Comparing two MGFs
$begingroup$
Suppose I have two MGFs given by
$MGF_1 = e^{-2 (p-1) p x^2}$
and
$MGF_2 = e^{8 (p-1)^2 p^2 x^2}$.
Is there an accepted way to compare the corresponding probability distributions to obtain a metric for similarity?
(How similar are the probability distributions in this case?)
probability-theory
asked Dec 30 '18 at 0:03
user120911user120911
231110
$endgroup$
add a comment |
$begingroup$
Suppose I have two MGFs given by
$MGF_1 = e^{-2 (p-1) p x^2}$
and
$MGF_2 = e^{8 (p-1)^2 p^2 x^2}$.
Is there an accepted way to compare the corresponding probability distributions to obtain a metric for similarity?
(How similar are the probability distributions in this case?)
probability-theory
asked Dec 30 '18 at 0:03
user120911user120911
231110
$endgroup$
add a comment |
$begingroup$
Suppose I have two MGFs given by
$MGF_1 = e^{-2 (p-1) p x^2}$
and
$MGF_2 = e^{8 (p-1)^2 p^2 x^2}$.
Is there an accepted way to compare the corresponding probability distributions to obtain a metric for similarity?
(How similar are the probability distributions in this case?)
probability-theory
asked Dec 30 '18 at 0:03
user120911user120911
231110
$endgroup$
Suppose I have two MGFs given by
$MGF_1 = e^{-2 (p-1) p x^2}$
and
$MGF_2 = e^{8 (p-1)^2 p^2 x^2}$.
Is there an accepted way to compare the corresponding probability distributions to obtain a metric for similarity?
(How similar are the probability distributions in this case?)
probability-theory
probability-theory
asked Dec 30 '18 at 0:03
user120911user120911
231110
asked Dec 30 '18 at 0:03
user120911user120911
231110
asked Dec 30 '18 at 0:03
user120911user120911
231110
asked Dec 30 '18 at 0:03
user120911user120911
231110
asked Dec 30 '18 at 0:03
user120911user120911
231110
231110
add a comment |
add a comment |
1 Answer
1
active
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$begingroup$
Both of your MGF's indicate the underlying distributions are Gaussian, with mean $0$, and variance in terms of $p$ (lazy to check MGF table).
Once you do this, you'd have $P_1:N(0,sigma_1^2)$ and $P_2:N(0,sigma_2^2)$. Now, you can compare them, one metric would be to measure the total variation (TV divergence between them). Another, would be Kullback-Leibler. In general, you can use any $f$-divergence.
answered Dec 30 '18 at 0:28
AaronAaron
1,912415
$endgroup$
$begingroup$
Are you sure they both are Gaussian?
$endgroup$
– user120911
Dec 30 '18 at 1:10
$begingroup$
If you don't like the table in wikipedia then how about Eq.(17) of Wolfram MathWorld? You can find this in most textbooks, or just calculate it yourself (it's pretty easy).
$endgroup$
– Lee David Chung Lin
Dec 30 '18 at 8:16
$begingroup$
@LeeDavidChungLin The two MGFs are equal when p = 0.5 and p = 1. If p sufficiently captured variance, and we are dealing with two Gaussians, then I would guess that other values of p would make these MGFs equal too. Can you explain that?
$endgroup$
– user120911
Dec 30 '18 at 8:32
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Both of your MGF's indicate the underlying distributions are Gaussian, with mean $0$, and variance in terms of $p$ (lazy to check MGF table).
Once you do this, you'd have $P_1:N(0,sigma_1^2)$ and $P_2:N(0,sigma_2^2)$. Now, you can compare them, one metric would be to measure the total variation (TV divergence between them). Another, would be Kullback-Leibler. In general, you can use any $f$-divergence.
answered Dec 30 '18 at 0:28
AaronAaron
1,912415
$endgroup$
$begingroup$
Are you sure they both are Gaussian?
$endgroup$
– user120911
Dec 30 '18 at 1:10
$begingroup$
If you don't like the table in wikipedia then how about Eq.(17) of Wolfram MathWorld? You can find this in most textbooks, or just calculate it yourself (it's pretty easy).
$endgroup$
– Lee David Chung Lin
Dec 30 '18 at 8:16
$begingroup$
@LeeDavidChungLin The two MGFs are equal when p = 0.5 and p = 1. If p sufficiently captured variance, and we are dealing with two Gaussians, then I would guess that other values of p would make these MGFs equal too. Can you explain that?
$endgroup$
– user120911
Dec 30 '18 at 8:32
add a comment |
$begingroup$
Both of your MGF's indicate the underlying distributions are Gaussian, with mean $0$, and variance in terms of $p$ (lazy to check MGF table).
Once you do this, you'd have $P_1:N(0,sigma_1^2)$ and $P_2:N(0,sigma_2^2)$. Now, you can compare them, one metric would be to measure the total variation (TV divergence between them). Another, would be Kullback-Leibler. In general, you can use any $f$-divergence.
answered Dec 30 '18 at 0:28
AaronAaron
1,912415
$endgroup$
$begingroup$
Are you sure they both are Gaussian?
$endgroup$
– user120911
Dec 30 '18 at 1:10
$begingroup$
If you don't like the table in wikipedia then how about Eq.(17) of Wolfram MathWorld? You can find this in most textbooks, or just calculate it yourself (it's pretty easy).
$endgroup$
– Lee David Chung Lin
Dec 30 '18 at 8:16
$begingroup$
@LeeDavidChungLin The two MGFs are equal when p = 0.5 and p = 1. If p sufficiently captured variance, and we are dealing with two Gaussians, then I would guess that other values of p would make these MGFs equal too. Can you explain that?
$endgroup$
– user120911
Dec 30 '18 at 8:32
add a comment |
$begingroup$
Both of your MGF's indicate the underlying distributions are Gaussian, with mean $0$, and variance in terms of $p$ (lazy to check MGF table).
Once you do this, you'd have $P_1:N(0,sigma_1^2)$ and $P_2:N(0,sigma_2^2)$. Now, you can compare them, one metric would be to measure the total variation (TV divergence between them). Another, would be Kullback-Leibler. In general, you can use any $f$-divergence.
answered Dec 30 '18 at 0:28
AaronAaron
1,912415
$endgroup$
Both of your MGF's indicate the underlying distributions are Gaussian, with mean $0$, and variance in terms of $p$ (lazy to check MGF table).
Once you do this, you'd have $P_1:N(0,sigma_1^2)$ and $P_2:N(0,sigma_2^2)$. Now, you can compare them, one metric would be to measure the total variation (TV divergence between them). Another, would be Kullback-Leibler. In general, you can use any $f$-divergence.
answered Dec 30 '18 at 0:28
AaronAaron
1,912415
answered Dec 30 '18 at 0:28
AaronAaron
1,912415
answered Dec 30 '18 at 0:28
AaronAaron
1,912415
answered Dec 30 '18 at 0:28
AaronAaron
1,912415
1,912415
$begingroup$
Are you sure they both are Gaussian?
$endgroup$
– user120911
Dec 30 '18 at 1:10
$begingroup$
If you don't like the table in wikipedia then how about Eq.(17) of Wolfram MathWorld? You can find this in most textbooks, or just calculate it yourself (it's pretty easy).
$endgroup$
– Lee David Chung Lin
Dec 30 '18 at 8:16
$begingroup$
@LeeDavidChungLin The two MGFs are equal when p = 0.5 and p = 1. If p sufficiently captured variance, and we are dealing with two Gaussians, then I would guess that other values of p would make these MGFs equal too. Can you explain that?
$endgroup$
– user120911
Dec 30 '18 at 8:32
add a comment |
$begingroup$
Are you sure they both are Gaussian?
$endgroup$
– user120911
Dec 30 '18 at 1:10
$begingroup$
If you don't like the table in wikipedia then how about Eq.(17) of Wolfram MathWorld? You can find this in most textbooks, or just calculate it yourself (it's pretty easy).
$endgroup$
– Lee David Chung Lin
Dec 30 '18 at 8:16
$begingroup$
@LeeDavidChungLin The two MGFs are equal when p = 0.5 and p = 1. If p sufficiently captured variance, and we are dealing with two Gaussians, then I would guess that other values of p would make these MGFs equal too. Can you explain that?
$endgroup$
– user120911
Dec 30 '18 at 8:32
$begingroup$
Are you sure they both are Gaussian?
$endgroup$
– user120911
Dec 30 '18 at 1:10
$begingroup$
Are you sure they both are Gaussian?
$endgroup$
– user120911
Dec 30 '18 at 1:10
$begingroup$
If you don't like the table in wikipedia then how about Eq.(17) of Wolfram MathWorld? You can find this in most textbooks, or just calculate it yourself (it's pretty easy).
$endgroup$
– Lee David Chung Lin
Dec 30 '18 at 8:16
$begingroup$
If you don't like the table in wikipedia then how about Eq.(17) of Wolfram MathWorld? You can find this in most textbooks, or just calculate it yourself (it's pretty easy).
$endgroup$
– Lee David Chung Lin
Dec 30 '18 at 8:16
$begingroup$
@LeeDavidChungLin The two MGFs are equal when p = 0.5 and p = 1. If p sufficiently captured variance, and we are dealing with two Gaussians, then I would guess that other values of p would make these MGFs equal too. Can you explain that?
$endgroup$
– user120911
Dec 30 '18 at 8:32
$begingroup$
@LeeDavidChungLin The two MGFs are equal when p = 0.5 and p = 1. If p sufficiently captured variance, and we are dealing with two Gaussians, then I would guess that other values of p would make these MGFs equal too. Can you explain that?
$endgroup$
– user120911
Dec 30 '18 at 8:32
add a comment |
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