How to solve for $mathbf{w}$ in this equation? [closed]
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I'm stuck in this problem. I don't know what to do next. The only thing that comes in my mind is to write this as M (M = order of x) equations and solve them. We have to solve for w (which is also a M dimensional vector).
The answer to the problem is:
where
P.S. The $mathbf{w}_{MAP}$ is the MAP estimate of the $mathbf{w}$ random variable. The solution to above equation gives $mathbf{w}_{MAP}$.
linear-algebra
asked Dec 24 '18 at 17:10
Rohan BhatiaRohan Bhatia
1014
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closed as off-topic by Saad, KReiser, Lee David Chung Lin, mrtaurho, Namaste Dec 25 '18 at 16:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, KReiser, Lee David Chung Lin, mrtaurho, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I'm stuck in this problem. I don't know what to do next. The only thing that comes in my mind is to write this as M (M = order of x) equations and solve them. We have to solve for w (which is also a M dimensional vector).
The answer to the problem is:
where
P.S. The $mathbf{w}_{MAP}$ is the MAP estimate of the $mathbf{w}$ random variable. The solution to above equation gives $mathbf{w}_{MAP}$.
linear-algebra
asked Dec 24 '18 at 17:10
Rohan BhatiaRohan Bhatia
1014
$endgroup$
closed as off-topic by Saad, KReiser, Lee David Chung Lin, mrtaurho, Namaste Dec 25 '18 at 16:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, KReiser, Lee David Chung Lin, mrtaurho, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
You write that $M$ is the order of $X$, but you never say what $X$ is.
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– Gerry Myerson
Dec 24 '18 at 18:32
$begingroup$
Sorry, I meant small $x_i$. It's a M dimensional vector. w is also a M dimensional vector
$endgroup$
– Rohan Bhatia
Dec 24 '18 at 18:42
add a comment |
$begingroup$
I'm stuck in this problem. I don't know what to do next. The only thing that comes in my mind is to write this as M (M = order of x) equations and solve them. We have to solve for w (which is also a M dimensional vector).
The answer to the problem is:
where
P.S. The $mathbf{w}_{MAP}$ is the MAP estimate of the $mathbf{w}$ random variable. The solution to above equation gives $mathbf{w}_{MAP}$.
linear-algebra
asked Dec 24 '18 at 17:10
Rohan BhatiaRohan Bhatia
1014
$endgroup$
I'm stuck in this problem. I don't know what to do next. The only thing that comes in my mind is to write this as M (M = order of x) equations and solve them. We have to solve for w (which is also a M dimensional vector).
The answer to the problem is:
where
P.S. The $mathbf{w}_{MAP}$ is the MAP estimate of the $mathbf{w}$ random variable. The solution to above equation gives $mathbf{w}_{MAP}$.
linear-algebra
linear-algebra
asked Dec 24 '18 at 17:10
Rohan BhatiaRohan Bhatia
1014
asked Dec 24 '18 at 17:10
Rohan BhatiaRohan Bhatia
1014
edited Dec 24 '18 at 18:42
Rohan Bhatia
asked Dec 24 '18 at 17:10
Rohan BhatiaRohan Bhatia
1014
asked Dec 24 '18 at 17:10
Rohan BhatiaRohan Bhatia
1014
asked Dec 24 '18 at 17:10
Rohan BhatiaRohan Bhatia
1014
1014
closed as off-topic by Saad, KReiser, Lee David Chung Lin, mrtaurho, Namaste Dec 25 '18 at 16:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, KReiser, Lee David Chung Lin, mrtaurho, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, KReiser, Lee David Chung Lin, mrtaurho, Namaste Dec 25 '18 at 16:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, KReiser, Lee David Chung Lin, mrtaurho, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
You write that $M$ is the order of $X$, but you never say what $X$ is.
$endgroup$
– Gerry Myerson
Dec 24 '18 at 18:32
$begingroup$
Sorry, I meant small $x_i$. It's a M dimensional vector. w is also a M dimensional vector
$endgroup$
– Rohan Bhatia
Dec 24 '18 at 18:42
add a comment |
$begingroup$
You write that $M$ is the order of $X$, but you never say what $X$ is.
$endgroup$
– Gerry Myerson
Dec 24 '18 at 18:32
$begingroup$
Sorry, I meant small $x_i$. It's a M dimensional vector. w is also a M dimensional vector
$endgroup$
– Rohan Bhatia
Dec 24 '18 at 18:42
$begingroup$
You write that $M$ is the order of $X$, but you never say what $X$ is.
$endgroup$
– Gerry Myerson
Dec 24 '18 at 18:32
$begingroup$
You write that $M$ is the order of $X$, but you never say what $X$ is.
$endgroup$
– Gerry Myerson
Dec 24 '18 at 18:32
$begingroup$
Sorry, I meant small $x_i$. It's a M dimensional vector. w is also a M dimensional vector
$endgroup$
– Rohan Bhatia
Dec 24 '18 at 18:42
$begingroup$
Sorry, I meant small $x_i$. It's a M dimensional vector. w is also a M dimensional vector
$endgroup$
– Rohan Bhatia
Dec 24 '18 at 18:42
add a comment |
1 Answer
1
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oldest
votes
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By writing $(w^Tx)x$ as $x(x^Tw) = (xx^T)w$ you can write the equation as
$$-r_{dx}(N) = lambda w - left(sum_{i=1}^N x_i x_i^Tright) w$$
$$-r_{dx}(N) = left(lambda I - sum_{i=1}^N x_i x_i^Tright) w$$
$$left( sum_{i=1}^N x_i x_i^T - lambda Iright) w = r_{dx}(N)$$
$$w = left( sum_{i=1}^N x_i x_i^T - lambda Iright)^{-1} r_{dx}(N)$$
answered Dec 24 '18 at 18:47
LinAlgLinAlg
10k1521
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By writing $(w^Tx)x$ as $x(x^Tw) = (xx^T)w$ you can write the equation as
$$-r_{dx}(N) = lambda w - left(sum_{i=1}^N x_i x_i^Tright) w$$
$$-r_{dx}(N) = left(lambda I - sum_{i=1}^N x_i x_i^Tright) w$$
$$left( sum_{i=1}^N x_i x_i^T - lambda Iright) w = r_{dx}(N)$$
$$w = left( sum_{i=1}^N x_i x_i^T - lambda Iright)^{-1} r_{dx}(N)$$
answered Dec 24 '18 at 18:47
LinAlgLinAlg
10k1521
$endgroup$
add a comment |
$begingroup$
By writing $(w^Tx)x$ as $x(x^Tw) = (xx^T)w$ you can write the equation as
$$-r_{dx}(N) = lambda w - left(sum_{i=1}^N x_i x_i^Tright) w$$
$$-r_{dx}(N) = left(lambda I - sum_{i=1}^N x_i x_i^Tright) w$$
$$left( sum_{i=1}^N x_i x_i^T - lambda Iright) w = r_{dx}(N)$$
$$w = left( sum_{i=1}^N x_i x_i^T - lambda Iright)^{-1} r_{dx}(N)$$
answered Dec 24 '18 at 18:47
LinAlgLinAlg
10k1521
$endgroup$
add a comment |
$begingroup$
By writing $(w^Tx)x$ as $x(x^Tw) = (xx^T)w$ you can write the equation as
$$-r_{dx}(N) = lambda w - left(sum_{i=1}^N x_i x_i^Tright) w$$
$$-r_{dx}(N) = left(lambda I - sum_{i=1}^N x_i x_i^Tright) w$$
$$left( sum_{i=1}^N x_i x_i^T - lambda Iright) w = r_{dx}(N)$$
$$w = left( sum_{i=1}^N x_i x_i^T - lambda Iright)^{-1} r_{dx}(N)$$
answered Dec 24 '18 at 18:47
LinAlgLinAlg
10k1521
$endgroup$
By writing $(w^Tx)x$ as $x(x^Tw) = (xx^T)w$ you can write the equation as
$$-r_{dx}(N) = lambda w - left(sum_{i=1}^N x_i x_i^Tright) w$$
$$-r_{dx}(N) = left(lambda I - sum_{i=1}^N x_i x_i^Tright) w$$
$$left( sum_{i=1}^N x_i x_i^T - lambda Iright) w = r_{dx}(N)$$
$$w = left( sum_{i=1}^N x_i x_i^T - lambda Iright)^{-1} r_{dx}(N)$$
answered Dec 24 '18 at 18:47
LinAlgLinAlg
10k1521
answered Dec 24 '18 at 18:47
LinAlgLinAlg
10k1521
answered Dec 24 '18 at 18:47
LinAlgLinAlg
10k1521
answered Dec 24 '18 at 18:47
LinAlgLinAlg
10k1521
10k1521
add a comment |
add a comment |
$begingroup$
You write that $M$ is the order of $X$, but you never say what $X$ is.
$endgroup$
– Gerry Myerson
Dec 24 '18 at 18:32
$begingroup$
Sorry, I meant small $x_i$. It's a M dimensional vector. w is also a M dimensional vector
$endgroup$
– Rohan Bhatia
Dec 24 '18 at 18:42