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How to estimate $alpha$ given a discrete sequence $a(n) = n^alpha+ n^1-alpha+b$.
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I have a data sequence of length $10^6$ which I know it can be approximately modeled as $a(n) = n^alpha+ n^1-alpha+b$ ($b$ is an unknown constant, $0<alpha <1$). But how to estimate $alpha$ here?
Update:
Sorry, I made some mistakes above. I sincerely apologize for the mistakes. I revise the problem and explain my concerns more in this update.
The data sequence is generated from simulations of one algorithm. It should have the order $O(n^max(alpha, 1-alpha))$. I know it can be approximately modeled as $a(n) = b_1n^alpha+ b_2n^1-alpha+b_3$ ($b_1,b_2,b_3$ are unknown constants) when $n$ is suffciently large.
I only care about the order $max(alpha, 1-alpha)$ to justify the performance of the algorithm.
I have tried using $frac1t_1-t_0sum_t=t_0+1^t_1log_2 fraca(2t)a(t)$. The problem is the graph I got does not get the minimum at $alpha = 0.5$. How to eliminate the numerical errors?
Here's a link of the new post of the problem.
numerical-methods
asked Jul 23 at 7:21
Jimmy Kang
647
add a comment |Â
up vote
0
down vote
favorite
I have a data sequence of length $10^6$ which I know it can be approximately modeled as $a(n) = n^alpha+ n^1-alpha+b$ ($b$ is an unknown constant, $0<alpha <1$). But how to estimate $alpha$ here?
Update:
Sorry, I made some mistakes above. I sincerely apologize for the mistakes. I revise the problem and explain my concerns more in this update.
The data sequence is generated from simulations of one algorithm. It should have the order $O(n^max(alpha, 1-alpha))$. I know it can be approximately modeled as $a(n) = b_1n^alpha+ b_2n^1-alpha+b_3$ ($b_1,b_2,b_3$ are unknown constants) when $n$ is suffciently large.
I only care about the order $max(alpha, 1-alpha)$ to justify the performance of the algorithm.
I have tried using $frac1t_1-t_0sum_t=t_0+1^t_1log_2 fraca(2t)a(t)$. The problem is the graph I got does not get the minimum at $alpha = 0.5$. How to eliminate the numerical errors?
Here's a link of the new post of the problem.
numerical-methods
asked Jul 23 at 7:21
Jimmy Kang
647
1
You can consider $0<alpha<frac12$.
– Yves Daoust
Jul 23 at 8:26
Is the data noisy or smooth ? Do the initial terms match the model ?
– Yves Daoust
Jul 23 at 9:06
What are you after, an estimate of the asymptotic behavior or just a best fit to the million points ?
– Yves Daoust
Jul 23 at 9:33
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a data sequence of length $10^6$ which I know it can be approximately modeled as $a(n) = n^alpha+ n^1-alpha+b$ ($b$ is an unknown constant, $0<alpha <1$). But how to estimate $alpha$ here?
Update:
Sorry, I made some mistakes above. I sincerely apologize for the mistakes. I revise the problem and explain my concerns more in this update.
The data sequence is generated from simulations of one algorithm. It should have the order $O(n^max(alpha, 1-alpha))$. I know it can be approximately modeled as $a(n) = b_1n^alpha+ b_2n^1-alpha+b_3$ ($b_1,b_2,b_3$ are unknown constants) when $n$ is suffciently large.
I only care about the order $max(alpha, 1-alpha)$ to justify the performance of the algorithm.
I have tried using $frac1t_1-t_0sum_t=t_0+1^t_1log_2 fraca(2t)a(t)$. The problem is the graph I got does not get the minimum at $alpha = 0.5$. How to eliminate the numerical errors?
Here's a link of the new post of the problem.
numerical-methods
asked Jul 23 at 7:21
Jimmy Kang
647
I have a data sequence of length $10^6$ which I know it can be approximately modeled as $a(n) = n^alpha+ n^1-alpha+b$ ($b$ is an unknown constant, $0<alpha <1$). But how to estimate $alpha$ here?
Update:
Sorry, I made some mistakes above. I sincerely apologize for the mistakes. I revise the problem and explain my concerns more in this update.
The data sequence is generated from simulations of one algorithm. It should have the order $O(n^max(alpha, 1-alpha))$. I know it can be approximately modeled as $a(n) = b_1n^alpha+ b_2n^1-alpha+b_3$ ($b_1,b_2,b_3$ are unknown constants) when $n$ is suffciently large.
I only care about the order $max(alpha, 1-alpha)$ to justify the performance of the algorithm.
I have tried using $frac1t_1-t_0sum_t=t_0+1^t_1log_2 fraca(2t)a(t)$. The problem is the graph I got does not get the minimum at $alpha = 0.5$. How to eliminate the numerical errors?
Here's a link of the new post of the problem.
numerical-methods
asked Jul 23 at 7:21
Jimmy Kang
647
edited Jul 24 at 15:59
asked Jul 23 at 7:21
Jimmy Kang
647
asked Jul 23 at 7:21
Jimmy Kang
647
asked Jul 23 at 7:21
Jimmy Kang
647
647
1
You can consider $0<alpha<frac12$.
– Yves Daoust
Jul 23 at 8:26
Is the data noisy or smooth ? Do the initial terms match the model ?
– Yves Daoust
Jul 23 at 9:06
What are you after, an estimate of the asymptotic behavior or just a best fit to the million points ?
– Yves Daoust
Jul 23 at 9:33
add a comment |Â
1
You can consider $0<alpha<frac12$.
– Yves Daoust
Jul 23 at 8:26
Is the data noisy or smooth ? Do the initial terms match the model ?
– Yves Daoust
Jul 23 at 9:06
What are you after, an estimate of the asymptotic behavior or just a best fit to the million points ?
– Yves Daoust
Jul 23 at 9:33
1
1
You can consider $0<alpha<frac12$.
– Yves Daoust
Jul 23 at 8:26
You can consider $0<alpha<frac12$.
– Yves Daoust
Jul 23 at 8:26
Is the data noisy or smooth ? Do the initial terms match the model ?
– Yves Daoust
Jul 23 at 9:06
Is the data noisy or smooth ? Do the initial terms match the model ?
– Yves Daoust
Jul 23 at 9:06
What are you after, an estimate of the asymptotic behavior or just a best fit to the million points ?
– Yves Daoust
Jul 23 at 9:33
What are you after, an estimate of the asymptotic behavior or just a best fit to the million points ?
– Yves Daoust
Jul 23 at 9:33
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
Rather than a two-parameter least-squares, you might note that your formula implies
$$a_i+1-a_i = n_i+1^alpha - n_i^alpha + n_i+1^1-alpha - n_i^1-alpha$$
Find the $alpha$ that minimizes the sum of squared errors here, then for $b$ take the average of $a_i - (n_i^alpha + n_i^1-alpha)$.
answered Jul 23 at 8:30
Robert Israel
304k22201441
For sure ! Much simpler than my junk !
– Claude Leibovici
Jul 23 at 8:38
If I am right, $n_i=i$, there is no need to index.
– Yves Daoust
Jul 23 at 8:40
@YvesDaoust. Not necessary since you can use, say, $n=10,20,30,cdots$.
– Claude Leibovici
Jul 23 at 8:53
@ClaudeLeibovici: right, but then the formula must be with $a_n_i$.
– Yves Daoust
Jul 23 at 8:56
Many thanks. I have tried your answer. I thought the data can be approximately modeled as $a(n) = n^alpha+ n^1-alpha+b$. Actually, I have $5$ sequences, each should be modeled with $alpha = 1/1.9,1/2.0, 1/2.1, 1/2.2, 1/2.5$ respectively. However, the strange thing is the partial derivative of the squared errors achieves $0$ at $alpha = 0.5$ for all sequences. I'm still trying to figure out what happens in the data.
– Jimmy Kang
Jul 23 at 20:20
add a comment |Â
up vote
2
down vote
If this helps:
If $a_0$ or $a_1-2$ are available and give reasonable estimates of $b$, for large $n$ we have
$$fraca_n-hat bsqrt nsim n^alpha-1/2+n^1/2-alpha=2coshleft(left(frac12-alpharight)log nright)$$
and
$$alphasimfrac12-dfractextarcoshdfraca_n-hat b2sqrt nlog n.$$
This can be used f.i. as a starting value of an iterative method.
answered Jul 23 at 9:00
Yves Daoust
111k665203
This seems to be a quite good idea, for sure !
– Claude Leibovici
Jul 23 at 9:17
Sorry for the late reply. Data is noisy. The initial terms do not match the model. My concern is just to estimate the asymptotic behavior. The data should be $O(n^max(alpha,1-alpha))$. I want to get the order of the sequence, i.e. $max(alpha,1-alpha)$.
– Jimmy Kang
Jul 23 at 20:01
@JimmyKang: why do you add a constant term ?
– Yves Daoust
Jul 23 at 20:41
I revised the problem just now. Sorry for the confusion caused.
– Jimmy Kang
Jul 23 at 21:21
add a comment |Â
up vote
1
down vote
If we look at the problem as a regression, you have $p$ data points $(n_i,a_i)$ and you want to adjust the model
$$a=n^alpha+n^1-alpha+b$$ In the least square sense, you need to minimize
$$SSQ=sum_i=1^p left(n_i^alpha+n_i^1-alpha+b -a_iright)^2$$ Computing the partial derivatives,we have
$$fracpartial SSQpartial b=2sum_i=1^p left(n_i^alpha+n_i^1-alpha+b -a_iright)=0tag 1$$
$$fracpartial SSQpartial a=2sum_i=1^p left(n_i^alpha+n_i^1-alpha+b -a_iright)left(n_i^alpha-n_i^1-alpharight)log(n_i)=0 tag 2$$ which is hard to solve.
However, from $(1)$, you can get $b$ as a function of $alpha$
$$b(alpha)=-frac 1p sum_i=1^p left(n_i^alpha+n_i^1-alpha -a_iright)$$ and then $(2)$ is "just" an equation in $alpha$.
The simplest would be to plot equation $(2)$ for $0 leq alpha leq 0.5$ and see where it does cancels. Zoom more and more to have more accurate results; when your accuracy has been reached, recompute $b$.
All of that can be done using Excel.
Edit
For illustration purposes, let us use the following data set
$$left(
beginarrayccc
i & n_i & a_n_i \
1 & 10 & 10 \
2 & 20 & 14 \
3 & 30 & 16 \
4 & 40 & 19 \
5 & 50 & 21 \
6 & 60 & 23 \
7 & 70 & 25 \
8 & 80 & 26 \
9 & 90 & 28 \
10 & 100 & 30
endarray
right)$$ A first run, using $Delta alpha=0.05$ would give
$$left(
beginarraycc
alpha & (2) \
0.05 & -36241.2 \
0.10 & -19913.1 \
0.15 & -10372.8 \
0.20 & -4943.28 \
0.25 & -1981.30 \
0.30 & -483.658 \
0.35 & 158.953
endarray
right)$$ So, the solution is between $0.30$ and $0.35$. Repeat using $Delta alpha=0.005$ to get
$$left(
beginarraycc
alpha & (2) \
0.315 & -221.502 \
0.320 & -148.821 \
0.325 & -82.8318 \
0.330 & -23.1689 \
0.335 & 30.5170
endarray
right)$$
So, the solution is between $0.330$ and $0.335$. Repeat using $Delta alpha=0.0005$ to get
$$left(
beginarraycc
alpha & (2) \
0.3305 & -17.537 \
0.3310 & -11.9644 \
0.3315 & -6.45098 \
0.3320 & -0.996267 \
0.3325 & 4.40004
endarray
right)$$ You will finish with $alpha=0.332092$ to which correspond $b=3.54869$.
Update
If the model is instead
$$a=b_1,n^alpha+b_2,n^1-alpha+b_3$$ consider that $alpha$ is fixed at a given value. For this value, define two parameters $x_i=n_i^ alpha$, $y_i=n_i^1-alpha$ and so, you face, for a given $alpha$ a linear regression
$$a=b_1,x+b_2,y+b_3$$ which is easy to solve using matrix calculation or simpler the normal equations
$$sum_i=1^p a_i=b_1 sum_i=1^p x_i+b_2 sum_i=1^p y_i+b_3,p$$
$$sum_i=1^p a_ix_i=b_1 sum_i=1^p x_i^2+b_2 sum_i=1^p x_iy_i+b_3sum_i=1^p x_i$$
$$sum_i=1^p a_iy_i=b_1 sum_i=1^p x_iy_i+b_2 sum_i=1^p y_i^2+b_3sum_i=1^p y_i$$ or just multilinear regression.
The resulting parameters are $b_1(alpha), b_2(alpha), b_3(alpha)$ and now consider
$$SSQ(alpha)=sum_i=1^p left(b_1(alpha),n_i^alpha+b_2(alpha),n_i^1-alpha+b_3(alpha) -a_iright)^2$$ which needs to be minimized with respect to $alpha$. As before, plot to locate more or less the minimum and zoom more and more until you reach the desired accuracy.
For example, considering the data set
$$left(
beginarrayccc
i & n_i & a_n_i \
1 & 10 & 254 \
2 & 20 & 357 \
3 & 30 & 442 \
4 & 40 & 516 \
5 & 50 & 584 \
6 & 60 & 646 \
7 & 70 & 705 \
8 & 80 & 761 \
9 & 90 & 814 \
10 & 100 & 865 \
11 & 110 & 914 \
12 & 120 & 961 \
13 & 130 & 1007 \
14 & 140 & 1052 \
15 & 150 & 1096
endarray
right)$$ we should have
$$left(
beginarraycc
alpha & SSQ(alpha) \
0.0 & 13251.5 \
0.1 & 126.579 \
0.2 & 33.3071 \
0.3 & 3.82528 \
0.4 & 1.38533 \
0.5 & 1745.69
endarray
right)$$ Continue zooming in the area of the minimum and finixh with
$$a=10.3547 ,n^0.363391+40.1755, n^0.636609+55.9928$$ which will give as final results
$$left(
beginarraycccc
i & n_i & a_n_i & textpredicted \
1 & 10 & 254 & 253.908 \
2 & 20 & 357 & 357.274 \
3 & 30 & 442 & 441.825 \
4 & 40 & 516 & 516.136 \
5 & 50 & 584 & 583.675 \
6 & 60 & 646 & 646.276 \
7 & 70 & 705 & 705.055 \
8 & 80 & 761 & 760.751 \
9 & 90 & 814 & 813.890 \
10 & 100 & 865 & 864.856 \
11 & 110 & 914 & 913.946 \
12 & 120 & 961 & 961.392 \
13 & 130 & 1007 & 1007.38 \
14 & 140 & 1052 & 1052.06 \
15 & 150 & 1096 & 1095.57
endarray
right)$$
answered Jul 23 at 7:53
Claude Leibovici
111k1055126
This answer gives me the basic idea of how to do the estimation here, especially the zoom operation. Thanks a lot.
– Jimmy Kang
Jul 23 at 19:56
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Rather than a two-parameter least-squares, you might note that your formula implies
$$a_i+1-a_i = n_i+1^alpha - n_i^alpha + n_i+1^1-alpha - n_i^1-alpha$$
Find the $alpha$ that minimizes the sum of squared errors here, then for $b$ take the average of $a_i - (n_i^alpha + n_i^1-alpha)$.
answered Jul 23 at 8:30
Robert Israel
304k22201441
For sure ! Much simpler than my junk !
– Claude Leibovici
Jul 23 at 8:38
If I am right, $n_i=i$, there is no need to index.
– Yves Daoust
Jul 23 at 8:40
@YvesDaoust. Not necessary since you can use, say, $n=10,20,30,cdots$.
– Claude Leibovici
Jul 23 at 8:53
@ClaudeLeibovici: right, but then the formula must be with $a_n_i$.
– Yves Daoust
Jul 23 at 8:56
Many thanks. I have tried your answer. I thought the data can be approximately modeled as $a(n) = n^alpha+ n^1-alpha+b$. Actually, I have $5$ sequences, each should be modeled with $alpha = 1/1.9,1/2.0, 1/2.1, 1/2.2, 1/2.5$ respectively. However, the strange thing is the partial derivative of the squared errors achieves $0$ at $alpha = 0.5$ for all sequences. I'm still trying to figure out what happens in the data.
– Jimmy Kang
Jul 23 at 20:20
add a comment |Â
up vote
2
down vote
Rather than a two-parameter least-squares, you might note that your formula implies
$$a_i+1-a_i = n_i+1^alpha - n_i^alpha + n_i+1^1-alpha - n_i^1-alpha$$
Find the $alpha$ that minimizes the sum of squared errors here, then for $b$ take the average of $a_i - (n_i^alpha + n_i^1-alpha)$.
answered Jul 23 at 8:30
Robert Israel
304k22201441
For sure ! Much simpler than my junk !
– Claude Leibovici
Jul 23 at 8:38
If I am right, $n_i=i$, there is no need to index.
– Yves Daoust
Jul 23 at 8:40
@YvesDaoust. Not necessary since you can use, say, $n=10,20,30,cdots$.
– Claude Leibovici
Jul 23 at 8:53
@ClaudeLeibovici: right, but then the formula must be with $a_n_i$.
– Yves Daoust
Jul 23 at 8:56
Many thanks. I have tried your answer. I thought the data can be approximately modeled as $a(n) = n^alpha+ n^1-alpha+b$. Actually, I have $5$ sequences, each should be modeled with $alpha = 1/1.9,1/2.0, 1/2.1, 1/2.2, 1/2.5$ respectively. However, the strange thing is the partial derivative of the squared errors achieves $0$ at $alpha = 0.5$ for all sequences. I'm still trying to figure out what happens in the data.
– Jimmy Kang
Jul 23 at 20:20
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Rather than a two-parameter least-squares, you might note that your formula implies
$$a_i+1-a_i = n_i+1^alpha - n_i^alpha + n_i+1^1-alpha - n_i^1-alpha$$
Find the $alpha$ that minimizes the sum of squared errors here, then for $b$ take the average of $a_i - (n_i^alpha + n_i^1-alpha)$.
answered Jul 23 at 8:30
Robert Israel
304k22201441
Rather than a two-parameter least-squares, you might note that your formula implies
$$a_i+1-a_i = n_i+1^alpha - n_i^alpha + n_i+1^1-alpha - n_i^1-alpha$$
Find the $alpha$ that minimizes the sum of squared errors here, then for $b$ take the average of $a_i - (n_i^alpha + n_i^1-alpha)$.
answered Jul 23 at 8:30
Robert Israel
304k22201441
answered Jul 23 at 8:30
Robert Israel
304k22201441
answered Jul 23 at 8:30
Robert Israel
304k22201441
answered Jul 23 at 8:30
Robert Israel
304k22201441
304k22201441
For sure ! Much simpler than my junk !
– Claude Leibovici
Jul 23 at 8:38
If I am right, $n_i=i$, there is no need to index.
– Yves Daoust
Jul 23 at 8:40
@YvesDaoust. Not necessary since you can use, say, $n=10,20,30,cdots$.
– Claude Leibovici
Jul 23 at 8:53
@ClaudeLeibovici: right, but then the formula must be with $a_n_i$.
– Yves Daoust
Jul 23 at 8:56
Many thanks. I have tried your answer. I thought the data can be approximately modeled as $a(n) = n^alpha+ n^1-alpha+b$. Actually, I have $5$ sequences, each should be modeled with $alpha = 1/1.9,1/2.0, 1/2.1, 1/2.2, 1/2.5$ respectively. However, the strange thing is the partial derivative of the squared errors achieves $0$ at $alpha = 0.5$ for all sequences. I'm still trying to figure out what happens in the data.
– Jimmy Kang
Jul 23 at 20:20
add a comment |Â
For sure ! Much simpler than my junk !
– Claude Leibovici
Jul 23 at 8:38
If I am right, $n_i=i$, there is no need to index.
– Yves Daoust
Jul 23 at 8:40
@YvesDaoust. Not necessary since you can use, say, $n=10,20,30,cdots$.
– Claude Leibovici
Jul 23 at 8:53
@ClaudeLeibovici: right, but then the formula must be with $a_n_i$.
– Yves Daoust
Jul 23 at 8:56
Many thanks. I have tried your answer. I thought the data can be approximately modeled as $a(n) = n^alpha+ n^1-alpha+b$. Actually, I have $5$ sequences, each should be modeled with $alpha = 1/1.9,1/2.0, 1/2.1, 1/2.2, 1/2.5$ respectively. However, the strange thing is the partial derivative of the squared errors achieves $0$ at $alpha = 0.5$ for all sequences. I'm still trying to figure out what happens in the data.
– Jimmy Kang
Jul 23 at 20:20
For sure ! Much simpler than my junk !
– Claude Leibovici
Jul 23 at 8:38
For sure ! Much simpler than my junk !
– Claude Leibovici
Jul 23 at 8:38
If I am right, $n_i=i$, there is no need to index.
– Yves Daoust
Jul 23 at 8:40
If I am right, $n_i=i$, there is no need to index.
– Yves Daoust
Jul 23 at 8:40
@YvesDaoust. Not necessary since you can use, say, $n=10,20,30,cdots$.
– Claude Leibovici
Jul 23 at 8:53
@YvesDaoust. Not necessary since you can use, say, $n=10,20,30,cdots$.
– Claude Leibovici
Jul 23 at 8:53
@ClaudeLeibovici: right, but then the formula must be with $a_n_i$.
– Yves Daoust
Jul 23 at 8:56
@ClaudeLeibovici: right, but then the formula must be with $a_n_i$.
– Yves Daoust
Jul 23 at 8:56
Many thanks. I have tried your answer. I thought the data can be approximately modeled as $a(n) = n^alpha+ n^1-alpha+b$. Actually, I have $5$ sequences, each should be modeled with $alpha = 1/1.9,1/2.0, 1/2.1, 1/2.2, 1/2.5$ respectively. However, the strange thing is the partial derivative of the squared errors achieves $0$ at $alpha = 0.5$ for all sequences. I'm still trying to figure out what happens in the data.
– Jimmy Kang
Jul 23 at 20:20
Many thanks. I have tried your answer. I thought the data can be approximately modeled as $a(n) = n^alpha+ n^1-alpha+b$. Actually, I have $5$ sequences, each should be modeled with $alpha = 1/1.9,1/2.0, 1/2.1, 1/2.2, 1/2.5$ respectively. However, the strange thing is the partial derivative of the squared errors achieves $0$ at $alpha = 0.5$ for all sequences. I'm still trying to figure out what happens in the data.
– Jimmy Kang
Jul 23 at 20:20
add a comment |Â
up vote
2
down vote
If this helps:
If $a_0$ or $a_1-2$ are available and give reasonable estimates of $b$, for large $n$ we have
$$fraca_n-hat bsqrt nsim n^alpha-1/2+n^1/2-alpha=2coshleft(left(frac12-alpharight)log nright)$$
and
$$alphasimfrac12-dfractextarcoshdfraca_n-hat b2sqrt nlog n.$$
This can be used f.i. as a starting value of an iterative method.
answered Jul 23 at 9:00
Yves Daoust
111k665203
This seems to be a quite good idea, for sure !
– Claude Leibovici
Jul 23 at 9:17
Sorry for the late reply. Data is noisy. The initial terms do not match the model. My concern is just to estimate the asymptotic behavior. The data should be $O(n^max(alpha,1-alpha))$. I want to get the order of the sequence, i.e. $max(alpha,1-alpha)$.
– Jimmy Kang
Jul 23 at 20:01
@JimmyKang: why do you add a constant term ?
– Yves Daoust
Jul 23 at 20:41
I revised the problem just now. Sorry for the confusion caused.
– Jimmy Kang
Jul 23 at 21:21
add a comment |Â
up vote
2
down vote
If this helps:
If $a_0$ or $a_1-2$ are available and give reasonable estimates of $b$, for large $n$ we have
$$fraca_n-hat bsqrt nsim n^alpha-1/2+n^1/2-alpha=2coshleft(left(frac12-alpharight)log nright)$$
and
$$alphasimfrac12-dfractextarcoshdfraca_n-hat b2sqrt nlog n.$$
This can be used f.i. as a starting value of an iterative method.
answered Jul 23 at 9:00
Yves Daoust
111k665203
This seems to be a quite good idea, for sure !
– Claude Leibovici
Jul 23 at 9:17
Sorry for the late reply. Data is noisy. The initial terms do not match the model. My concern is just to estimate the asymptotic behavior. The data should be $O(n^max(alpha,1-alpha))$. I want to get the order of the sequence, i.e. $max(alpha,1-alpha)$.
– Jimmy Kang
Jul 23 at 20:01
@JimmyKang: why do you add a constant term ?
– Yves Daoust
Jul 23 at 20:41
I revised the problem just now. Sorry for the confusion caused.
– Jimmy Kang
Jul 23 at 21:21
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If this helps:
If $a_0$ or $a_1-2$ are available and give reasonable estimates of $b$, for large $n$ we have
$$fraca_n-hat bsqrt nsim n^alpha-1/2+n^1/2-alpha=2coshleft(left(frac12-alpharight)log nright)$$
and
$$alphasimfrac12-dfractextarcoshdfraca_n-hat b2sqrt nlog n.$$
This can be used f.i. as a starting value of an iterative method.
answered Jul 23 at 9:00
Yves Daoust
111k665203
If this helps:
If $a_0$ or $a_1-2$ are available and give reasonable estimates of $b$, for large $n$ we have
$$fraca_n-hat bsqrt nsim n^alpha-1/2+n^1/2-alpha=2coshleft(left(frac12-alpharight)log nright)$$
and
$$alphasimfrac12-dfractextarcoshdfraca_n-hat b2sqrt nlog n.$$
This can be used f.i. as a starting value of an iterative method.
answered Jul 23 at 9:00
Yves Daoust
111k665203
edited Jul 23 at 9:08
answered Jul 23 at 9:00
Yves Daoust
111k665203
answered Jul 23 at 9:00
Yves Daoust
111k665203
answered Jul 23 at 9:00
Yves Daoust
111k665203
111k665203
This seems to be a quite good idea, for sure !
– Claude Leibovici
Jul 23 at 9:17
Sorry for the late reply. Data is noisy. The initial terms do not match the model. My concern is just to estimate the asymptotic behavior. The data should be $O(n^max(alpha,1-alpha))$. I want to get the order of the sequence, i.e. $max(alpha,1-alpha)$.
– Jimmy Kang
Jul 23 at 20:01
@JimmyKang: why do you add a constant term ?
– Yves Daoust
Jul 23 at 20:41
I revised the problem just now. Sorry for the confusion caused.
– Jimmy Kang
Jul 23 at 21:21
add a comment |Â
This seems to be a quite good idea, for sure !
– Claude Leibovici
Jul 23 at 9:17
Sorry for the late reply. Data is noisy. The initial terms do not match the model. My concern is just to estimate the asymptotic behavior. The data should be $O(n^max(alpha,1-alpha))$. I want to get the order of the sequence, i.e. $max(alpha,1-alpha)$.
– Jimmy Kang
Jul 23 at 20:01
@JimmyKang: why do you add a constant term ?
– Yves Daoust
Jul 23 at 20:41
I revised the problem just now. Sorry for the confusion caused.
– Jimmy Kang
Jul 23 at 21:21
This seems to be a quite good idea, for sure !
– Claude Leibovici
Jul 23 at 9:17
This seems to be a quite good idea, for sure !
– Claude Leibovici
Jul 23 at 9:17
Sorry for the late reply. Data is noisy. The initial terms do not match the model. My concern is just to estimate the asymptotic behavior. The data should be $O(n^max(alpha,1-alpha))$. I want to get the order of the sequence, i.e. $max(alpha,1-alpha)$.
– Jimmy Kang
Jul 23 at 20:01
Sorry for the late reply. Data is noisy. The initial terms do not match the model. My concern is just to estimate the asymptotic behavior. The data should be $O(n^max(alpha,1-alpha))$. I want to get the order of the sequence, i.e. $max(alpha,1-alpha)$.
– Jimmy Kang
Jul 23 at 20:01
@JimmyKang: why do you add a constant term ?
– Yves Daoust
Jul 23 at 20:41
@JimmyKang: why do you add a constant term ?
– Yves Daoust
Jul 23 at 20:41
I revised the problem just now. Sorry for the confusion caused.
– Jimmy Kang
Jul 23 at 21:21
I revised the problem just now. Sorry for the confusion caused.
– Jimmy Kang
Jul 23 at 21:21
add a comment |Â
up vote
1
down vote
If we look at the problem as a regression, you have $p$ data points $(n_i,a_i)$ and you want to adjust the model
$$a=n^alpha+n^1-alpha+b$$ In the least square sense, you need to minimize
$$SSQ=sum_i=1^p left(n_i^alpha+n_i^1-alpha+b -a_iright)^2$$ Computing the partial derivatives,we have
$$fracpartial SSQpartial b=2sum_i=1^p left(n_i^alpha+n_i^1-alpha+b -a_iright)=0tag 1$$
$$fracpartial SSQpartial a=2sum_i=1^p left(n_i^alpha+n_i^1-alpha+b -a_iright)left(n_i^alpha-n_i^1-alpharight)log(n_i)=0 tag 2$$ which is hard to solve.
However, from $(1)$, you can get $b$ as a function of $alpha$
$$b(alpha)=-frac 1p sum_i=1^p left(n_i^alpha+n_i^1-alpha -a_iright)$$ and then $(2)$ is "just" an equation in $alpha$.
The simplest would be to plot equation $(2)$ for $0 leq alpha leq 0.5$ and see where it does cancels. Zoom more and more to have more accurate results; when your accuracy has been reached, recompute $b$.
All of that can be done using Excel.
Edit
For illustration purposes, let us use the following data set
$$left(
beginarrayccc
i & n_i & a_n_i \
1 & 10 & 10 \
2 & 20 & 14 \
3 & 30 & 16 \
4 & 40 & 19 \
5 & 50 & 21 \
6 & 60 & 23 \
7 & 70 & 25 \
8 & 80 & 26 \
9 & 90 & 28 \
10 & 100 & 30
endarray
right)$$ A first run, using $Delta alpha=0.05$ would give
$$left(
beginarraycc
alpha & (2) \
0.05 & -36241.2 \
0.10 & -19913.1 \
0.15 & -10372.8 \
0.20 & -4943.28 \
0.25 & -1981.30 \
0.30 & -483.658 \
0.35 & 158.953
endarray
right)$$ So, the solution is between $0.30$ and $0.35$. Repeat using $Delta alpha=0.005$ to get
$$left(
beginarraycc
alpha & (2) \
0.315 & -221.502 \
0.320 & -148.821 \
0.325 & -82.8318 \
0.330 & -23.1689 \
0.335 & 30.5170
endarray
right)$$
So, the solution is between $0.330$ and $0.335$. Repeat using $Delta alpha=0.0005$ to get
$$left(
beginarraycc
alpha & (2) \
0.3305 & -17.537 \
0.3310 & -11.9644 \
0.3315 & -6.45098 \
0.3320 & -0.996267 \
0.3325 & 4.40004
endarray
right)$$ You will finish with $alpha=0.332092$ to which correspond $b=3.54869$.
Update
If the model is instead
$$a=b_1,n^alpha+b_2,n^1-alpha+b_3$$ consider that $alpha$ is fixed at a given value. For this value, define two parameters $x_i=n_i^ alpha$, $y_i=n_i^1-alpha$ and so, you face, for a given $alpha$ a linear regression
$$a=b_1,x+b_2,y+b_3$$ which is easy to solve using matrix calculation or simpler the normal equations
$$sum_i=1^p a_i=b_1 sum_i=1^p x_i+b_2 sum_i=1^p y_i+b_3,p$$
$$sum_i=1^p a_ix_i=b_1 sum_i=1^p x_i^2+b_2 sum_i=1^p x_iy_i+b_3sum_i=1^p x_i$$
$$sum_i=1^p a_iy_i=b_1 sum_i=1^p x_iy_i+b_2 sum_i=1^p y_i^2+b_3sum_i=1^p y_i$$ or just multilinear regression.
The resulting parameters are $b_1(alpha), b_2(alpha), b_3(alpha)$ and now consider
$$SSQ(alpha)=sum_i=1^p left(b_1(alpha),n_i^alpha+b_2(alpha),n_i^1-alpha+b_3(alpha) -a_iright)^2$$ which needs to be minimized with respect to $alpha$. As before, plot to locate more or less the minimum and zoom more and more until you reach the desired accuracy.
For example, considering the data set
$$left(
beginarrayccc
i & n_i & a_n_i \
1 & 10 & 254 \
2 & 20 & 357 \
3 & 30 & 442 \
4 & 40 & 516 \
5 & 50 & 584 \
6 & 60 & 646 \
7 & 70 & 705 \
8 & 80 & 761 \
9 & 90 & 814 \
10 & 100 & 865 \
11 & 110 & 914 \
12 & 120 & 961 \
13 & 130 & 1007 \
14 & 140 & 1052 \
15 & 150 & 1096
endarray
right)$$ we should have
$$left(
beginarraycc
alpha & SSQ(alpha) \
0.0 & 13251.5 \
0.1 & 126.579 \
0.2 & 33.3071 \
0.3 & 3.82528 \
0.4 & 1.38533 \
0.5 & 1745.69
endarray
right)$$ Continue zooming in the area of the minimum and finixh with
$$a=10.3547 ,n^0.363391+40.1755, n^0.636609+55.9928$$ which will give as final results
$$left(
beginarraycccc
i & n_i & a_n_i & textpredicted \
1 & 10 & 254 & 253.908 \
2 & 20 & 357 & 357.274 \
3 & 30 & 442 & 441.825 \
4 & 40 & 516 & 516.136 \
5 & 50 & 584 & 583.675 \
6 & 60 & 646 & 646.276 \
7 & 70 & 705 & 705.055 \
8 & 80 & 761 & 760.751 \
9 & 90 & 814 & 813.890 \
10 & 100 & 865 & 864.856 \
11 & 110 & 914 & 913.946 \
12 & 120 & 961 & 961.392 \
13 & 130 & 1007 & 1007.38 \
14 & 140 & 1052 & 1052.06 \
15 & 150 & 1096 & 1095.57
endarray
right)$$
answered Jul 23 at 7:53
Claude Leibovici
111k1055126
This answer gives me the basic idea of how to do the estimation here, especially the zoom operation. Thanks a lot.
– Jimmy Kang
Jul 23 at 19:56
add a comment |Â
up vote
1
down vote
If we look at the problem as a regression, you have $p$ data points $(n_i,a_i)$ and you want to adjust the model
$$a=n^alpha+n^1-alpha+b$$ In the least square sense, you need to minimize
$$SSQ=sum_i=1^p left(n_i^alpha+n_i^1-alpha+b -a_iright)^2$$ Computing the partial derivatives,we have
$$fracpartial SSQpartial b=2sum_i=1^p left(n_i^alpha+n_i^1-alpha+b -a_iright)=0tag 1$$
$$fracpartial SSQpartial a=2sum_i=1^p left(n_i^alpha+n_i^1-alpha+b -a_iright)left(n_i^alpha-n_i^1-alpharight)log(n_i)=0 tag 2$$ which is hard to solve.
However, from $(1)$, you can get $b$ as a function of $alpha$
$$b(alpha)=-frac 1p sum_i=1^p left(n_i^alpha+n_i^1-alpha -a_iright)$$ and then $(2)$ is "just" an equation in $alpha$.
The simplest would be to plot equation $(2)$ for $0 leq alpha leq 0.5$ and see where it does cancels. Zoom more and more to have more accurate results; when your accuracy has been reached, recompute $b$.
All of that can be done using Excel.
Edit
For illustration purposes, let us use the following data set
$$left(
beginarrayccc
i & n_i & a_n_i \
1 & 10 & 10 \
2 & 20 & 14 \
3 & 30 & 16 \
4 & 40 & 19 \
5 & 50 & 21 \
6 & 60 & 23 \
7 & 70 & 25 \
8 & 80 & 26 \
9 & 90 & 28 \
10 & 100 & 30
endarray
right)$$ A first run, using $Delta alpha=0.05$ would give
$$left(
beginarraycc
alpha & (2) \
0.05 & -36241.2 \
0.10 & -19913.1 \
0.15 & -10372.8 \
0.20 & -4943.28 \
0.25 & -1981.30 \
0.30 & -483.658 \
0.35 & 158.953
endarray
right)$$ So, the solution is between $0.30$ and $0.35$. Repeat using $Delta alpha=0.005$ to get
$$left(
beginarraycc
alpha & (2) \
0.315 & -221.502 \
0.320 & -148.821 \
0.325 & -82.8318 \
0.330 & -23.1689 \
0.335 & 30.5170
endarray
right)$$
So, the solution is between $0.330$ and $0.335$. Repeat using $Delta alpha=0.0005$ to get
$$left(
beginarraycc
alpha & (2) \
0.3305 & -17.537 \
0.3310 & -11.9644 \
0.3315 & -6.45098 \
0.3320 & -0.996267 \
0.3325 & 4.40004
endarray
right)$$ You will finish with $alpha=0.332092$ to which correspond $b=3.54869$.
Update
If the model is instead
$$a=b_1,n^alpha+b_2,n^1-alpha+b_3$$ consider that $alpha$ is fixed at a given value. For this value, define two parameters $x_i=n_i^ alpha$, $y_i=n_i^1-alpha$ and so, you face, for a given $alpha$ a linear regression
$$a=b_1,x+b_2,y+b_3$$ which is easy to solve using matrix calculation or simpler the normal equations
$$sum_i=1^p a_i=b_1 sum_i=1^p x_i+b_2 sum_i=1^p y_i+b_3,p$$
$$sum_i=1^p a_ix_i=b_1 sum_i=1^p x_i^2+b_2 sum_i=1^p x_iy_i+b_3sum_i=1^p x_i$$
$$sum_i=1^p a_iy_i=b_1 sum_i=1^p x_iy_i+b_2 sum_i=1^p y_i^2+b_3sum_i=1^p y_i$$ or just multilinear regression.
The resulting parameters are $b_1(alpha), b_2(alpha), b_3(alpha)$ and now consider
$$SSQ(alpha)=sum_i=1^p left(b_1(alpha),n_i^alpha+b_2(alpha),n_i^1-alpha+b_3(alpha) -a_iright)^2$$ which needs to be minimized with respect to $alpha$. As before, plot to locate more or less the minimum and zoom more and more until you reach the desired accuracy.
For example, considering the data set
$$left(
beginarrayccc
i & n_i & a_n_i \
1 & 10 & 254 \
2 & 20 & 357 \
3 & 30 & 442 \
4 & 40 & 516 \
5 & 50 & 584 \
6 & 60 & 646 \
7 & 70 & 705 \
8 & 80 & 761 \
9 & 90 & 814 \
10 & 100 & 865 \
11 & 110 & 914 \
12 & 120 & 961 \
13 & 130 & 1007 \
14 & 140 & 1052 \
15 & 150 & 1096
endarray
right)$$ we should have
$$left(
beginarraycc
alpha & SSQ(alpha) \
0.0 & 13251.5 \
0.1 & 126.579 \
0.2 & 33.3071 \
0.3 & 3.82528 \
0.4 & 1.38533 \
0.5 & 1745.69
endarray
right)$$ Continue zooming in the area of the minimum and finixh with
$$a=10.3547 ,n^0.363391+40.1755, n^0.636609+55.9928$$ which will give as final results
$$left(
beginarraycccc
i & n_i & a_n_i & textpredicted \
1 & 10 & 254 & 253.908 \
2 & 20 & 357 & 357.274 \
3 & 30 & 442 & 441.825 \
4 & 40 & 516 & 516.136 \
5 & 50 & 584 & 583.675 \
6 & 60 & 646 & 646.276 \
7 & 70 & 705 & 705.055 \
8 & 80 & 761 & 760.751 \
9 & 90 & 814 & 813.890 \
10 & 100 & 865 & 864.856 \
11 & 110 & 914 & 913.946 \
12 & 120 & 961 & 961.392 \
13 & 130 & 1007 & 1007.38 \
14 & 140 & 1052 & 1052.06 \
15 & 150 & 1096 & 1095.57
endarray
right)$$
answered Jul 23 at 7:53
Claude Leibovici
111k1055126
This answer gives me the basic idea of how to do the estimation here, especially the zoom operation. Thanks a lot.
– Jimmy Kang
Jul 23 at 19:56
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If we look at the problem as a regression, you have $p$ data points $(n_i,a_i)$ and you want to adjust the model
$$a=n^alpha+n^1-alpha+b$$ In the least square sense, you need to minimize
$$SSQ=sum_i=1^p left(n_i^alpha+n_i^1-alpha+b -a_iright)^2$$ Computing the partial derivatives,we have
$$fracpartial SSQpartial b=2sum_i=1^p left(n_i^alpha+n_i^1-alpha+b -a_iright)=0tag 1$$
$$fracpartial SSQpartial a=2sum_i=1^p left(n_i^alpha+n_i^1-alpha+b -a_iright)left(n_i^alpha-n_i^1-alpharight)log(n_i)=0 tag 2$$ which is hard to solve.
However, from $(1)$, you can get $b$ as a function of $alpha$
$$b(alpha)=-frac 1p sum_i=1^p left(n_i^alpha+n_i^1-alpha -a_iright)$$ and then $(2)$ is "just" an equation in $alpha$.
The simplest would be to plot equation $(2)$ for $0 leq alpha leq 0.5$ and see where it does cancels. Zoom more and more to have more accurate results; when your accuracy has been reached, recompute $b$.
All of that can be done using Excel.
Edit
For illustration purposes, let us use the following data set
$$left(
beginarrayccc
i & n_i & a_n_i \
1 & 10 & 10 \
2 & 20 & 14 \
3 & 30 & 16 \
4 & 40 & 19 \
5 & 50 & 21 \
6 & 60 & 23 \
7 & 70 & 25 \
8 & 80 & 26 \
9 & 90 & 28 \
10 & 100 & 30
endarray
right)$$ A first run, using $Delta alpha=0.05$ would give
$$left(
beginarraycc
alpha & (2) \
0.05 & -36241.2 \
0.10 & -19913.1 \
0.15 & -10372.8 \
0.20 & -4943.28 \
0.25 & -1981.30 \
0.30 & -483.658 \
0.35 & 158.953
endarray
right)$$ So, the solution is between $0.30$ and $0.35$. Repeat using $Delta alpha=0.005$ to get
$$left(
beginarraycc
alpha & (2) \
0.315 & -221.502 \
0.320 & -148.821 \
0.325 & -82.8318 \
0.330 & -23.1689 \
0.335 & 30.5170
endarray
right)$$
So, the solution is between $0.330$ and $0.335$. Repeat using $Delta alpha=0.0005$ to get
$$left(
beginarraycc
alpha & (2) \
0.3305 & -17.537 \
0.3310 & -11.9644 \
0.3315 & -6.45098 \
0.3320 & -0.996267 \
0.3325 & 4.40004
endarray
right)$$ You will finish with $alpha=0.332092$ to which correspond $b=3.54869$.
Update
If the model is instead
$$a=b_1,n^alpha+b_2,n^1-alpha+b_3$$ consider that $alpha$ is fixed at a given value. For this value, define two parameters $x_i=n_i^ alpha$, $y_i=n_i^1-alpha$ and so, you face, for a given $alpha$ a linear regression
$$a=b_1,x+b_2,y+b_3$$ which is easy to solve using matrix calculation or simpler the normal equations
$$sum_i=1^p a_i=b_1 sum_i=1^p x_i+b_2 sum_i=1^p y_i+b_3,p$$
$$sum_i=1^p a_ix_i=b_1 sum_i=1^p x_i^2+b_2 sum_i=1^p x_iy_i+b_3sum_i=1^p x_i$$
$$sum_i=1^p a_iy_i=b_1 sum_i=1^p x_iy_i+b_2 sum_i=1^p y_i^2+b_3sum_i=1^p y_i$$ or just multilinear regression.
The resulting parameters are $b_1(alpha), b_2(alpha), b_3(alpha)$ and now consider
$$SSQ(alpha)=sum_i=1^p left(b_1(alpha),n_i^alpha+b_2(alpha),n_i^1-alpha+b_3(alpha) -a_iright)^2$$ which needs to be minimized with respect to $alpha$. As before, plot to locate more or less the minimum and zoom more and more until you reach the desired accuracy.
For example, considering the data set
$$left(
beginarrayccc
i & n_i & a_n_i \
1 & 10 & 254 \
2 & 20 & 357 \
3 & 30 & 442 \
4 & 40 & 516 \
5 & 50 & 584 \
6 & 60 & 646 \
7 & 70 & 705 \
8 & 80 & 761 \
9 & 90 & 814 \
10 & 100 & 865 \
11 & 110 & 914 \
12 & 120 & 961 \
13 & 130 & 1007 \
14 & 140 & 1052 \
15 & 150 & 1096
endarray
right)$$ we should have
$$left(
beginarraycc
alpha & SSQ(alpha) \
0.0 & 13251.5 \
0.1 & 126.579 \
0.2 & 33.3071 \
0.3 & 3.82528 \
0.4 & 1.38533 \
0.5 & 1745.69
endarray
right)$$ Continue zooming in the area of the minimum and finixh with
$$a=10.3547 ,n^0.363391+40.1755, n^0.636609+55.9928$$ which will give as final results
$$left(
beginarraycccc
i & n_i & a_n_i & textpredicted \
1 & 10 & 254 & 253.908 \
2 & 20 & 357 & 357.274 \
3 & 30 & 442 & 441.825 \
4 & 40 & 516 & 516.136 \
5 & 50 & 584 & 583.675 \
6 & 60 & 646 & 646.276 \
7 & 70 & 705 & 705.055 \
8 & 80 & 761 & 760.751 \
9 & 90 & 814 & 813.890 \
10 & 100 & 865 & 864.856 \
11 & 110 & 914 & 913.946 \
12 & 120 & 961 & 961.392 \
13 & 130 & 1007 & 1007.38 \
14 & 140 & 1052 & 1052.06 \
15 & 150 & 1096 & 1095.57
endarray
right)$$
answered Jul 23 at 7:53
Claude Leibovici
111k1055126
If we look at the problem as a regression, you have $p$ data points $(n_i,a_i)$ and you want to adjust the model
$$a=n^alpha+n^1-alpha+b$$ In the least square sense, you need to minimize
$$SSQ=sum_i=1^p left(n_i^alpha+n_i^1-alpha+b -a_iright)^2$$ Computing the partial derivatives,we have
$$fracpartial SSQpartial b=2sum_i=1^p left(n_i^alpha+n_i^1-alpha+b -a_iright)=0tag 1$$
$$fracpartial SSQpartial a=2sum_i=1^p left(n_i^alpha+n_i^1-alpha+b -a_iright)left(n_i^alpha-n_i^1-alpharight)log(n_i)=0 tag 2$$ which is hard to solve.
However, from $(1)$, you can get $b$ as a function of $alpha$
$$b(alpha)=-frac 1p sum_i=1^p left(n_i^alpha+n_i^1-alpha -a_iright)$$ and then $(2)$ is "just" an equation in $alpha$.
The simplest would be to plot equation $(2)$ for $0 leq alpha leq 0.5$ and see where it does cancels. Zoom more and more to have more accurate results; when your accuracy has been reached, recompute $b$.
All of that can be done using Excel.
Edit
For illustration purposes, let us use the following data set
$$left(
beginarrayccc
i & n_i & a_n_i \
1 & 10 & 10 \
2 & 20 & 14 \
3 & 30 & 16 \
4 & 40 & 19 \
5 & 50 & 21 \
6 & 60 & 23 \
7 & 70 & 25 \
8 & 80 & 26 \
9 & 90 & 28 \
10 & 100 & 30
endarray
right)$$ A first run, using $Delta alpha=0.05$ would give
$$left(
beginarraycc
alpha & (2) \
0.05 & -36241.2 \
0.10 & -19913.1 \
0.15 & -10372.8 \
0.20 & -4943.28 \
0.25 & -1981.30 \
0.30 & -483.658 \
0.35 & 158.953
endarray
right)$$ So, the solution is between $0.30$ and $0.35$. Repeat using $Delta alpha=0.005$ to get
$$left(
beginarraycc
alpha & (2) \
0.315 & -221.502 \
0.320 & -148.821 \
0.325 & -82.8318 \
0.330 & -23.1689 \
0.335 & 30.5170
endarray
right)$$
So, the solution is between $0.330$ and $0.335$. Repeat using $Delta alpha=0.0005$ to get
$$left(
beginarraycc
alpha & (2) \
0.3305 & -17.537 \
0.3310 & -11.9644 \
0.3315 & -6.45098 \
0.3320 & -0.996267 \
0.3325 & 4.40004
endarray
right)$$ You will finish with $alpha=0.332092$ to which correspond $b=3.54869$.
Update
If the model is instead
$$a=b_1,n^alpha+b_2,n^1-alpha+b_3$$ consider that $alpha$ is fixed at a given value. For this value, define two parameters $x_i=n_i^ alpha$, $y_i=n_i^1-alpha$ and so, you face, for a given $alpha$ a linear regression
$$a=b_1,x+b_2,y+b_3$$ which is easy to solve using matrix calculation or simpler the normal equations
$$sum_i=1^p a_i=b_1 sum_i=1^p x_i+b_2 sum_i=1^p y_i+b_3,p$$
$$sum_i=1^p a_ix_i=b_1 sum_i=1^p x_i^2+b_2 sum_i=1^p x_iy_i+b_3sum_i=1^p x_i$$
$$sum_i=1^p a_iy_i=b_1 sum_i=1^p x_iy_i+b_2 sum_i=1^p y_i^2+b_3sum_i=1^p y_i$$ or just multilinear regression.
The resulting parameters are $b_1(alpha), b_2(alpha), b_3(alpha)$ and now consider
$$SSQ(alpha)=sum_i=1^p left(b_1(alpha),n_i^alpha+b_2(alpha),n_i^1-alpha+b_3(alpha) -a_iright)^2$$ which needs to be minimized with respect to $alpha$. As before, plot to locate more or less the minimum and zoom more and more until you reach the desired accuracy.
For example, considering the data set
$$left(
beginarrayccc
i & n_i & a_n_i \
1 & 10 & 254 \
2 & 20 & 357 \
3 & 30 & 442 \
4 & 40 & 516 \
5 & 50 & 584 \
6 & 60 & 646 \
7 & 70 & 705 \
8 & 80 & 761 \
9 & 90 & 814 \
10 & 100 & 865 \
11 & 110 & 914 \
12 & 120 & 961 \
13 & 130 & 1007 \
14 & 140 & 1052 \
15 & 150 & 1096
endarray
right)$$ we should have
$$left(
beginarraycc
alpha & SSQ(alpha) \
0.0 & 13251.5 \
0.1 & 126.579 \
0.2 & 33.3071 \
0.3 & 3.82528 \
0.4 & 1.38533 \
0.5 & 1745.69
endarray
right)$$ Continue zooming in the area of the minimum and finixh with
$$a=10.3547 ,n^0.363391+40.1755, n^0.636609+55.9928$$ which will give as final results
$$left(
beginarraycccc
i & n_i & a_n_i & textpredicted \
1 & 10 & 254 & 253.908 \
2 & 20 & 357 & 357.274 \
3 & 30 & 442 & 441.825 \
4 & 40 & 516 & 516.136 \
5 & 50 & 584 & 583.675 \
6 & 60 & 646 & 646.276 \
7 & 70 & 705 & 705.055 \
8 & 80 & 761 & 760.751 \
9 & 90 & 814 & 813.890 \
10 & 100 & 865 & 864.856 \
11 & 110 & 914 & 913.946 \
12 & 120 & 961 & 961.392 \
13 & 130 & 1007 & 1007.38 \
14 & 140 & 1052 & 1052.06 \
15 & 150 & 1096 & 1095.57
endarray
right)$$
answered Jul 23 at 7:53
Claude Leibovici
111k1055126
edited Jul 24 at 3:52
answered Jul 23 at 7:53
Claude Leibovici
111k1055126
answered Jul 23 at 7:53
Claude Leibovici
111k1055126
answered Jul 23 at 7:53
Claude Leibovici
111k1055126
111k1055126
This answer gives me the basic idea of how to do the estimation here, especially the zoom operation. Thanks a lot.
– Jimmy Kang
Jul 23 at 19:56
add a comment |Â
This answer gives me the basic idea of how to do the estimation here, especially the zoom operation. Thanks a lot.
– Jimmy Kang
Jul 23 at 19:56
This answer gives me the basic idea of how to do the estimation here, especially the zoom operation. Thanks a lot.
– Jimmy Kang
Jul 23 at 19:56
This answer gives me the basic idea of how to do the estimation here, especially the zoom operation. Thanks a lot.
– Jimmy Kang
Jul 23 at 19:56
add a comment |Â
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1
You can consider $0<alpha<frac12$.
– Yves Daoust
Jul 23 at 8:26
Is the data noisy or smooth ? Do the initial terms match the model ?
– Yves Daoust
Jul 23 at 9:06
What are you after, an estimate of the asymptotic behavior or just a best fit to the million points ?
– Yves Daoust
Jul 23 at 9:33