Can two different distributions have the same value of mean, variance, skewness, and kurtosis?
Assuming that you have two discrete population distributions.
Can they have identical values of mean ,variance, skewness and kurtosis while being different in shape visually ?
Do these four values act like a fingerprint of any distribution?
descriptive-statistics skewness kurtosis
edited 2 days ago
kjetil b halvorsen
28.9k980208
asked 2 days ago
Adurthi Ashwin SwarupAdurthi Ashwin Swarup
1315
add a comment |
Assuming that you have two discrete population distributions.
Can they have identical values of mean ,variance, skewness and kurtosis while being different in shape visually ?
Do these four values act like a fingerprint of any distribution?
descriptive-statistics skewness kurtosis
edited 2 days ago
kjetil b halvorsen
28.9k980208
asked 2 days ago
Adurthi Ashwin SwarupAdurthi Ashwin Swarup
1315
6
Yes. In fact, one can construct two different discrete distributions with all moments equal, yet do not agree in distribution. Also check out this question: Two random variables with same moments.
– Francis
2 days ago
You may already know this but when in such a situation where you have some statistics which reduce the space of possible distributions but do not identify a single distribution then you should generally choose the probability distribution with the maximum entropy.
– Pace
yesterday
add a comment |
Assuming that you have two discrete population distributions.
Can they have identical values of mean ,variance, skewness and kurtosis while being different in shape visually ?
Do these four values act like a fingerprint of any distribution?
descriptive-statistics skewness kurtosis
edited 2 days ago
kjetil b halvorsen
28.9k980208
asked 2 days ago
Adurthi Ashwin SwarupAdurthi Ashwin Swarup
1315
Assuming that you have two discrete population distributions.
Can they have identical values of mean ,variance, skewness and kurtosis while being different in shape visually ?
Do these four values act like a fingerprint of any distribution?
descriptive-statistics skewness kurtosis
descriptive-statistics skewness kurtosis
edited 2 days ago
kjetil b halvorsen
28.9k980208
asked 2 days ago
Adurthi Ashwin SwarupAdurthi Ashwin Swarup
1315
edited 2 days ago
kjetil b halvorsen
28.9k980208
asked 2 days ago
Adurthi Ashwin SwarupAdurthi Ashwin Swarup
1315
edited 2 days ago
kjetil b halvorsen
28.9k980208
edited 2 days ago
kjetil b halvorsen
28.9k980208
edited 2 days ago
kjetil b halvorsen
28.9k980208
28.9k980208
asked 2 days ago
Adurthi Ashwin SwarupAdurthi Ashwin Swarup
1315
asked 2 days ago
Adurthi Ashwin SwarupAdurthi Ashwin Swarup
1315
asked 2 days ago
Adurthi Ashwin SwarupAdurthi Ashwin Swarup
1315
1315
6
Yes. In fact, one can construct two different discrete distributions with all moments equal, yet do not agree in distribution. Also check out this question: Two random variables with same moments.
– Francis
2 days ago
You may already know this but when in such a situation where you have some statistics which reduce the space of possible distributions but do not identify a single distribution then you should generally choose the probability distribution with the maximum entropy.
– Pace
yesterday
add a comment |
6
Yes. In fact, one can construct two different discrete distributions with all moments equal, yet do not agree in distribution. Also check out this question: Two random variables with same moments.
– Francis
2 days ago
You may already know this but when in such a situation where you have some statistics which reduce the space of possible distributions but do not identify a single distribution then you should generally choose the probability distribution with the maximum entropy.
– Pace
yesterday
6
6
Yes. In fact, one can construct two different discrete distributions with all moments equal, yet do not agree in distribution. Also check out this question: Two random variables with same moments.
– Francis
2 days ago
Yes. In fact, one can construct two different discrete distributions with all moments equal, yet do not agree in distribution. Also check out this question: Two random variables with same moments.
– Francis
2 days ago
You may already know this but when in such a situation where you have some statistics which reduce the space of possible distributions but do not identify a single distribution then you should generally choose the probability distribution with the maximum entropy.
– Pace
yesterday
You may already know this but when in such a situation where you have some statistics which reduce the space of possible distributions but do not identify a single distribution then you should generally choose the probability distribution with the maximum entropy.
– Pace
yesterday
add a comment |
2 Answers
2
active
oldest
votes
Take a mixture of two Normal distributions with density
$$f(x|mu_1,mu_2,sigma_1,sigma_2,omega)=
frac{omega}{sqrt{2pi}sigma_1}exp{-(x-mu_1)^2/2sigma_1^2}+
frac{1-omega}{sqrt{2pi}sigma_2}exp{-(x-mu_2)^2/2sigma_2^2}$$
This distribution has five parameters constrained by four equations
begin{align*}
mathbb{E}[X]&=omegamu_1+(1-omega)mu_2\
text{var}(X)&=omegasigma_1^2+(1-omega)sigma_2^2+omega(mu_1-mathbb{E}[X])^2+(1-omega)(mu_2-mathbb{E}[X])^2\
mathbb{E}[X^3]&=ldots\
mathbb{E}[X^4]&=ldots
end{align*}
Assuming these equations are compatible, there is therefore an infinite number of solutions $(mu_1,mu_2,sigma_1,sigma_2,omega)$.
answered 2 days ago
Xi'anXi'an
54k690348
How come 4 equations ? Wont there be 4 equations to be solved for each distribution ? Assuming 4 eq for mean , variance , skewness and kurtosis - And is this applicable to discrete distributions as well ?
– Adurthi Ashwin Swarup
2 days ago
You can use a discrete version of this mixture distribution, with no difference in the conclusion.
– Xi'an
2 days ago
add a comment |
Xi'an's answer proved (or at least hinted a proof) that there are different distributions with the same mean, variance, skewness and kurtosis. I just want to show an example of three visually distinct discrete distributions with the same moments (mean=skewness=0, variance=1 and kurtosis=2):
The code to generate them is:
library(moments)
n <- 1e6
x <- c(-sqrt(2), 0, +sqrt(2))
p <- c(1,2,1)
mostra1 <- sample(x, size=n, prob=p, replace=TRUE)
x <- c(-1.4629338416371, -0.350630832572269, 0.350630832573386, 1.46293384163564)
p <- c(1, 1.3, 1.3, 1)
mostra2 <- sample(x, size=n, prob=p, replace=TRUE)
x <- c(-1.5049621442915, -0.457635862316285, 0.457635862316022, 1.50496214429192)
p <- c(1, 1.6, 1.6, 1)
mostra3 <- sample(x, size=n, prob=p, replace=TRUE)
mostra <- rbind(data.frame(x=mostra1, grup="a"),
data.frame(x=mostra2, grup="b"),
data.frame(x=mostra3, grup="c"))
aggregate(x~grup, data=mostra, mean)
aggregate(x~grup, data=mostra, var)
aggregate(x~grup, data=mostra, skewness)
aggregate(x~grup, data=mostra, kurtosis)
library(ggplot2)
ggplot(mostra)+
geom_histogram(aes(x, fill=grup), bins=100)
answered yesterday
PerePere
4,0491718
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Take a mixture of two Normal distributions with density
$$f(x|mu_1,mu_2,sigma_1,sigma_2,omega)=
frac{omega}{sqrt{2pi}sigma_1}exp{-(x-mu_1)^2/2sigma_1^2}+
frac{1-omega}{sqrt{2pi}sigma_2}exp{-(x-mu_2)^2/2sigma_2^2}$$
This distribution has five parameters constrained by four equations
begin{align*}
mathbb{E}[X]&=omegamu_1+(1-omega)mu_2\
text{var}(X)&=omegasigma_1^2+(1-omega)sigma_2^2+omega(mu_1-mathbb{E}[X])^2+(1-omega)(mu_2-mathbb{E}[X])^2\
mathbb{E}[X^3]&=ldots\
mathbb{E}[X^4]&=ldots
end{align*}
Assuming these equations are compatible, there is therefore an infinite number of solutions $(mu_1,mu_2,sigma_1,sigma_2,omega)$.
answered 2 days ago
Xi'anXi'an
54k690348
How come 4 equations ? Wont there be 4 equations to be solved for each distribution ? Assuming 4 eq for mean , variance , skewness and kurtosis - And is this applicable to discrete distributions as well ?
– Adurthi Ashwin Swarup
2 days ago
You can use a discrete version of this mixture distribution, with no difference in the conclusion.
– Xi'an
2 days ago
add a comment |
Take a mixture of two Normal distributions with density
$$f(x|mu_1,mu_2,sigma_1,sigma_2,omega)=
frac{omega}{sqrt{2pi}sigma_1}exp{-(x-mu_1)^2/2sigma_1^2}+
frac{1-omega}{sqrt{2pi}sigma_2}exp{-(x-mu_2)^2/2sigma_2^2}$$
This distribution has five parameters constrained by four equations
begin{align*}
mathbb{E}[X]&=omegamu_1+(1-omega)mu_2\
text{var}(X)&=omegasigma_1^2+(1-omega)sigma_2^2+omega(mu_1-mathbb{E}[X])^2+(1-omega)(mu_2-mathbb{E}[X])^2\
mathbb{E}[X^3]&=ldots\
mathbb{E}[X^4]&=ldots
end{align*}
Assuming these equations are compatible, there is therefore an infinite number of solutions $(mu_1,mu_2,sigma_1,sigma_2,omega)$.
answered 2 days ago
Xi'anXi'an
54k690348
How come 4 equations ? Wont there be 4 equations to be solved for each distribution ? Assuming 4 eq for mean , variance , skewness and kurtosis - And is this applicable to discrete distributions as well ?
– Adurthi Ashwin Swarup
2 days ago
You can use a discrete version of this mixture distribution, with no difference in the conclusion.
– Xi'an
2 days ago
add a comment |
Take a mixture of two Normal distributions with density
$$f(x|mu_1,mu_2,sigma_1,sigma_2,omega)=
frac{omega}{sqrt{2pi}sigma_1}exp{-(x-mu_1)^2/2sigma_1^2}+
frac{1-omega}{sqrt{2pi}sigma_2}exp{-(x-mu_2)^2/2sigma_2^2}$$
This distribution has five parameters constrained by four equations
begin{align*}
mathbb{E}[X]&=omegamu_1+(1-omega)mu_2\
text{var}(X)&=omegasigma_1^2+(1-omega)sigma_2^2+omega(mu_1-mathbb{E}[X])^2+(1-omega)(mu_2-mathbb{E}[X])^2\
mathbb{E}[X^3]&=ldots\
mathbb{E}[X^4]&=ldots
end{align*}
Assuming these equations are compatible, there is therefore an infinite number of solutions $(mu_1,mu_2,sigma_1,sigma_2,omega)$.
answered 2 days ago
Xi'anXi'an
54k690348
Take a mixture of two Normal distributions with density
$$f(x|mu_1,mu_2,sigma_1,sigma_2,omega)=
frac{omega}{sqrt{2pi}sigma_1}exp{-(x-mu_1)^2/2sigma_1^2}+
frac{1-omega}{sqrt{2pi}sigma_2}exp{-(x-mu_2)^2/2sigma_2^2}$$
This distribution has five parameters constrained by four equations
begin{align*}
mathbb{E}[X]&=omegamu_1+(1-omega)mu_2\
text{var}(X)&=omegasigma_1^2+(1-omega)sigma_2^2+omega(mu_1-mathbb{E}[X])^2+(1-omega)(mu_2-mathbb{E}[X])^2\
mathbb{E}[X^3]&=ldots\
mathbb{E}[X^4]&=ldots
end{align*}
Assuming these equations are compatible, there is therefore an infinite number of solutions $(mu_1,mu_2,sigma_1,sigma_2,omega)$.
answered 2 days ago
Xi'anXi'an
54k690348
edited 2 days ago
answered 2 days ago
Xi'anXi'an
54k690348
answered 2 days ago
Xi'anXi'an
54k690348
answered 2 days ago
Xi'anXi'an
54k690348
54k690348
How come 4 equations ? Wont there be 4 equations to be solved for each distribution ? Assuming 4 eq for mean , variance , skewness and kurtosis - And is this applicable to discrete distributions as well ?
– Adurthi Ashwin Swarup
2 days ago
You can use a discrete version of this mixture distribution, with no difference in the conclusion.
– Xi'an
2 days ago
add a comment |
How come 4 equations ? Wont there be 4 equations to be solved for each distribution ? Assuming 4 eq for mean , variance , skewness and kurtosis - And is this applicable to discrete distributions as well ?
– Adurthi Ashwin Swarup
2 days ago
You can use a discrete version of this mixture distribution, with no difference in the conclusion.
– Xi'an
2 days ago
How come 4 equations ? Wont there be 4 equations to be solved for each distribution ? Assuming 4 eq for mean , variance , skewness and kurtosis - And is this applicable to discrete distributions as well ?
– Adurthi Ashwin Swarup
2 days ago
How come 4 equations ? Wont there be 4 equations to be solved for each distribution ? Assuming 4 eq for mean , variance , skewness and kurtosis - And is this applicable to discrete distributions as well ?
– Adurthi Ashwin Swarup
2 days ago
You can use a discrete version of this mixture distribution, with no difference in the conclusion.
– Xi'an
2 days ago
You can use a discrete version of this mixture distribution, with no difference in the conclusion.
– Xi'an
2 days ago
add a comment |
Xi'an's answer proved (or at least hinted a proof) that there are different distributions with the same mean, variance, skewness and kurtosis. I just want to show an example of three visually distinct discrete distributions with the same moments (mean=skewness=0, variance=1 and kurtosis=2):
The code to generate them is:
library(moments)
n <- 1e6
x <- c(-sqrt(2), 0, +sqrt(2))
p <- c(1,2,1)
mostra1 <- sample(x, size=n, prob=p, replace=TRUE)
x <- c(-1.4629338416371, -0.350630832572269, 0.350630832573386, 1.46293384163564)
p <- c(1, 1.3, 1.3, 1)
mostra2 <- sample(x, size=n, prob=p, replace=TRUE)
x <- c(-1.5049621442915, -0.457635862316285, 0.457635862316022, 1.50496214429192)
p <- c(1, 1.6, 1.6, 1)
mostra3 <- sample(x, size=n, prob=p, replace=TRUE)
mostra <- rbind(data.frame(x=mostra1, grup="a"),
data.frame(x=mostra2, grup="b"),
data.frame(x=mostra3, grup="c"))
aggregate(x~grup, data=mostra, mean)
aggregate(x~grup, data=mostra, var)
aggregate(x~grup, data=mostra, skewness)
aggregate(x~grup, data=mostra, kurtosis)
library(ggplot2)
ggplot(mostra)+
geom_histogram(aes(x, fill=grup), bins=100)
answered yesterday
PerePere
4,0491718
add a comment |
Xi'an's answer proved (or at least hinted a proof) that there are different distributions with the same mean, variance, skewness and kurtosis. I just want to show an example of three visually distinct discrete distributions with the same moments (mean=skewness=0, variance=1 and kurtosis=2):
The code to generate them is:
library(moments)
n <- 1e6
x <- c(-sqrt(2), 0, +sqrt(2))
p <- c(1,2,1)
mostra1 <- sample(x, size=n, prob=p, replace=TRUE)
x <- c(-1.4629338416371, -0.350630832572269, 0.350630832573386, 1.46293384163564)
p <- c(1, 1.3, 1.3, 1)
mostra2 <- sample(x, size=n, prob=p, replace=TRUE)
x <- c(-1.5049621442915, -0.457635862316285, 0.457635862316022, 1.50496214429192)
p <- c(1, 1.6, 1.6, 1)
mostra3 <- sample(x, size=n, prob=p, replace=TRUE)
mostra <- rbind(data.frame(x=mostra1, grup="a"),
data.frame(x=mostra2, grup="b"),
data.frame(x=mostra3, grup="c"))
aggregate(x~grup, data=mostra, mean)
aggregate(x~grup, data=mostra, var)
aggregate(x~grup, data=mostra, skewness)
aggregate(x~grup, data=mostra, kurtosis)
library(ggplot2)
ggplot(mostra)+
geom_histogram(aes(x, fill=grup), bins=100)
answered yesterday
PerePere
4,0491718
add a comment |
Xi'an's answer proved (or at least hinted a proof) that there are different distributions with the same mean, variance, skewness and kurtosis. I just want to show an example of three visually distinct discrete distributions with the same moments (mean=skewness=0, variance=1 and kurtosis=2):
The code to generate them is:
library(moments)
n <- 1e6
x <- c(-sqrt(2), 0, +sqrt(2))
p <- c(1,2,1)
mostra1 <- sample(x, size=n, prob=p, replace=TRUE)
x <- c(-1.4629338416371, -0.350630832572269, 0.350630832573386, 1.46293384163564)
p <- c(1, 1.3, 1.3, 1)
mostra2 <- sample(x, size=n, prob=p, replace=TRUE)
x <- c(-1.5049621442915, -0.457635862316285, 0.457635862316022, 1.50496214429192)
p <- c(1, 1.6, 1.6, 1)
mostra3 <- sample(x, size=n, prob=p, replace=TRUE)
mostra <- rbind(data.frame(x=mostra1, grup="a"),
data.frame(x=mostra2, grup="b"),
data.frame(x=mostra3, grup="c"))
aggregate(x~grup, data=mostra, mean)
aggregate(x~grup, data=mostra, var)
aggregate(x~grup, data=mostra, skewness)
aggregate(x~grup, data=mostra, kurtosis)
library(ggplot2)
ggplot(mostra)+
geom_histogram(aes(x, fill=grup), bins=100)
answered yesterday
PerePere
4,0491718
Xi'an's answer proved (or at least hinted a proof) that there are different distributions with the same mean, variance, skewness and kurtosis. I just want to show an example of three visually distinct discrete distributions with the same moments (mean=skewness=0, variance=1 and kurtosis=2):
The code to generate them is:
library(moments)
n <- 1e6
x <- c(-sqrt(2), 0, +sqrt(2))
p <- c(1,2,1)
mostra1 <- sample(x, size=n, prob=p, replace=TRUE)
x <- c(-1.4629338416371, -0.350630832572269, 0.350630832573386, 1.46293384163564)
p <- c(1, 1.3, 1.3, 1)
mostra2 <- sample(x, size=n, prob=p, replace=TRUE)
x <- c(-1.5049621442915, -0.457635862316285, 0.457635862316022, 1.50496214429192)
p <- c(1, 1.6, 1.6, 1)
mostra3 <- sample(x, size=n, prob=p, replace=TRUE)
mostra <- rbind(data.frame(x=mostra1, grup="a"),
data.frame(x=mostra2, grup="b"),
data.frame(x=mostra3, grup="c"))
aggregate(x~grup, data=mostra, mean)
aggregate(x~grup, data=mostra, var)
aggregate(x~grup, data=mostra, skewness)
aggregate(x~grup, data=mostra, kurtosis)
library(ggplot2)
ggplot(mostra)+
geom_histogram(aes(x, fill=grup), bins=100)
answered yesterday
PerePere
4,0491718
answered yesterday
PerePere
4,0491718
answered yesterday
PerePere
4,0491718
answered yesterday
PerePere
4,0491718
4,0491718
add a comment |
add a comment |
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6
Yes. In fact, one can construct two different discrete distributions with all moments equal, yet do not agree in distribution. Also check out this question: Two random variables with same moments.
– Francis
2 days ago
You may already know this but when in such a situation where you have some statistics which reduce the space of possible distributions but do not identify a single distribution then you should generally choose the probability distribution with the maximum entropy.
– Pace
yesterday