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Integral involving l1-norm
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I am trying to figure out how to evaluate an integral of the following form. Assume that $textbfx$ is a $P times 1$ vector, where $||.||_1$ reprents the l-1 norm, i.e. $sum_j=1^P |x_j|$.
The simplest version to evaluate is this first integral as you can just re-write the l1-norm and integrate directly.
$$int_-infty^infty exp(- ||textbfx||_1) dtextbfx = 2^P$$
The trickier part is I'm now trying to figure out how to do this when I left multiply $textbfx$ by some matrix $textbfA$, i.e:
$$int_-infty^infty exp(- ||textbfAtextbfx||_1) dtextbfx = ?$$
My thought was to do some change of variables on the integral, i.e. $textbfy = textbfA textbfx$, but as $textbfA$ isn't necessarily square, I wasn't really sure how to do this in a general case.
Any help would be greatly appreciated!
integration norm indefinite-integrals change-of-variable
asked Jul 26 at 23:15
MGA
413
 |Â
show 5 more comments
up vote
1
down vote
favorite
I am trying to figure out how to evaluate an integral of the following form. Assume that $textbfx$ is a $P times 1$ vector, where $||.||_1$ reprents the l-1 norm, i.e. $sum_j=1^P |x_j|$.
The simplest version to evaluate is this first integral as you can just re-write the l1-norm and integrate directly.
$$int_-infty^infty exp(- ||textbfx||_1) dtextbfx = 2^P$$
The trickier part is I'm now trying to figure out how to do this when I left multiply $textbfx$ by some matrix $textbfA$, i.e:
$$int_-infty^infty exp(- ||textbfAtextbfx||_1) dtextbfx = ?$$
My thought was to do some change of variables on the integral, i.e. $textbfy = textbfA textbfx$, but as $textbfA$ isn't necessarily square, I wasn't really sure how to do this in a general case.
Any help would be greatly appreciated!
integration norm indefinite-integrals change-of-variable
asked Jul 26 at 23:15
MGA
413
2
Do, thin QR factorization of $A$. When $R_1$ is not invertible the integral diverges. When $R_1$ is invertible, do your changes of variable.
– user578878
Jul 27 at 0:28
Thanks! Could you say more about how this helps? It seems like when I do the QR decomposition, I'm left with the l1 norm of Q times the transformed matrix and thus I still have the same problem. Maybe I'm missing something obvious here?
– MGA
Jul 27 at 0:52
Also I am still confused about how to do the change of variables with $R_1$?
– MGA
Jul 27 at 1:18
1
The answer is not $|R_1|^-1cdot 2^m$. For the example of vectors $q_1,q_2$ that I wrote above, the result is $10/9$, instead of $2^2$. Notice that although I denoted it $Id$, that matrix is not the identity. It is the identity with extra zero rows at the bottom.
– user578878
Jul 27 at 2:52
1
Yes, there is a closed form, which you would obtain by doing what I said above.
– user578878
Jul 27 at 3:00
 |Â
show 5 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to figure out how to evaluate an integral of the following form. Assume that $textbfx$ is a $P times 1$ vector, where $||.||_1$ reprents the l-1 norm, i.e. $sum_j=1^P |x_j|$.
The simplest version to evaluate is this first integral as you can just re-write the l1-norm and integrate directly.
$$int_-infty^infty exp(- ||textbfx||_1) dtextbfx = 2^P$$
The trickier part is I'm now trying to figure out how to do this when I left multiply $textbfx$ by some matrix $textbfA$, i.e:
$$int_-infty^infty exp(- ||textbfAtextbfx||_1) dtextbfx = ?$$
My thought was to do some change of variables on the integral, i.e. $textbfy = textbfA textbfx$, but as $textbfA$ isn't necessarily square, I wasn't really sure how to do this in a general case.
Any help would be greatly appreciated!
integration norm indefinite-integrals change-of-variable
asked Jul 26 at 23:15
MGA
413
I am trying to figure out how to evaluate an integral of the following form. Assume that $textbfx$ is a $P times 1$ vector, where $||.||_1$ reprents the l-1 norm, i.e. $sum_j=1^P |x_j|$.
The simplest version to evaluate is this first integral as you can just re-write the l1-norm and integrate directly.
$$int_-infty^infty exp(- ||textbfx||_1) dtextbfx = 2^P$$
The trickier part is I'm now trying to figure out how to do this when I left multiply $textbfx$ by some matrix $textbfA$, i.e:
$$int_-infty^infty exp(- ||textbfAtextbfx||_1) dtextbfx = ?$$
My thought was to do some change of variables on the integral, i.e. $textbfy = textbfA textbfx$, but as $textbfA$ isn't necessarily square, I wasn't really sure how to do this in a general case.
Any help would be greatly appreciated!
integration norm indefinite-integrals change-of-variable
asked Jul 26 at 23:15
MGA
413
asked Jul 26 at 23:15
MGA
413
asked Jul 26 at 23:15
MGA
413
asked Jul 26 at 23:15
MGA
413
413
2
Do, thin QR factorization of $A$. When $R_1$ is not invertible the integral diverges. When $R_1$ is invertible, do your changes of variable.
– user578878
Jul 27 at 0:28
Thanks! Could you say more about how this helps? It seems like when I do the QR decomposition, I'm left with the l1 norm of Q times the transformed matrix and thus I still have the same problem. Maybe I'm missing something obvious here?
– MGA
Jul 27 at 0:52
Also I am still confused about how to do the change of variables with $R_1$?
– MGA
Jul 27 at 1:18
1
The answer is not $|R_1|^-1cdot 2^m$. For the example of vectors $q_1,q_2$ that I wrote above, the result is $10/9$, instead of $2^2$. Notice that although I denoted it $Id$, that matrix is not the identity. It is the identity with extra zero rows at the bottom.
– user578878
Jul 27 at 2:52
1
Yes, there is a closed form, which you would obtain by doing what I said above.
– user578878
Jul 27 at 3:00
 |Â
show 5 more comments
2
Do, thin QR factorization of $A$. When $R_1$ is not invertible the integral diverges. When $R_1$ is invertible, do your changes of variable.
– user578878
Jul 27 at 0:28
Thanks! Could you say more about how this helps? It seems like when I do the QR decomposition, I'm left with the l1 norm of Q times the transformed matrix and thus I still have the same problem. Maybe I'm missing something obvious here?
– MGA
Jul 27 at 0:52
Also I am still confused about how to do the change of variables with $R_1$?
– MGA
Jul 27 at 1:18
1
The answer is not $|R_1|^-1cdot 2^m$. For the example of vectors $q_1,q_2$ that I wrote above, the result is $10/9$, instead of $2^2$. Notice that although I denoted it $Id$, that matrix is not the identity. It is the identity with extra zero rows at the bottom.
– user578878
Jul 27 at 2:52
1
Yes, there is a closed form, which you would obtain by doing what I said above.
– user578878
Jul 27 at 3:00
2
2
Do, thin QR factorization of $A$. When $R_1$ is not invertible the integral diverges. When $R_1$ is invertible, do your changes of variable.
– user578878
Jul 27 at 0:28
Do, thin QR factorization of $A$. When $R_1$ is not invertible the integral diverges. When $R_1$ is invertible, do your changes of variable.
– user578878
Jul 27 at 0:28
Thanks! Could you say more about how this helps? It seems like when I do the QR decomposition, I'm left with the l1 norm of Q times the transformed matrix and thus I still have the same problem. Maybe I'm missing something obvious here?
– MGA
Jul 27 at 0:52
Thanks! Could you say more about how this helps? It seems like when I do the QR decomposition, I'm left with the l1 norm of Q times the transformed matrix and thus I still have the same problem. Maybe I'm missing something obvious here?
– MGA
Jul 27 at 0:52
Also I am still confused about how to do the change of variables with $R_1$?
– MGA
Jul 27 at 1:18
Also I am still confused about how to do the change of variables with $R_1$?
– MGA
Jul 27 at 1:18
1
1
The answer is not $|R_1|^-1cdot 2^m$. For the example of vectors $q_1,q_2$ that I wrote above, the result is $10/9$, instead of $2^2$. Notice that although I denoted it $Id$, that matrix is not the identity. It is the identity with extra zero rows at the bottom.
– user578878
Jul 27 at 2:52
The answer is not $|R_1|^-1cdot 2^m$. For the example of vectors $q_1,q_2$ that I wrote above, the result is $10/9$, instead of $2^2$. Notice that although I denoted it $Id$, that matrix is not the identity. It is the identity with extra zero rows at the bottom.
– user578878
Jul 27 at 2:52
1
1
Yes, there is a closed form, which you would obtain by doing what I said above.
– user578878
Jul 27 at 3:00
Yes, there is a closed form, which you would obtain by doing what I said above.
– user578878
Jul 27 at 3:00
 |Â
show 5 more comments
1 Answer
1
active
oldest
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up vote
0
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Let $n$ be dimension of our space (sorry, I change $P$ with $n$) and let $A$ be a $k times n$ matrix ($k$ rows and $n$ columns). Let $k leq n$. Write $A$ in the block form
$$
A = beginpmatrix
A_1 &A_2
endpmatrix
$$
where $A_1$ consists of the first $ktimes k$ elements and $A_2$ consists of the last $ktimes (n-k)$ elements. Now form $n times n$ matrix $B$ as the following
$$
B = beginpmatrix
A_1 &A_2 \
0 & 1
endpmatrix
$$
(here $1$ stands for the identity matrix).
This $B$ serves your desired change of variables. Note that if $k<n$ the integral diverges.
answered Jul 27 at 0:43
dsw
963
Hi, I'm trying to think through this answer in the k > n case but I think it can't be true that if k < n the integral diverges. Consider the case where A = [1,0,0,0] and x is a 4 dimensional vector. A * x = x_1 and the integral evaluates to '2'. Am I missing something?
– MGA
Jul 27 at 0:53
1
@MGA in this case the integral becomes $int e^ dx_1 dotsm dx_4 = infty$. The case $k>n$ is lost in my answer, indeed. In this case we want to reduce $A$ to an upper triangular matrix (something with zero rows), as in the comment by nextpuzzle (note $operatornamerk A leq n$).
– dsw
Jul 27 at 1:28
Thanks. Could you say more about where $B$ comes from? Why can't I flip the order of the identity matrix and 0 in the augmented matrix?
– MGA
Jul 27 at 2:44
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let $n$ be dimension of our space (sorry, I change $P$ with $n$) and let $A$ be a $k times n$ matrix ($k$ rows and $n$ columns). Let $k leq n$. Write $A$ in the block form
$$
A = beginpmatrix
A_1 &A_2
endpmatrix
$$
where $A_1$ consists of the first $ktimes k$ elements and $A_2$ consists of the last $ktimes (n-k)$ elements. Now form $n times n$ matrix $B$ as the following
$$
B = beginpmatrix
A_1 &A_2 \
0 & 1
endpmatrix
$$
(here $1$ stands for the identity matrix).
This $B$ serves your desired change of variables. Note that if $k<n$ the integral diverges.
answered Jul 27 at 0:43
dsw
963
Hi, I'm trying to think through this answer in the k > n case but I think it can't be true that if k < n the integral diverges. Consider the case where A = [1,0,0,0] and x is a 4 dimensional vector. A * x = x_1 and the integral evaluates to '2'. Am I missing something?
– MGA
Jul 27 at 0:53
1
@MGA in this case the integral becomes $int e^ dx_1 dotsm dx_4 = infty$. The case $k>n$ is lost in my answer, indeed. In this case we want to reduce $A$ to an upper triangular matrix (something with zero rows), as in the comment by nextpuzzle (note $operatornamerk A leq n$).
– dsw
Jul 27 at 1:28
Thanks. Could you say more about where $B$ comes from? Why can't I flip the order of the identity matrix and 0 in the augmented matrix?
– MGA
Jul 27 at 2:44
add a comment |Â
up vote
0
down vote
Let $n$ be dimension of our space (sorry, I change $P$ with $n$) and let $A$ be a $k times n$ matrix ($k$ rows and $n$ columns). Let $k leq n$. Write $A$ in the block form
$$
A = beginpmatrix
A_1 &A_2
endpmatrix
$$
where $A_1$ consists of the first $ktimes k$ elements and $A_2$ consists of the last $ktimes (n-k)$ elements. Now form $n times n$ matrix $B$ as the following
$$
B = beginpmatrix
A_1 &A_2 \
0 & 1
endpmatrix
$$
(here $1$ stands for the identity matrix).
This $B$ serves your desired change of variables. Note that if $k<n$ the integral diverges.
answered Jul 27 at 0:43
dsw
963
Hi, I'm trying to think through this answer in the k > n case but I think it can't be true that if k < n the integral diverges. Consider the case where A = [1,0,0,0] and x is a 4 dimensional vector. A * x = x_1 and the integral evaluates to '2'. Am I missing something?
– MGA
Jul 27 at 0:53
1
@MGA in this case the integral becomes $int e^ dx_1 dotsm dx_4 = infty$. The case $k>n$ is lost in my answer, indeed. In this case we want to reduce $A$ to an upper triangular matrix (something with zero rows), as in the comment by nextpuzzle (note $operatornamerk A leq n$).
– dsw
Jul 27 at 1:28
Thanks. Could you say more about where $B$ comes from? Why can't I flip the order of the identity matrix and 0 in the augmented matrix?
– MGA
Jul 27 at 2:44
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $n$ be dimension of our space (sorry, I change $P$ with $n$) and let $A$ be a $k times n$ matrix ($k$ rows and $n$ columns). Let $k leq n$. Write $A$ in the block form
$$
A = beginpmatrix
A_1 &A_2
endpmatrix
$$
where $A_1$ consists of the first $ktimes k$ elements and $A_2$ consists of the last $ktimes (n-k)$ elements. Now form $n times n$ matrix $B$ as the following
$$
B = beginpmatrix
A_1 &A_2 \
0 & 1
endpmatrix
$$
(here $1$ stands for the identity matrix).
This $B$ serves your desired change of variables. Note that if $k<n$ the integral diverges.
answered Jul 27 at 0:43
dsw
963
Let $n$ be dimension of our space (sorry, I change $P$ with $n$) and let $A$ be a $k times n$ matrix ($k$ rows and $n$ columns). Let $k leq n$. Write $A$ in the block form
$$
A = beginpmatrix
A_1 &A_2
endpmatrix
$$
where $A_1$ consists of the first $ktimes k$ elements and $A_2$ consists of the last $ktimes (n-k)$ elements. Now form $n times n$ matrix $B$ as the following
$$
B = beginpmatrix
A_1 &A_2 \
0 & 1
endpmatrix
$$
(here $1$ stands for the identity matrix).
This $B$ serves your desired change of variables. Note that if $k<n$ the integral diverges.
answered Jul 27 at 0:43
dsw
963
edited Jul 27 at 1:28
answered Jul 27 at 0:43
dsw
963
answered Jul 27 at 0:43
dsw
963
answered Jul 27 at 0:43
dsw
963
963
Hi, I'm trying to think through this answer in the k > n case but I think it can't be true that if k < n the integral diverges. Consider the case where A = [1,0,0,0] and x is a 4 dimensional vector. A * x = x_1 and the integral evaluates to '2'. Am I missing something?
– MGA
Jul 27 at 0:53
1
@MGA in this case the integral becomes $int e^ dx_1 dotsm dx_4 = infty$. The case $k>n$ is lost in my answer, indeed. In this case we want to reduce $A$ to an upper triangular matrix (something with zero rows), as in the comment by nextpuzzle (note $operatornamerk A leq n$).
– dsw
Jul 27 at 1:28
Thanks. Could you say more about where $B$ comes from? Why can't I flip the order of the identity matrix and 0 in the augmented matrix?
– MGA
Jul 27 at 2:44
add a comment |Â
Hi, I'm trying to think through this answer in the k > n case but I think it can't be true that if k < n the integral diverges. Consider the case where A = [1,0,0,0] and x is a 4 dimensional vector. A * x = x_1 and the integral evaluates to '2'. Am I missing something?
– MGA
Jul 27 at 0:53
1
@MGA in this case the integral becomes $int e^ dx_1 dotsm dx_4 = infty$. The case $k>n$ is lost in my answer, indeed. In this case we want to reduce $A$ to an upper triangular matrix (something with zero rows), as in the comment by nextpuzzle (note $operatornamerk A leq n$).
– dsw
Jul 27 at 1:28
Thanks. Could you say more about where $B$ comes from? Why can't I flip the order of the identity matrix and 0 in the augmented matrix?
– MGA
Jul 27 at 2:44
Hi, I'm trying to think through this answer in the k > n case but I think it can't be true that if k < n the integral diverges. Consider the case where A = [1,0,0,0] and x is a 4 dimensional vector. A * x = x_1 and the integral evaluates to '2'. Am I missing something?
– MGA
Jul 27 at 0:53
Hi, I'm trying to think through this answer in the k > n case but I think it can't be true that if k < n the integral diverges. Consider the case where A = [1,0,0,0] and x is a 4 dimensional vector. A * x = x_1 and the integral evaluates to '2'. Am I missing something?
– MGA
Jul 27 at 0:53
1
1
@MGA in this case the integral becomes $int e^ dx_1 dotsm dx_4 = infty$. The case $k>n$ is lost in my answer, indeed. In this case we want to reduce $A$ to an upper triangular matrix (something with zero rows), as in the comment by nextpuzzle (note $operatornamerk A leq n$).
– dsw
Jul 27 at 1:28
@MGA in this case the integral becomes $int e^ dx_1 dotsm dx_4 = infty$. The case $k>n$ is lost in my answer, indeed. In this case we want to reduce $A$ to an upper triangular matrix (something with zero rows), as in the comment by nextpuzzle (note $operatornamerk A leq n$).
– dsw
Jul 27 at 1:28
Thanks. Could you say more about where $B$ comes from? Why can't I flip the order of the identity matrix and 0 in the augmented matrix?
– MGA
Jul 27 at 2:44
Thanks. Could you say more about where $B$ comes from? Why can't I flip the order of the identity matrix and 0 in the augmented matrix?
– MGA
Jul 27 at 2:44
add a comment |Â
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2
Do, thin QR factorization of $A$. When $R_1$ is not invertible the integral diverges. When $R_1$ is invertible, do your changes of variable.
– user578878
Jul 27 at 0:28
Thanks! Could you say more about how this helps? It seems like when I do the QR decomposition, I'm left with the l1 norm of Q times the transformed matrix and thus I still have the same problem. Maybe I'm missing something obvious here?
– MGA
Jul 27 at 0:52
Also I am still confused about how to do the change of variables with $R_1$?
– MGA
Jul 27 at 1:18
1
The answer is not $|R_1|^-1cdot 2^m$. For the example of vectors $q_1,q_2$ that I wrote above, the result is $10/9$, instead of $2^2$. Notice that although I denoted it $Id$, that matrix is not the identity. It is the identity with extra zero rows at the bottom.
– user578878
Jul 27 at 2:52
1
Yes, there is a closed form, which you would obtain by doing what I said above.
– user578878
Jul 27 at 3:00