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Linear combination of power law distributions
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I am exploring whether a linear combination of power law distributions is also a power law distribution. Specifically, if $X sim (alpha -1) x_min^(alpha -1) x^-alpha$ and $Y sim (beta -1) y_min^(beta -1)y^-beta$, we are interested in finding the distribution of $Z = rho X + (1-rho) Y, rho in (0, 1)$? ($X$ and $Y$ are independent)
I tried with the moment generating function approach:
If $M_Z(t)$ is the moment generating function of $Z$, $$M_Z(t) = E[e^tZ]=E[e^t(rho X + (1-rho)Y]=E[e^trho X]cdot E[e^t(1-rho)Y] = M_X(trho) M_Y(t(1-rho)).$$
This brings us to the moment generating functions of the power law distributions themselves:
$$M_X(s) = int_x_min^infty e^sx(alpha -1)x_min^(alpha-1) x^-alpha , dx = (alpha - 1)x_min^(alpha-1)int_x_min^infty e^sxx^-alpha , dx. $$
Like it is mentioned here, this resembles the incomplete gamma distribution, but I cannot complete the derivation. And, more importantly, I am not sure if this will help me answer the original question, i.e., is the linear combination of two power law distributions also a power distribution?
probability probability-theory probability-distributions moment-generating-functions
edited Jul 20 at 4:13
Michael Hardy
204k23186462
asked Jul 20 at 4:10
buzaku
3091212
add a comment |Â
up vote
0
down vote
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I am exploring whether a linear combination of power law distributions is also a power law distribution. Specifically, if $X sim (alpha -1) x_min^(alpha -1) x^-alpha$ and $Y sim (beta -1) y_min^(beta -1)y^-beta$, we are interested in finding the distribution of $Z = rho X + (1-rho) Y, rho in (0, 1)$? ($X$ and $Y$ are independent)
I tried with the moment generating function approach:
If $M_Z(t)$ is the moment generating function of $Z$, $$M_Z(t) = E[e^tZ]=E[e^t(rho X + (1-rho)Y]=E[e^trho X]cdot E[e^t(1-rho)Y] = M_X(trho) M_Y(t(1-rho)).$$
This brings us to the moment generating functions of the power law distributions themselves:
$$M_X(s) = int_x_min^infty e^sx(alpha -1)x_min^(alpha-1) x^-alpha , dx = (alpha - 1)x_min^(alpha-1)int_x_min^infty e^sxx^-alpha , dx. $$
Like it is mentioned here, this resembles the incomplete gamma distribution, but I cannot complete the derivation. And, more importantly, I am not sure if this will help me answer the original question, i.e., is the linear combination of two power law distributions also a power distribution?
probability probability-theory probability-distributions moment-generating-functions
edited Jul 20 at 4:13
Michael Hardy
204k23186462
asked Jul 20 at 4:10
buzaku
3091212
1
It depends on what you mean by power law distribution. If you mean Pareto distribution then I would guess the answer would be no. If you mean particular behaviour in the tail (on a log-log scale roughly linear on the right) then I would guess the answer might be yes.
– Henry
Jul 21 at 10:39
Incidentally, power law distributions have heavy tails and some of their moments are infinite, as are their moment generating functions for all $t gt 0$
– Henry
Jul 21 at 10:47
@Henry In this case, I do mean the log-log linear behavior in the tail. And you are correct, the moments are defined only for $t<0$ above. However, I am not able to complete the proof using this or through a differentiation of the cumulative distribution function.
– buzaku
Jul 21 at 12:05
add a comment |Â
up vote
0
down vote
favorite
up vote
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down vote
favorite
I am exploring whether a linear combination of power law distributions is also a power law distribution. Specifically, if $X sim (alpha -1) x_min^(alpha -1) x^-alpha$ and $Y sim (beta -1) y_min^(beta -1)y^-beta$, we are interested in finding the distribution of $Z = rho X + (1-rho) Y, rho in (0, 1)$? ($X$ and $Y$ are independent)
I tried with the moment generating function approach:
If $M_Z(t)$ is the moment generating function of $Z$, $$M_Z(t) = E[e^tZ]=E[e^t(rho X + (1-rho)Y]=E[e^trho X]cdot E[e^t(1-rho)Y] = M_X(trho) M_Y(t(1-rho)).$$
This brings us to the moment generating functions of the power law distributions themselves:
$$M_X(s) = int_x_min^infty e^sx(alpha -1)x_min^(alpha-1) x^-alpha , dx = (alpha - 1)x_min^(alpha-1)int_x_min^infty e^sxx^-alpha , dx. $$
Like it is mentioned here, this resembles the incomplete gamma distribution, but I cannot complete the derivation. And, more importantly, I am not sure if this will help me answer the original question, i.e., is the linear combination of two power law distributions also a power distribution?
probability probability-theory probability-distributions moment-generating-functions
edited Jul 20 at 4:13
Michael Hardy
204k23186462
asked Jul 20 at 4:10
buzaku
3091212
I am exploring whether a linear combination of power law distributions is also a power law distribution. Specifically, if $X sim (alpha -1) x_min^(alpha -1) x^-alpha$ and $Y sim (beta -1) y_min^(beta -1)y^-beta$, we are interested in finding the distribution of $Z = rho X + (1-rho) Y, rho in (0, 1)$? ($X$ and $Y$ are independent)
I tried with the moment generating function approach:
If $M_Z(t)$ is the moment generating function of $Z$, $$M_Z(t) = E[e^tZ]=E[e^t(rho X + (1-rho)Y]=E[e^trho X]cdot E[e^t(1-rho)Y] = M_X(trho) M_Y(t(1-rho)).$$
This brings us to the moment generating functions of the power law distributions themselves:
$$M_X(s) = int_x_min^infty e^sx(alpha -1)x_min^(alpha-1) x^-alpha , dx = (alpha - 1)x_min^(alpha-1)int_x_min^infty e^sxx^-alpha , dx. $$
Like it is mentioned here, this resembles the incomplete gamma distribution, but I cannot complete the derivation. And, more importantly, I am not sure if this will help me answer the original question, i.e., is the linear combination of two power law distributions also a power distribution?
probability probability-theory probability-distributions moment-generating-functions
edited Jul 20 at 4:13
Michael Hardy
204k23186462
asked Jul 20 at 4:10
buzaku
3091212
edited Jul 20 at 4:13
Michael Hardy
204k23186462
edited Jul 20 at 4:13
Michael Hardy
204k23186462
edited Jul 20 at 4:13
Michael Hardy
204k23186462
204k23186462
asked Jul 20 at 4:10
buzaku
3091212
asked Jul 20 at 4:10
buzaku
3091212
asked Jul 20 at 4:10
buzaku
3091212
3091212
1
It depends on what you mean by power law distribution. If you mean Pareto distribution then I would guess the answer would be no. If you mean particular behaviour in the tail (on a log-log scale roughly linear on the right) then I would guess the answer might be yes.
– Henry
Jul 21 at 10:39
Incidentally, power law distributions have heavy tails and some of their moments are infinite, as are their moment generating functions for all $t gt 0$
– Henry
Jul 21 at 10:47
@Henry In this case, I do mean the log-log linear behavior in the tail. And you are correct, the moments are defined only for $t<0$ above. However, I am not able to complete the proof using this or through a differentiation of the cumulative distribution function.
– buzaku
Jul 21 at 12:05
add a comment |Â
1
It depends on what you mean by power law distribution. If you mean Pareto distribution then I would guess the answer would be no. If you mean particular behaviour in the tail (on a log-log scale roughly linear on the right) then I would guess the answer might be yes.
– Henry
Jul 21 at 10:39
Incidentally, power law distributions have heavy tails and some of their moments are infinite, as are their moment generating functions for all $t gt 0$
– Henry
Jul 21 at 10:47
@Henry In this case, I do mean the log-log linear behavior in the tail. And you are correct, the moments are defined only for $t<0$ above. However, I am not able to complete the proof using this or through a differentiation of the cumulative distribution function.
– buzaku
Jul 21 at 12:05
1
1
It depends on what you mean by power law distribution. If you mean Pareto distribution then I would guess the answer would be no. If you mean particular behaviour in the tail (on a log-log scale roughly linear on the right) then I would guess the answer might be yes.
– Henry
Jul 21 at 10:39
It depends on what you mean by power law distribution. If you mean Pareto distribution then I would guess the answer would be no. If you mean particular behaviour in the tail (on a log-log scale roughly linear on the right) then I would guess the answer might be yes.
– Henry
Jul 21 at 10:39
Incidentally, power law distributions have heavy tails and some of their moments are infinite, as are their moment generating functions for all $t gt 0$
– Henry
Jul 21 at 10:47
Incidentally, power law distributions have heavy tails and some of their moments are infinite, as are their moment generating functions for all $t gt 0$
– Henry
Jul 21 at 10:47
@Henry In this case, I do mean the log-log linear behavior in the tail. And you are correct, the moments are defined only for $t<0$ above. However, I am not able to complete the proof using this or through a differentiation of the cumulative distribution function.
– buzaku
Jul 21 at 12:05
@Henry In this case, I do mean the log-log linear behavior in the tail. And you are correct, the moments are defined only for $t<0$ above. However, I am not able to complete the proof using this or through a differentiation of the cumulative distribution function.
– buzaku
Jul 21 at 12:05
add a comment |Â
1 Answer
1
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oldest
votes
up vote
0
down vote
accepted
The moment generating function doesn't exist for many heavy tailed distributions. In general if you take suitably normalized linear combinations of power-law distributed random variables with infinite variances (ie with tail parameter $alpha<2$, the generalized version of central limit theorem shows weak convergence to a stable distribution with the same tail parameter. If $alphageq 2$ the linear combinations will tend to converge toward a Gaussian. Note that when $alpha=2$ the variance may still be infinite but convergence to Gaussian is still possible. An example of a heavy tailed stable distribution is the Cauchy. All stable distributions except Gaussian have power law tails. John P. Nolan has some useful notes on his website.
answered Aug 1 at 14:10
Will Townes
1534
Thank you for pointing towards the stable distributions...
– buzaku
Aug 3 at 4:27
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
accepted
The moment generating function doesn't exist for many heavy tailed distributions. In general if you take suitably normalized linear combinations of power-law distributed random variables with infinite variances (ie with tail parameter $alpha<2$, the generalized version of central limit theorem shows weak convergence to a stable distribution with the same tail parameter. If $alphageq 2$ the linear combinations will tend to converge toward a Gaussian. Note that when $alpha=2$ the variance may still be infinite but convergence to Gaussian is still possible. An example of a heavy tailed stable distribution is the Cauchy. All stable distributions except Gaussian have power law tails. John P. Nolan has some useful notes on his website.
answered Aug 1 at 14:10
Will Townes
1534
Thank you for pointing towards the stable distributions...
– buzaku
Aug 3 at 4:27
add a comment |Â
up vote
0
down vote
accepted
The moment generating function doesn't exist for many heavy tailed distributions. In general if you take suitably normalized linear combinations of power-law distributed random variables with infinite variances (ie with tail parameter $alpha<2$, the generalized version of central limit theorem shows weak convergence to a stable distribution with the same tail parameter. If $alphageq 2$ the linear combinations will tend to converge toward a Gaussian. Note that when $alpha=2$ the variance may still be infinite but convergence to Gaussian is still possible. An example of a heavy tailed stable distribution is the Cauchy. All stable distributions except Gaussian have power law tails. John P. Nolan has some useful notes on his website.
answered Aug 1 at 14:10
Will Townes
1534
Thank you for pointing towards the stable distributions...
– buzaku
Aug 3 at 4:27
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The moment generating function doesn't exist for many heavy tailed distributions. In general if you take suitably normalized linear combinations of power-law distributed random variables with infinite variances (ie with tail parameter $alpha<2$, the generalized version of central limit theorem shows weak convergence to a stable distribution with the same tail parameter. If $alphageq 2$ the linear combinations will tend to converge toward a Gaussian. Note that when $alpha=2$ the variance may still be infinite but convergence to Gaussian is still possible. An example of a heavy tailed stable distribution is the Cauchy. All stable distributions except Gaussian have power law tails. John P. Nolan has some useful notes on his website.
answered Aug 1 at 14:10
Will Townes
1534
The moment generating function doesn't exist for many heavy tailed distributions. In general if you take suitably normalized linear combinations of power-law distributed random variables with infinite variances (ie with tail parameter $alpha<2$, the generalized version of central limit theorem shows weak convergence to a stable distribution with the same tail parameter. If $alphageq 2$ the linear combinations will tend to converge toward a Gaussian. Note that when $alpha=2$ the variance may still be infinite but convergence to Gaussian is still possible. An example of a heavy tailed stable distribution is the Cauchy. All stable distributions except Gaussian have power law tails. John P. Nolan has some useful notes on his website.
answered Aug 1 at 14:10
Will Townes
1534
answered Aug 1 at 14:10
Will Townes
1534
answered Aug 1 at 14:10
Will Townes
1534
answered Aug 1 at 14:10
Will Townes
1534
1534
Thank you for pointing towards the stable distributions...
– buzaku
Aug 3 at 4:27
add a comment |Â
Thank you for pointing towards the stable distributions...
– buzaku
Aug 3 at 4:27
Thank you for pointing towards the stable distributions...
– buzaku
Aug 3 at 4:27
Thank you for pointing towards the stable distributions...
– buzaku
Aug 3 at 4:27
add a comment |Â
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1
It depends on what you mean by power law distribution. If you mean Pareto distribution then I would guess the answer would be no. If you mean particular behaviour in the tail (on a log-log scale roughly linear on the right) then I would guess the answer might be yes.
– Henry
Jul 21 at 10:39
Incidentally, power law distributions have heavy tails and some of their moments are infinite, as are their moment generating functions for all $t gt 0$
– Henry
Jul 21 at 10:47
@Henry In this case, I do mean the log-log linear behavior in the tail. And you are correct, the moments are defined only for $t<0$ above. However, I am not able to complete the proof using this or through a differentiation of the cumulative distribution function.
– buzaku
Jul 21 at 12:05