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?
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asked Jan 5, 2019 at 7:44
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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. $\endgroup$– FrancisCommented Jan 5, 2019 at 8:22
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$\begingroup$ 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. $\endgroup$– PaceCommented Jan 5, 2019 at 20:36
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2 Answers
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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)
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$\begingroup$ 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 ? $\endgroup$ Commented Jan 10, 2019 at 8:47
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$\begingroup$ 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. $\endgroup$– PereCommented Jan 10, 2019 at 8:54
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Take a mixture of two Normal distributions with density $$f(x|\mu_1,\mu_2,\sigma_1,\sigma_2,\omega)= \frac{\omega}{\sqrt{2\pi}\sigma_1}\exp\{-(x-\mu_1)^2/2\sigma_1^2\}+ \frac{1-\omega}{\sqrt{2\pi}\sigma_2}\exp\{-(x-\mu_2)^2/2\sigma_2^2\}$$ This distribution has five parameters constrained by four equations \begin{align*} \mathbb{E}[X]&=\omega\mu_1+(1-\omega)\mu_2\\ \text{var}(X)&=\omega\sigma_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 $(\mu_1,\mu_2,\sigma_1,\sigma_2,\omega)$.
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$\begingroup$ 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 ? $\endgroup$ Commented Jan 5, 2019 at 14:02
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$\begingroup$ You can use a discrete version of this mixture distribution, with no difference in the conclusion. $\endgroup$– Xi'anCommented Jan 5, 2019 at 14:14