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 yesterday
kjetil b halvorsen
28.8k980208
asked yesterday
Adurthi 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 yesterday
kjetil b halvorsen
28.8k980208
asked yesterday
Adurthi 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
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 |
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 yesterday
kjetil b halvorsen
28.8k980208
asked yesterday
Adurthi 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 yesterday
kjetil b halvorsen
28.8k980208
asked yesterday
Adurthi Ashwin Swarup
1315
edited yesterday
kjetil b halvorsen
28.8k980208
asked yesterday
Adurthi Ashwin Swarup
1315
edited yesterday
kjetil b halvorsen
28.8k980208
edited yesterday
kjetil b halvorsen
28.8k980208
edited yesterday
kjetil b halvorsen
28.8k980208
28.8k980208
asked yesterday
Adurthi Ashwin Swarup
1315
asked yesterday
Adurthi Ashwin Swarup
1315
asked yesterday
Adurthi 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
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 |
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
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
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
yesterday
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
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
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 yesterday
Xi'an
53.8k689347
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
yesterday
You can use a discrete version of this mixture distribution, with no difference in the conclusion.
– Xi'an
yesterday
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
Pere
4,0391718
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 yesterday
Xi'an
53.8k689347
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
yesterday
You can use a discrete version of this mixture distribution, with no difference in the conclusion.
– Xi'an
yesterday
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 yesterday
Xi'an
53.8k689347
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
yesterday
You can use a discrete version of this mixture distribution, with no difference in the conclusion.
– Xi'an
yesterday
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 yesterday
Xi'an
53.8k689347
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 yesterday
Xi'an
53.8k689347
edited yesterday
answered yesterday
Xi'an
53.8k689347
answered yesterday
Xi'an
53.8k689347
answered yesterday
Xi'an
53.8k689347
53.8k689347
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
yesterday
You can use a discrete version of this mixture distribution, with no difference in the conclusion.
– Xi'an
yesterday
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
yesterday
You can use a discrete version of this mixture distribution, with no difference in the conclusion.
– Xi'an
yesterday
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
yesterday
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
yesterday
You can use a discrete version of this mixture distribution, with no difference in the conclusion.
– Xi'an
yesterday
You can use a discrete version of this mixture distribution, with no difference in the conclusion.
– Xi'an
yesterday
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
Pere
4,0391718
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
Pere
4,0391718
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
Pere
4,0391718
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
Pere
4,0391718
answered yesterday
Pere
4,0391718
answered yesterday
Pere
4,0391718
answered yesterday
Pere
4,0391718
4,0391718
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
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