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Help in numerical verification of Chen's identity in rough path theory.
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Would anyone please help me to verify Chen's identity as claimed in https://en.wikipedia.org/wiki/Rough_path#Signature for a simple two dimensional path? Consider a path between $a$ to $d$ and the paths goes through the points say $a(1,4)$, $b(2,9)$, $c(3,3)$, $d(4,7)$. Further, consider that the path between the points are linear. Now one can calculate the signature using iterated integrals of the whole path from $a$ to $d$ and also using Chen's identity, considering the whole path as concatenation of three individual path, $a$ to $b$ then $b$ to $c$ and $c$ to $d$. The components of signature of each individual path can be expressed as $frac1k!prod_1^k(y_2-y_1)$ where $k$ is the level of signature and $y_2$ and $y_1$ represents the $y$ coordinate of ending and starting point respectively for each path. I am interested only up-to level 3. I am having very hard time to calculate the third iterated integrals of each path. I know the second iterated integral corresponds to area and I can calculate it using Trapezoid Rule, but could not get the concepts of iterated integrals, therefore failed to evaluate the triple iterated integrals. In addition I am not sure how to use the Chen's identity. Any help would be highly appreciated.
linear-algebra integration iterated-integrals rough-path-theory
asked Aug 6 at 23:48
Creator
1,0971826
add a comment |Â
up vote
0
down vote
favorite
Would anyone please help me to verify Chen's identity as claimed in https://en.wikipedia.org/wiki/Rough_path#Signature for a simple two dimensional path? Consider a path between $a$ to $d$ and the paths goes through the points say $a(1,4)$, $b(2,9)$, $c(3,3)$, $d(4,7)$. Further, consider that the path between the points are linear. Now one can calculate the signature using iterated integrals of the whole path from $a$ to $d$ and also using Chen's identity, considering the whole path as concatenation of three individual path, $a$ to $b$ then $b$ to $c$ and $c$ to $d$. The components of signature of each individual path can be expressed as $frac1k!prod_1^k(y_2-y_1)$ where $k$ is the level of signature and $y_2$ and $y_1$ represents the $y$ coordinate of ending and starting point respectively for each path. I am interested only up-to level 3. I am having very hard time to calculate the third iterated integrals of each path. I know the second iterated integral corresponds to area and I can calculate it using Trapezoid Rule, but could not get the concepts of iterated integrals, therefore failed to evaluate the triple iterated integrals. In addition I am not sure how to use the Chen's identity. Any help would be highly appreciated.
linear-algebra integration iterated-integrals rough-path-theory
asked Aug 6 at 23:48
Creator
1,0971826
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Would anyone please help me to verify Chen's identity as claimed in https://en.wikipedia.org/wiki/Rough_path#Signature for a simple two dimensional path? Consider a path between $a$ to $d$ and the paths goes through the points say $a(1,4)$, $b(2,9)$, $c(3,3)$, $d(4,7)$. Further, consider that the path between the points are linear. Now one can calculate the signature using iterated integrals of the whole path from $a$ to $d$ and also using Chen's identity, considering the whole path as concatenation of three individual path, $a$ to $b$ then $b$ to $c$ and $c$ to $d$. The components of signature of each individual path can be expressed as $frac1k!prod_1^k(y_2-y_1)$ where $k$ is the level of signature and $y_2$ and $y_1$ represents the $y$ coordinate of ending and starting point respectively for each path. I am interested only up-to level 3. I am having very hard time to calculate the third iterated integrals of each path. I know the second iterated integral corresponds to area and I can calculate it using Trapezoid Rule, but could not get the concepts of iterated integrals, therefore failed to evaluate the triple iterated integrals. In addition I am not sure how to use the Chen's identity. Any help would be highly appreciated.
linear-algebra integration iterated-integrals rough-path-theory
asked Aug 6 at 23:48
Creator
1,0971826
Would anyone please help me to verify Chen's identity as claimed in https://en.wikipedia.org/wiki/Rough_path#Signature for a simple two dimensional path? Consider a path between $a$ to $d$ and the paths goes through the points say $a(1,4)$, $b(2,9)$, $c(3,3)$, $d(4,7)$. Further, consider that the path between the points are linear. Now one can calculate the signature using iterated integrals of the whole path from $a$ to $d$ and also using Chen's identity, considering the whole path as concatenation of three individual path, $a$ to $b$ then $b$ to $c$ and $c$ to $d$. The components of signature of each individual path can be expressed as $frac1k!prod_1^k(y_2-y_1)$ where $k$ is the level of signature and $y_2$ and $y_1$ represents the $y$ coordinate of ending and starting point respectively for each path. I am interested only up-to level 3. I am having very hard time to calculate the third iterated integrals of each path. I know the second iterated integral corresponds to area and I can calculate it using Trapezoid Rule, but could not get the concepts of iterated integrals, therefore failed to evaluate the triple iterated integrals. In addition I am not sure how to use the Chen's identity. Any help would be highly appreciated.
linear-algebra integration iterated-integrals rough-path-theory
asked Aug 6 at 23:48
Creator
1,0971826
edited Aug 7 at 1:37
asked Aug 6 at 23:48
Creator
1,0971826
asked Aug 6 at 23:48
Creator
1,0971826
asked Aug 6 at 23:48
Creator
1,0971826
1,0971826
add a comment |Â
add a comment |Â
1 Answer
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up vote
1
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I think you are asking for an explicit calculation of the third level of the signature of a three-segment path - not using Chen's formula but just doing the integration according to the definition of the signature. This would be a comparatively simple but very long calculation. Here I will do the explicit calculation for only a single element of level two of the signature of a two-segment two-dimensional path with all the integrals in detail. I hope this will illustrate the general pattern usefully. I will not mention anything specific to level two, and I will not use the trapezium rule. I am also not going to use the points you have picked, which have one of their coordinates being in arithmetic progression, in order to avoid lucky simplifications being evident.
Consider calculating element 12 of the signature of the path linearly interpolating the points (1,4), (2,9) and (4,3). It lives on the second level.
First we need to define a formula for the path, a parametrisation. (It turns out that we will get the same answer whatever parametrisation we choose). We will use the parametrisation of the curve given on $0<t<2$ by
$$
x_1(t)=begincasest+1&tle1\2t&t>1endcases
$$
and
$$
x_2(t)=begincases5t+4&tle1\15-6t&t>1endcases.
$$
The definition of the signature element 12 of our curve on $[0,2]$ is
$$
X_0,2^x(12)=int_0^2int_0^t dx_1(s),dx_2(t)=int_0^2left[int_0^t dx_1(s)right],dx_2(t).
$$
These integrals are Riemann-Stieltjes integrals. [Note: If we were calculating the signature of a smooth curve given by a simple formula, we would typically use integration by substitution to get them into a normal-looking Riemann integral - this element would look like
$int_0^2int_0^t x_1'(s),ds,x_2'(t),dt$. We cannot do that here because $x_1$ and $x_2$ are not differentiable at 1. The curve, indeed any curve which is piecewise linear but not a straight line, turns a corner and so no parametrisation will have derivatives at the turning point.]
Let's proceed with the calculation of the integral. The innermost integral is always elementary, as it is given by a difference in the endpoints. Our signature element is therefore
$$
X_0,2^x(12)=int_0^2 [x_1(t)-x_1(0)] , dx_2(t)
$$
All integrals except the innermost will need to be split separately into parts according to the piecewise definition of the curve. We then get something we can evaluate explicitly, using integration by substitution.
beginalign
X_0,2^x(12)&=int_0^1 [x_1(t)-x_1(0)] , dx_2(t) + int_1^2 [x_1(t)-x_1(0)] , dx_2(t)
\&=int_0^1 [(t+1)-1] , dx_2(t) + int_1^2 [(2t)-1] , dx_2(t)
\&=int_0^1 [(t+1)-1] , fracdx_2(t)dt,dt + int_1^2 [(2t)-1] , fracdx_2(t)dt,dt
\&=int_0^1 [(t+1)-1] , 5,dt + int_1^2 [(2t)-1] , (-6),dt
\&=frac52-12=frac-192
endalign
This concludes the explicit calculation of the element.
But there's an even more verbose way to split up the definition
beginalign
X_0,2^x(12)&=int_0^1 [x_1(t)-x_1(0)] , dx_2(t) + int_1^2 [x_1(t)-x_1(0)] , dx_2(t)
\&=int_0^1 [x_1(t)-x_1(0)] , dx_2(t) + int_1^2 [x_1(1)-x_1(0)] , dx_2(t)+ int_1^2 [x_1(t)-x_1(1)] , dx_2(t)
\&=int_0^1 [x_1(t)-x_1(0)] , dx_2(t) + [x_1(1)-x_1(0)] int_1^2 , dx_2(t)+ int_1^2 [x_1(t)-x_1(1)] , dx_2(t)
endalign
We recognise each part of this as a signature of parts of the path, explicitly:
beginalign
X_0,2^x(12)&=X_0,1^x(12)+X_0,1^x(1)X_1,2^x(2)+X_1,2^x(12)
endalign
This is exactly what Chen's formula for a signature element of a concatenation of two paths would tell us. In this case the two paths are straight so we can use the formula you mention (explicitly that element $i_1dots i_k$ of the signature of the straight path from $a$ to $b$ is $frac1k!prod_j=1^k(b_i_j-a_i_j)$) to evaluate them, giving
beginalign
X_0,2^x(12)&=X_0,1^x(12)+X_0,1^x(1)X_1,2^x(2)+X_1,2^x(12)
\&=[tfrac12(2-1)(9-4)]+[(2-1)][(3-9)]+[tfrac12(4-2)(3-9)]
\&=tfrac52-6-6=frac-192.
endalign
In this case we see that explicit calculation and the use of Chen's formula agree.
answered Aug 14 at 22:44
JeremyFR
20617
Thanks a lot. Just to let know, why I could not do. I never did the use the parameter 't', so never formed an equation. BIG thanks.
– Creator
Aug 15 at 0:00
It seems I need to implement only two equations to get the signatures for linear path between points. One is Chen's formula and the is $frac1k!prod_j=1^k(b_i_j-a_i_j)$. I am only interested in speech like signals where one dimension is arithmetic progression (constant sampling period). Is there a shot cut formula for such case?
– Creator
Aug 17 at 2:59
Implementing those two formulae can be very little code, little hassle and can often run pretty fast - it's linear in the length of the input. In particular cases (e.g. low dimension like you have, low level, very long signals, a language like python or matlab where you want to avoid loops) you may be able to be clever and come up with something which runs faster. By "short cut" do you mean saving coding/testing time or saving run time? What platform/programming language?
– JeremyFR
Aug 17 at 11:05
By short cut, I mean saving run time and I use MATLAB. My background is signal processing.
– Creator
2 days ago
I am not very familiar with matlab. I had a go at writing a general signature calculation function for matlab based on these formulae. It is at github.com/bottler/iisignature/blob/master/examples/matlab/… . I suspect it has plenty of room for optimization. It runs instantaneously in the simple examples I tried. For heavy use, I would expect calling out to a compiled library - like the way iisignature works in Python - to be the fastest.
– JeremyFR
22 hours ago
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I think you are asking for an explicit calculation of the third level of the signature of a three-segment path - not using Chen's formula but just doing the integration according to the definition of the signature. This would be a comparatively simple but very long calculation. Here I will do the explicit calculation for only a single element of level two of the signature of a two-segment two-dimensional path with all the integrals in detail. I hope this will illustrate the general pattern usefully. I will not mention anything specific to level two, and I will not use the trapezium rule. I am also not going to use the points you have picked, which have one of their coordinates being in arithmetic progression, in order to avoid lucky simplifications being evident.
Consider calculating element 12 of the signature of the path linearly interpolating the points (1,4), (2,9) and (4,3). It lives on the second level.
First we need to define a formula for the path, a parametrisation. (It turns out that we will get the same answer whatever parametrisation we choose). We will use the parametrisation of the curve given on $0<t<2$ by
$$
x_1(t)=begincasest+1&tle1\2t&t>1endcases
$$
and
$$
x_2(t)=begincases5t+4&tle1\15-6t&t>1endcases.
$$
The definition of the signature element 12 of our curve on $[0,2]$ is
$$
X_0,2^x(12)=int_0^2int_0^t dx_1(s),dx_2(t)=int_0^2left[int_0^t dx_1(s)right],dx_2(t).
$$
These integrals are Riemann-Stieltjes integrals. [Note: If we were calculating the signature of a smooth curve given by a simple formula, we would typically use integration by substitution to get them into a normal-looking Riemann integral - this element would look like
$int_0^2int_0^t x_1'(s),ds,x_2'(t),dt$. We cannot do that here because $x_1$ and $x_2$ are not differentiable at 1. The curve, indeed any curve which is piecewise linear but not a straight line, turns a corner and so no parametrisation will have derivatives at the turning point.]
Let's proceed with the calculation of the integral. The innermost integral is always elementary, as it is given by a difference in the endpoints. Our signature element is therefore
$$
X_0,2^x(12)=int_0^2 [x_1(t)-x_1(0)] , dx_2(t)
$$
All integrals except the innermost will need to be split separately into parts according to the piecewise definition of the curve. We then get something we can evaluate explicitly, using integration by substitution.
beginalign
X_0,2^x(12)&=int_0^1 [x_1(t)-x_1(0)] , dx_2(t) + int_1^2 [x_1(t)-x_1(0)] , dx_2(t)
\&=int_0^1 [(t+1)-1] , dx_2(t) + int_1^2 [(2t)-1] , dx_2(t)
\&=int_0^1 [(t+1)-1] , fracdx_2(t)dt,dt + int_1^2 [(2t)-1] , fracdx_2(t)dt,dt
\&=int_0^1 [(t+1)-1] , 5,dt + int_1^2 [(2t)-1] , (-6),dt
\&=frac52-12=frac-192
endalign
This concludes the explicit calculation of the element.
But there's an even more verbose way to split up the definition
beginalign
X_0,2^x(12)&=int_0^1 [x_1(t)-x_1(0)] , dx_2(t) + int_1^2 [x_1(t)-x_1(0)] , dx_2(t)
\&=int_0^1 [x_1(t)-x_1(0)] , dx_2(t) + int_1^2 [x_1(1)-x_1(0)] , dx_2(t)+ int_1^2 [x_1(t)-x_1(1)] , dx_2(t)
\&=int_0^1 [x_1(t)-x_1(0)] , dx_2(t) + [x_1(1)-x_1(0)] int_1^2 , dx_2(t)+ int_1^2 [x_1(t)-x_1(1)] , dx_2(t)
endalign
We recognise each part of this as a signature of parts of the path, explicitly:
beginalign
X_0,2^x(12)&=X_0,1^x(12)+X_0,1^x(1)X_1,2^x(2)+X_1,2^x(12)
endalign
This is exactly what Chen's formula for a signature element of a concatenation of two paths would tell us. In this case the two paths are straight so we can use the formula you mention (explicitly that element $i_1dots i_k$ of the signature of the straight path from $a$ to $b$ is $frac1k!prod_j=1^k(b_i_j-a_i_j)$) to evaluate them, giving
beginalign
X_0,2^x(12)&=X_0,1^x(12)+X_0,1^x(1)X_1,2^x(2)+X_1,2^x(12)
\&=[tfrac12(2-1)(9-4)]+[(2-1)][(3-9)]+[tfrac12(4-2)(3-9)]
\&=tfrac52-6-6=frac-192.
endalign
In this case we see that explicit calculation and the use of Chen's formula agree.
answered Aug 14 at 22:44
JeremyFR
20617
Thanks a lot. Just to let know, why I could not do. I never did the use the parameter 't', so never formed an equation. BIG thanks.
– Creator
Aug 15 at 0:00
It seems I need to implement only two equations to get the signatures for linear path between points. One is Chen's formula and the is $frac1k!prod_j=1^k(b_i_j-a_i_j)$. I am only interested in speech like signals where one dimension is arithmetic progression (constant sampling period). Is there a shot cut formula for such case?
– Creator
Aug 17 at 2:59
Implementing those two formulae can be very little code, little hassle and can often run pretty fast - it's linear in the length of the input. In particular cases (e.g. low dimension like you have, low level, very long signals, a language like python or matlab where you want to avoid loops) you may be able to be clever and come up with something which runs faster. By "short cut" do you mean saving coding/testing time or saving run time? What platform/programming language?
– JeremyFR
Aug 17 at 11:05
By short cut, I mean saving run time and I use MATLAB. My background is signal processing.
– Creator
2 days ago
I am not very familiar with matlab. I had a go at writing a general signature calculation function for matlab based on these formulae. It is at github.com/bottler/iisignature/blob/master/examples/matlab/… . I suspect it has plenty of room for optimization. It runs instantaneously in the simple examples I tried. For heavy use, I would expect calling out to a compiled library - like the way iisignature works in Python - to be the fastest.
– JeremyFR
22 hours ago
 |Â
show 1 more comment
up vote
1
down vote
accepted
I think you are asking for an explicit calculation of the third level of the signature of a three-segment path - not using Chen's formula but just doing the integration according to the definition of the signature. This would be a comparatively simple but very long calculation. Here I will do the explicit calculation for only a single element of level two of the signature of a two-segment two-dimensional path with all the integrals in detail. I hope this will illustrate the general pattern usefully. I will not mention anything specific to level two, and I will not use the trapezium rule. I am also not going to use the points you have picked, which have one of their coordinates being in arithmetic progression, in order to avoid lucky simplifications being evident.
Consider calculating element 12 of the signature of the path linearly interpolating the points (1,4), (2,9) and (4,3). It lives on the second level.
First we need to define a formula for the path, a parametrisation. (It turns out that we will get the same answer whatever parametrisation we choose). We will use the parametrisation of the curve given on $0<t<2$ by
$$
x_1(t)=begincasest+1&tle1\2t&t>1endcases
$$
and
$$
x_2(t)=begincases5t+4&tle1\15-6t&t>1endcases.
$$
The definition of the signature element 12 of our curve on $[0,2]$ is
$$
X_0,2^x(12)=int_0^2int_0^t dx_1(s),dx_2(t)=int_0^2left[int_0^t dx_1(s)right],dx_2(t).
$$
These integrals are Riemann-Stieltjes integrals. [Note: If we were calculating the signature of a smooth curve given by a simple formula, we would typically use integration by substitution to get them into a normal-looking Riemann integral - this element would look like
$int_0^2int_0^t x_1'(s),ds,x_2'(t),dt$. We cannot do that here because $x_1$ and $x_2$ are not differentiable at 1. The curve, indeed any curve which is piecewise linear but not a straight line, turns a corner and so no parametrisation will have derivatives at the turning point.]
Let's proceed with the calculation of the integral. The innermost integral is always elementary, as it is given by a difference in the endpoints. Our signature element is therefore
$$
X_0,2^x(12)=int_0^2 [x_1(t)-x_1(0)] , dx_2(t)
$$
All integrals except the innermost will need to be split separately into parts according to the piecewise definition of the curve. We then get something we can evaluate explicitly, using integration by substitution.
beginalign
X_0,2^x(12)&=int_0^1 [x_1(t)-x_1(0)] , dx_2(t) + int_1^2 [x_1(t)-x_1(0)] , dx_2(t)
\&=int_0^1 [(t+1)-1] , dx_2(t) + int_1^2 [(2t)-1] , dx_2(t)
\&=int_0^1 [(t+1)-1] , fracdx_2(t)dt,dt + int_1^2 [(2t)-1] , fracdx_2(t)dt,dt
\&=int_0^1 [(t+1)-1] , 5,dt + int_1^2 [(2t)-1] , (-6),dt
\&=frac52-12=frac-192
endalign
This concludes the explicit calculation of the element.
But there's an even more verbose way to split up the definition
beginalign
X_0,2^x(12)&=int_0^1 [x_1(t)-x_1(0)] , dx_2(t) + int_1^2 [x_1(t)-x_1(0)] , dx_2(t)
\&=int_0^1 [x_1(t)-x_1(0)] , dx_2(t) + int_1^2 [x_1(1)-x_1(0)] , dx_2(t)+ int_1^2 [x_1(t)-x_1(1)] , dx_2(t)
\&=int_0^1 [x_1(t)-x_1(0)] , dx_2(t) + [x_1(1)-x_1(0)] int_1^2 , dx_2(t)+ int_1^2 [x_1(t)-x_1(1)] , dx_2(t)
endalign
We recognise each part of this as a signature of parts of the path, explicitly:
beginalign
X_0,2^x(12)&=X_0,1^x(12)+X_0,1^x(1)X_1,2^x(2)+X_1,2^x(12)
endalign
This is exactly what Chen's formula for a signature element of a concatenation of two paths would tell us. In this case the two paths are straight so we can use the formula you mention (explicitly that element $i_1dots i_k$ of the signature of the straight path from $a$ to $b$ is $frac1k!prod_j=1^k(b_i_j-a_i_j)$) to evaluate them, giving
beginalign
X_0,2^x(12)&=X_0,1^x(12)+X_0,1^x(1)X_1,2^x(2)+X_1,2^x(12)
\&=[tfrac12(2-1)(9-4)]+[(2-1)][(3-9)]+[tfrac12(4-2)(3-9)]
\&=tfrac52-6-6=frac-192.
endalign
In this case we see that explicit calculation and the use of Chen's formula agree.
answered Aug 14 at 22:44
JeremyFR
20617
Thanks a lot. Just to let know, why I could not do. I never did the use the parameter 't', so never formed an equation. BIG thanks.
– Creator
Aug 15 at 0:00
It seems I need to implement only two equations to get the signatures for linear path between points. One is Chen's formula and the is $frac1k!prod_j=1^k(b_i_j-a_i_j)$. I am only interested in speech like signals where one dimension is arithmetic progression (constant sampling period). Is there a shot cut formula for such case?
– Creator
Aug 17 at 2:59
Implementing those two formulae can be very little code, little hassle and can often run pretty fast - it's linear in the length of the input. In particular cases (e.g. low dimension like you have, low level, very long signals, a language like python or matlab where you want to avoid loops) you may be able to be clever and come up with something which runs faster. By "short cut" do you mean saving coding/testing time or saving run time? What platform/programming language?
– JeremyFR
Aug 17 at 11:05
By short cut, I mean saving run time and I use MATLAB. My background is signal processing.
– Creator
2 days ago
I am not very familiar with matlab. I had a go at writing a general signature calculation function for matlab based on these formulae. It is at github.com/bottler/iisignature/blob/master/examples/matlab/… . I suspect it has plenty of room for optimization. It runs instantaneously in the simple examples I tried. For heavy use, I would expect calling out to a compiled library - like the way iisignature works in Python - to be the fastest.
– JeremyFR
22 hours ago
 |Â
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I think you are asking for an explicit calculation of the third level of the signature of a three-segment path - not using Chen's formula but just doing the integration according to the definition of the signature. This would be a comparatively simple but very long calculation. Here I will do the explicit calculation for only a single element of level two of the signature of a two-segment two-dimensional path with all the integrals in detail. I hope this will illustrate the general pattern usefully. I will not mention anything specific to level two, and I will not use the trapezium rule. I am also not going to use the points you have picked, which have one of their coordinates being in arithmetic progression, in order to avoid lucky simplifications being evident.
Consider calculating element 12 of the signature of the path linearly interpolating the points (1,4), (2,9) and (4,3). It lives on the second level.
First we need to define a formula for the path, a parametrisation. (It turns out that we will get the same answer whatever parametrisation we choose). We will use the parametrisation of the curve given on $0<t<2$ by
$$
x_1(t)=begincasest+1&tle1\2t&t>1endcases
$$
and
$$
x_2(t)=begincases5t+4&tle1\15-6t&t>1endcases.
$$
The definition of the signature element 12 of our curve on $[0,2]$ is
$$
X_0,2^x(12)=int_0^2int_0^t dx_1(s),dx_2(t)=int_0^2left[int_0^t dx_1(s)right],dx_2(t).
$$
These integrals are Riemann-Stieltjes integrals. [Note: If we were calculating the signature of a smooth curve given by a simple formula, we would typically use integration by substitution to get them into a normal-looking Riemann integral - this element would look like
$int_0^2int_0^t x_1'(s),ds,x_2'(t),dt$. We cannot do that here because $x_1$ and $x_2$ are not differentiable at 1. The curve, indeed any curve which is piecewise linear but not a straight line, turns a corner and so no parametrisation will have derivatives at the turning point.]
Let's proceed with the calculation of the integral. The innermost integral is always elementary, as it is given by a difference in the endpoints. Our signature element is therefore
$$
X_0,2^x(12)=int_0^2 [x_1(t)-x_1(0)] , dx_2(t)
$$
All integrals except the innermost will need to be split separately into parts according to the piecewise definition of the curve. We then get something we can evaluate explicitly, using integration by substitution.
beginalign
X_0,2^x(12)&=int_0^1 [x_1(t)-x_1(0)] , dx_2(t) + int_1^2 [x_1(t)-x_1(0)] , dx_2(t)
\&=int_0^1 [(t+1)-1] , dx_2(t) + int_1^2 [(2t)-1] , dx_2(t)
\&=int_0^1 [(t+1)-1] , fracdx_2(t)dt,dt + int_1^2 [(2t)-1] , fracdx_2(t)dt,dt
\&=int_0^1 [(t+1)-1] , 5,dt + int_1^2 [(2t)-1] , (-6),dt
\&=frac52-12=frac-192
endalign
This concludes the explicit calculation of the element.
But there's an even more verbose way to split up the definition
beginalign
X_0,2^x(12)&=int_0^1 [x_1(t)-x_1(0)] , dx_2(t) + int_1^2 [x_1(t)-x_1(0)] , dx_2(t)
\&=int_0^1 [x_1(t)-x_1(0)] , dx_2(t) + int_1^2 [x_1(1)-x_1(0)] , dx_2(t)+ int_1^2 [x_1(t)-x_1(1)] , dx_2(t)
\&=int_0^1 [x_1(t)-x_1(0)] , dx_2(t) + [x_1(1)-x_1(0)] int_1^2 , dx_2(t)+ int_1^2 [x_1(t)-x_1(1)] , dx_2(t)
endalign
We recognise each part of this as a signature of parts of the path, explicitly:
beginalign
X_0,2^x(12)&=X_0,1^x(12)+X_0,1^x(1)X_1,2^x(2)+X_1,2^x(12)
endalign
This is exactly what Chen's formula for a signature element of a concatenation of two paths would tell us. In this case the two paths are straight so we can use the formula you mention (explicitly that element $i_1dots i_k$ of the signature of the straight path from $a$ to $b$ is $frac1k!prod_j=1^k(b_i_j-a_i_j)$) to evaluate them, giving
beginalign
X_0,2^x(12)&=X_0,1^x(12)+X_0,1^x(1)X_1,2^x(2)+X_1,2^x(12)
\&=[tfrac12(2-1)(9-4)]+[(2-1)][(3-9)]+[tfrac12(4-2)(3-9)]
\&=tfrac52-6-6=frac-192.
endalign
In this case we see that explicit calculation and the use of Chen's formula agree.
answered Aug 14 at 22:44
JeremyFR
20617
I think you are asking for an explicit calculation of the third level of the signature of a three-segment path - not using Chen's formula but just doing the integration according to the definition of the signature. This would be a comparatively simple but very long calculation. Here I will do the explicit calculation for only a single element of level two of the signature of a two-segment two-dimensional path with all the integrals in detail. I hope this will illustrate the general pattern usefully. I will not mention anything specific to level two, and I will not use the trapezium rule. I am also not going to use the points you have picked, which have one of their coordinates being in arithmetic progression, in order to avoid lucky simplifications being evident.
Consider calculating element 12 of the signature of the path linearly interpolating the points (1,4), (2,9) and (4,3). It lives on the second level.
First we need to define a formula for the path, a parametrisation. (It turns out that we will get the same answer whatever parametrisation we choose). We will use the parametrisation of the curve given on $0<t<2$ by
$$
x_1(t)=begincasest+1&tle1\2t&t>1endcases
$$
and
$$
x_2(t)=begincases5t+4&tle1\15-6t&t>1endcases.
$$
The definition of the signature element 12 of our curve on $[0,2]$ is
$$
X_0,2^x(12)=int_0^2int_0^t dx_1(s),dx_2(t)=int_0^2left[int_0^t dx_1(s)right],dx_2(t).
$$
These integrals are Riemann-Stieltjes integrals. [Note: If we were calculating the signature of a smooth curve given by a simple formula, we would typically use integration by substitution to get them into a normal-looking Riemann integral - this element would look like
$int_0^2int_0^t x_1'(s),ds,x_2'(t),dt$. We cannot do that here because $x_1$ and $x_2$ are not differentiable at 1. The curve, indeed any curve which is piecewise linear but not a straight line, turns a corner and so no parametrisation will have derivatives at the turning point.]
Let's proceed with the calculation of the integral. The innermost integral is always elementary, as it is given by a difference in the endpoints. Our signature element is therefore
$$
X_0,2^x(12)=int_0^2 [x_1(t)-x_1(0)] , dx_2(t)
$$
All integrals except the innermost will need to be split separately into parts according to the piecewise definition of the curve. We then get something we can evaluate explicitly, using integration by substitution.
beginalign
X_0,2^x(12)&=int_0^1 [x_1(t)-x_1(0)] , dx_2(t) + int_1^2 [x_1(t)-x_1(0)] , dx_2(t)
\&=int_0^1 [(t+1)-1] , dx_2(t) + int_1^2 [(2t)-1] , dx_2(t)
\&=int_0^1 [(t+1)-1] , fracdx_2(t)dt,dt + int_1^2 [(2t)-1] , fracdx_2(t)dt,dt
\&=int_0^1 [(t+1)-1] , 5,dt + int_1^2 [(2t)-1] , (-6),dt
\&=frac52-12=frac-192
endalign
This concludes the explicit calculation of the element.
But there's an even more verbose way to split up the definition
beginalign
X_0,2^x(12)&=int_0^1 [x_1(t)-x_1(0)] , dx_2(t) + int_1^2 [x_1(t)-x_1(0)] , dx_2(t)
\&=int_0^1 [x_1(t)-x_1(0)] , dx_2(t) + int_1^2 [x_1(1)-x_1(0)] , dx_2(t)+ int_1^2 [x_1(t)-x_1(1)] , dx_2(t)
\&=int_0^1 [x_1(t)-x_1(0)] , dx_2(t) + [x_1(1)-x_1(0)] int_1^2 , dx_2(t)+ int_1^2 [x_1(t)-x_1(1)] , dx_2(t)
endalign
We recognise each part of this as a signature of parts of the path, explicitly:
beginalign
X_0,2^x(12)&=X_0,1^x(12)+X_0,1^x(1)X_1,2^x(2)+X_1,2^x(12)
endalign
This is exactly what Chen's formula for a signature element of a concatenation of two paths would tell us. In this case the two paths are straight so we can use the formula you mention (explicitly that element $i_1dots i_k$ of the signature of the straight path from $a$ to $b$ is $frac1k!prod_j=1^k(b_i_j-a_i_j)$) to evaluate them, giving
beginalign
X_0,2^x(12)&=X_0,1^x(12)+X_0,1^x(1)X_1,2^x(2)+X_1,2^x(12)
\&=[tfrac12(2-1)(9-4)]+[(2-1)][(3-9)]+[tfrac12(4-2)(3-9)]
\&=tfrac52-6-6=frac-192.
endalign
In this case we see that explicit calculation and the use of Chen's formula agree.
answered Aug 14 at 22:44
JeremyFR
20617
answered Aug 14 at 22:44
JeremyFR
20617
answered Aug 14 at 22:44
JeremyFR
20617
answered Aug 14 at 22:44
JeremyFR
20617
20617
Thanks a lot. Just to let know, why I could not do. I never did the use the parameter 't', so never formed an equation. BIG thanks.
– Creator
Aug 15 at 0:00
It seems I need to implement only two equations to get the signatures for linear path between points. One is Chen's formula and the is $frac1k!prod_j=1^k(b_i_j-a_i_j)$. I am only interested in speech like signals where one dimension is arithmetic progression (constant sampling period). Is there a shot cut formula for such case?
– Creator
Aug 17 at 2:59
Implementing those two formulae can be very little code, little hassle and can often run pretty fast - it's linear in the length of the input. In particular cases (e.g. low dimension like you have, low level, very long signals, a language like python or matlab where you want to avoid loops) you may be able to be clever and come up with something which runs faster. By "short cut" do you mean saving coding/testing time or saving run time? What platform/programming language?
– JeremyFR
Aug 17 at 11:05
By short cut, I mean saving run time and I use MATLAB. My background is signal processing.
– Creator
2 days ago
I am not very familiar with matlab. I had a go at writing a general signature calculation function for matlab based on these formulae. It is at github.com/bottler/iisignature/blob/master/examples/matlab/… . I suspect it has plenty of room for optimization. It runs instantaneously in the simple examples I tried. For heavy use, I would expect calling out to a compiled library - like the way iisignature works in Python - to be the fastest.
– JeremyFR
22 hours ago
 |Â
show 1 more comment
Thanks a lot. Just to let know, why I could not do. I never did the use the parameter 't', so never formed an equation. BIG thanks.
– Creator
Aug 15 at 0:00
It seems I need to implement only two equations to get the signatures for linear path between points. One is Chen's formula and the is $frac1k!prod_j=1^k(b_i_j-a_i_j)$. I am only interested in speech like signals where one dimension is arithmetic progression (constant sampling period). Is there a shot cut formula for such case?
– Creator
Aug 17 at 2:59
Implementing those two formulae can be very little code, little hassle and can often run pretty fast - it's linear in the length of the input. In particular cases (e.g. low dimension like you have, low level, very long signals, a language like python or matlab where you want to avoid loops) you may be able to be clever and come up with something which runs faster. By "short cut" do you mean saving coding/testing time or saving run time? What platform/programming language?
– JeremyFR
Aug 17 at 11:05
By short cut, I mean saving run time and I use MATLAB. My background is signal processing.
– Creator
2 days ago
I am not very familiar with matlab. I had a go at writing a general signature calculation function for matlab based on these formulae. It is at github.com/bottler/iisignature/blob/master/examples/matlab/… . I suspect it has plenty of room for optimization. It runs instantaneously in the simple examples I tried. For heavy use, I would expect calling out to a compiled library - like the way iisignature works in Python - to be the fastest.
– JeremyFR
22 hours ago
Thanks a lot. Just to let know, why I could not do. I never did the use the parameter 't', so never formed an equation. BIG thanks.
– Creator
Aug 15 at 0:00
Thanks a lot. Just to let know, why I could not do. I never did the use the parameter 't', so never formed an equation. BIG thanks.
– Creator
Aug 15 at 0:00
It seems I need to implement only two equations to get the signatures for linear path between points. One is Chen's formula and the is $frac1k!prod_j=1^k(b_i_j-a_i_j)$. I am only interested in speech like signals where one dimension is arithmetic progression (constant sampling period). Is there a shot cut formula for such case?
– Creator
Aug 17 at 2:59
It seems I need to implement only two equations to get the signatures for linear path between points. One is Chen's formula and the is $frac1k!prod_j=1^k(b_i_j-a_i_j)$. I am only interested in speech like signals where one dimension is arithmetic progression (constant sampling period). Is there a shot cut formula for such case?
– Creator
Aug 17 at 2:59
Implementing those two formulae can be very little code, little hassle and can often run pretty fast - it's linear in the length of the input. In particular cases (e.g. low dimension like you have, low level, very long signals, a language like python or matlab where you want to avoid loops) you may be able to be clever and come up with something which runs faster. By "short cut" do you mean saving coding/testing time or saving run time? What platform/programming language?
– JeremyFR
Aug 17 at 11:05
Implementing those two formulae can be very little code, little hassle and can often run pretty fast - it's linear in the length of the input. In particular cases (e.g. low dimension like you have, low level, very long signals, a language like python or matlab where you want to avoid loops) you may be able to be clever and come up with something which runs faster. By "short cut" do you mean saving coding/testing time or saving run time? What platform/programming language?
– JeremyFR
Aug 17 at 11:05
By short cut, I mean saving run time and I use MATLAB. My background is signal processing.
– Creator
2 days ago
By short cut, I mean saving run time and I use MATLAB. My background is signal processing.
– Creator
2 days ago
I am not very familiar with matlab. I had a go at writing a general signature calculation function for matlab based on these formulae. It is at github.com/bottler/iisignature/blob/master/examples/matlab/… . I suspect it has plenty of room for optimization. It runs instantaneously in the simple examples I tried. For heavy use, I would expect calling out to a compiled library - like the way iisignature works in Python - to be the fastest.
– JeremyFR
22 hours ago
I am not very familiar with matlab. I had a go at writing a general signature calculation function for matlab based on these formulae. It is at github.com/bottler/iisignature/blob/master/examples/matlab/… . I suspect it has plenty of room for optimization. It runs instantaneously in the simple examples I tried. For heavy use, I would expect calling out to a compiled library - like the way iisignature works in Python - to be the fastest.
– JeremyFR
22 hours ago
 |Â
show 1 more comment
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