Can two different distributions have the same value of mean, variance, skewness, and kurtosis?
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$begingroup$
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 Jan 5 at 10:30
kjetil b halvorsen
32k985237
asked Jan 5 at 7:44
Adurthi Ashwin SwarupAdurthi Ashwin Swarup
1336
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 10:30
kjetil b halvorsen
32k985237
asked Jan 5 at 7:44
Adurthi Ashwin SwarupAdurthi Ashwin Swarup
1336
$endgroup$
6
$begingroup$
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.
$endgroup$
– Francis
Jan 5 at 8:22
$begingroup$
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.
$endgroup$
– Pace
Jan 5 at 20:36
add a comment |
$begingroup$
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 Jan 5 at 10:30
kjetil b halvorsen
32k985237
asked Jan 5 at 7:44
Adurthi Ashwin SwarupAdurthi Ashwin Swarup
1336
$endgroup$
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 Jan 5 at 10:30
kjetil b halvorsen
32k985237
asked Jan 5 at 7:44
Adurthi Ashwin SwarupAdurthi Ashwin Swarup
1336
edited Jan 5 at 10:30
kjetil b halvorsen
32k985237
asked Jan 5 at 7:44
Adurthi Ashwin SwarupAdurthi Ashwin Swarup
1336
edited Jan 5 at 10:30
kjetil b halvorsen
32k985237
edited Jan 5 at 10:30
kjetil b halvorsen
32k985237
edited Jan 5 at 10:30
kjetil b halvorsen
32k985237
32k985237
asked Jan 5 at 7:44
Adurthi Ashwin SwarupAdurthi Ashwin Swarup
1336
asked Jan 5 at 7:44
Adurthi Ashwin SwarupAdurthi Ashwin Swarup
1336
asked Jan 5 at 7:44
Adurthi Ashwin SwarupAdurthi Ashwin Swarup
1336
1336
6
$begingroup$
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.
$endgroup$
– Francis
Jan 5 at 8:22
$begingroup$
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.
$endgroup$
– Pace
Jan 5 at 20:36
add a comment |
6
$begingroup$
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.
$endgroup$
– Francis
Jan 5 at 8:22
$begingroup$
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.
$endgroup$
– Pace
Jan 5 at 20:36
6
6
$begingroup$
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.
$endgroup$
– Francis
Jan 5 at 8:22
$begingroup$
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.
$endgroup$
– Francis
Jan 5 at 8:22
$begingroup$
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.
$endgroup$
– Pace
Jan 5 at 20:36
$begingroup$
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.
$endgroup$
– Pace
Jan 5 at 20:36
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 Jan 5 at 18:04
PerePere
4,6831820
$endgroup$
$begingroup$
This is really good mate. The question I wanted to is that are the shapes significantly different ? can you plot the shape of the distributions well ?
$endgroup$
– Adurthi Ashwin Swarup
Jan 10 at 8:47
$begingroup$
I did. These are discrete distributions and the bar diagram above shows the shape of the three distributions (one different color each). Probability is zero for all values except for those where a bar is. The same exercise could be done with a continuous distribution - for example, plotting a case of Xi'an's answer.
$endgroup$
– Pere
Jan 10 at 8:54
add a comment |
$begingroup$
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 Jan 5 at 10:22
Xi'anXi'an
59.2k897366
$endgroup$
$begingroup$
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 ?
$endgroup$
– Adurthi Ashwin Swarup
Jan 5 at 14:02
$begingroup$
You can use a discrete version of this mixture distribution, with no difference in the conclusion.
$endgroup$
– Xi'an
Jan 5 at 14:14
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
$begingroup$
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 Jan 5 at 18:04
PerePere
4,6831820
$endgroup$
$begingroup$
This is really good mate. The question I wanted to is that are the shapes significantly different ? can you plot the shape of the distributions well ?
$endgroup$
– Adurthi Ashwin Swarup
Jan 10 at 8:47
$begingroup$
I did. These are discrete distributions and the bar diagram above shows the shape of the three distributions (one different color each). Probability is zero for all values except for those where a bar is. The same exercise could be done with a continuous distribution - for example, plotting a case of Xi'an's answer.
$endgroup$
– Pere
Jan 10 at 8:54
add a comment |
$begingroup$
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 Jan 5 at 18:04
PerePere
4,6831820
$endgroup$
$begingroup$
This is really good mate. The question I wanted to is that are the shapes significantly different ? can you plot the shape of the distributions well ?
$endgroup$
– Adurthi Ashwin Swarup
Jan 10 at 8:47
$begingroup$
I did. These are discrete distributions and the bar diagram above shows the shape of the three distributions (one different color each). Probability is zero for all values except for those where a bar is. The same exercise could be done with a continuous distribution - for example, plotting a case of Xi'an's answer.
$endgroup$
– Pere
Jan 10 at 8:54
add a comment |
$begingroup$
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 Jan 5 at 18:04
PerePere
4,6831820
$endgroup$
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 Jan 5 at 18:04
PerePere
4,6831820
answered Jan 5 at 18:04
PerePere
4,6831820
answered Jan 5 at 18:04
PerePere
4,6831820
answered Jan 5 at 18:04
PerePere
4,6831820
4,6831820
$begingroup$
This is really good mate. The question I wanted to is that are the shapes significantly different ? can you plot the shape of the distributions well ?
$endgroup$
– Adurthi Ashwin Swarup
Jan 10 at 8:47
$begingroup$
I did. These are discrete distributions and the bar diagram above shows the shape of the three distributions (one different color each). Probability is zero for all values except for those where a bar is. The same exercise could be done with a continuous distribution - for example, plotting a case of Xi'an's answer.
$endgroup$
– Pere
Jan 10 at 8:54
add a comment |
$begingroup$
This is really good mate. The question I wanted to is that are the shapes significantly different ? can you plot the shape of the distributions well ?
$endgroup$
– Adurthi Ashwin Swarup
Jan 10 at 8:47
$begingroup$
I did. These are discrete distributions and the bar diagram above shows the shape of the three distributions (one different color each). Probability is zero for all values except for those where a bar is. The same exercise could be done with a continuous distribution - for example, plotting a case of Xi'an's answer.
$endgroup$
– Pere
Jan 10 at 8:54
$begingroup$
This is really good mate. The question I wanted to is that are the shapes significantly different ? can you plot the shape of the distributions well ?
$endgroup$
– Adurthi Ashwin Swarup
Jan 10 at 8:47
$begingroup$
This is really good mate. The question I wanted to is that are the shapes significantly different ? can you plot the shape of the distributions well ?
$endgroup$
– Adurthi Ashwin Swarup
Jan 10 at 8:47
$begingroup$
I did. These are discrete distributions and the bar diagram above shows the shape of the three distributions (one different color each). Probability is zero for all values except for those where a bar is. The same exercise could be done with a continuous distribution - for example, plotting a case of Xi'an's answer.
$endgroup$
– Pere
Jan 10 at 8:54
$begingroup$
I did. These are discrete distributions and the bar diagram above shows the shape of the three distributions (one different color each). Probability is zero for all values except for those where a bar is. The same exercise could be done with a continuous distribution - for example, plotting a case of Xi'an's answer.
$endgroup$
– Pere
Jan 10 at 8:54
add a comment |
$begingroup$
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 Jan 5 at 10:22
Xi'anXi'an
59.2k897366
$endgroup$
$begingroup$
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 ?
$endgroup$
– Adurthi Ashwin Swarup
Jan 5 at 14:02
$begingroup$
You can use a discrete version of this mixture distribution, with no difference in the conclusion.
$endgroup$
– Xi'an
Jan 5 at 14:14
add a comment |
$begingroup$
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 Jan 5 at 10:22
Xi'anXi'an
59.2k897366
$endgroup$
$begingroup$
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 ?
$endgroup$
– Adurthi Ashwin Swarup
Jan 5 at 14:02
$begingroup$
You can use a discrete version of this mixture distribution, with no difference in the conclusion.
$endgroup$
– Xi'an
Jan 5 at 14:14
add a comment |
$begingroup$
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 Jan 5 at 10:22
Xi'anXi'an
59.2k897366
$endgroup$
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 Jan 5 at 10:22
Xi'anXi'an
59.2k897366
edited Jan 5 at 14:12
answered Jan 5 at 10:22
Xi'anXi'an
59.2k897366
answered Jan 5 at 10:22
Xi'anXi'an
59.2k897366
answered Jan 5 at 10:22
Xi'anXi'an
59.2k897366
59.2k897366
$begingroup$
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 ?
$endgroup$
– Adurthi Ashwin Swarup
Jan 5 at 14:02
$begingroup$
You can use a discrete version of this mixture distribution, with no difference in the conclusion.
$endgroup$
– Xi'an
Jan 5 at 14:14
add a comment |
$begingroup$
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 ?
$endgroup$
– Adurthi Ashwin Swarup
Jan 5 at 14:02
$begingroup$
You can use a discrete version of this mixture distribution, with no difference in the conclusion.
$endgroup$
– Xi'an
Jan 5 at 14:14
$begingroup$
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 ?
$endgroup$
– Adurthi Ashwin Swarup
Jan 5 at 14:02
$begingroup$
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 ?
$endgroup$
– Adurthi Ashwin Swarup
Jan 5 at 14:02
$begingroup$
You can use a discrete version of this mixture distribution, with no difference in the conclusion.
$endgroup$
– Xi'an
Jan 5 at 14:14
$begingroup$
You can use a discrete version of this mixture distribution, with no difference in the conclusion.
$endgroup$
– Xi'an
Jan 5 at 14:14
add a comment |
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$begingroup$
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.
$endgroup$
– Francis
Jan 5 at 8:22
$begingroup$
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.
$endgroup$
– Pace
Jan 5 at 20:36