + γ ≤ 0$$
Assuming Matrix $Q$ is positive definite, there exists an invertible matrix, say $R$, satisfying $Q = RR^T$
This allows us to rearrange the equation to:
$$(R^Tx)^T(R^Tx) + 2p^Tx + γ ≤ 0 $$
This is fine and makes sense. However the next step I have had issues understanding:
Define $$ y = (y_1, . . . , y_k)^T = R^T x + R^{−1}p $$
Then we have:
$$ y^Ty = (R^Tx)^T(R^Tx) + 2p^Tx + p^TQ^{-1}p $$
Pictures from the text below show the entire derivation. Any suggestions or clarifications on how the last two lines are derived? Perhaps some linear algebra trick I am not aware of? Thanks!
part1
part2
optimization
optimization
asked Nov 25 '18 at 22:23
user3547551user3547551
12
asked Nov 25 '18 at 22:23
user3547551user3547551
12
edited Nov 25 '18 at 22:32
user3547551
asked Nov 25 '18 at 22:23
user3547551user3547551
12
asked Nov 25 '18 at 22:23
user3547551user3547551
12
asked Nov 25 '18 at 22:23
user3547551user3547551
12
12
4
$begingroup$
No tricks involved. Just the use of basic identities and the fact that $p^Tx=x^Tp$ (since that’s just the dot product of $x$ and $p$).
$endgroup$
– amd
Nov 26 '18 at 0:04
$begingroup$
Thank you, using the dot product rule of x and p did not come to mind!
$endgroup$
– user3547551
Nov 27 '18 at 15:48
add a comment |
4
$begingroup$
No tricks involved. Just the use of basic identities and the fact that $p^Tx=x^Tp$ (since that’s just the dot product of $x$ and $p$).
$endgroup$
– amd
Nov 26 '18 at 0:04
$begingroup$
Thank you, using the dot product rule of x and p did not come to mind!
$endgroup$
– user3547551
Nov 27 '18 at 15:48
4
4
$begingroup$
No tricks involved. Just the use of basic identities and the fact that $p^Tx=x^Tp$ (since that’s just the dot product of $x$ and $p$).
$endgroup$
– amd
Nov 26 '18 at 0:04
$begingroup$
No tricks involved. Just the use of basic identities and the fact that $p^Tx=x^Tp$ (since that’s just the dot product of $x$ and $p$).
$endgroup$
– amd
Nov 26 '18 at 0:04
$begingroup$
Thank you, using the dot product rule of x and p did not come to mind!
$endgroup$
– user3547551
Nov 27 '18 at 15:48
$begingroup$
Thank you, using the dot product rule of x and p did not come to mind!
$endgroup$
– user3547551
Nov 27 '18 at 15:48
add a comment |
1 Answer
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$begingroup$
If $y=R^Tx$ then
$$|y|_2^2=y^Ty=(R^Tx)^T(R^Tx)=x^TRR^Tx=x^TQx$$
so you can equivalently write your quadratic problem in conic form as
$$t+2p^Tx+gammaleq 0,quad tgeq |y|_2^2.$$
The second constraint is a rotated quadratic cone.
answered Nov 26 '18 at 6:59
Michal AdamaszekMichal Adamaszek
2,08148
$endgroup$
add a comment |
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1 Answer
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$begingroup$
If $y=R^Tx$ then
$$|y|_2^2=y^Ty=(R^Tx)^T(R^Tx)=x^TRR^Tx=x^TQx$$
so you can equivalently write your quadratic problem in conic form as
$$t+2p^Tx+gammaleq 0,quad tgeq |y|_2^2.$$
The second constraint is a rotated quadratic cone.
answered Nov 26 '18 at 6:59
Michal AdamaszekMichal Adamaszek
2,08148
$endgroup$
add a comment |
$begingroup$
If $y=R^Tx$ then
$$|y|_2^2=y^Ty=(R^Tx)^T(R^Tx)=x^TRR^Tx=x^TQx$$
so you can equivalently write your quadratic problem in conic form as
$$t+2p^Tx+gammaleq 0,quad tgeq |y|_2^2.$$
The second constraint is a rotated quadratic cone.
answered Nov 26 '18 at 6:59
Michal AdamaszekMichal Adamaszek
2,08148
$endgroup$
add a comment |
$begingroup$
If $y=R^Tx$ then
$$|y|_2^2=y^Ty=(R^Tx)^T(R^Tx)=x^TRR^Tx=x^TQx$$
so you can equivalently write your quadratic problem in conic form as
$$t+2p^Tx+gammaleq 0,quad tgeq |y|_2^2.$$
The second constraint is a rotated quadratic cone.
answered Nov 26 '18 at 6:59
Michal AdamaszekMichal Adamaszek
2,08148
$endgroup$
If $y=R^Tx$ then
$$|y|_2^2=y^Ty=(R^Tx)^T(R^Tx)=x^TRR^Tx=x^TQx$$
so you can equivalently write your quadratic problem in conic form as
$$t+2p^Tx+gammaleq 0,quad tgeq |y|_2^2.$$
The second constraint is a rotated quadratic cone.
answered Nov 26 '18 at 6:59
Michal AdamaszekMichal Adamaszek
2,08148
answered Nov 26 '18 at 6:59
Michal AdamaszekMichal Adamaszek
2,08148
answered Nov 26 '18 at 6:59
Michal AdamaszekMichal Adamaszek
2,08148
answered Nov 26 '18 at 6:59
Michal AdamaszekMichal Adamaszek
2,08148
2,08148
add a comment |
add a comment |
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$begingroup$
No tricks involved. Just the use of basic identities and the fact that $p^Tx=x^Tp$ (since that’s just the dot product of $x$ and $p$).
$endgroup$
– amd
Nov 26 '18 at 0:04
$begingroup$
Thank you, using the dot product rule of x and p did not come to mind!
$endgroup$
– user3547551
Nov 27 '18 at 15:48