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If two random variables $x$ and $y$ are identical and independently distributed, do they need to have the same mean and variance? Can there exist a case where they are iid and still have different parameters?

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up vote 4 down vote accepted

If two variables are iid , then they must have the same distribution.Which means they have the same parameters namely mean, std dev.

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The mean and variance are determined by the distribution. Thus, if they have the same distribution, they must have the same mean and variance.

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If $X$ and $Y$ are iid random variables with cumulative distribution function $F(t)=\mathbb{P}(X\leq t)=\mathbb{P}(Y\leq t)$ and if mean and variance of the variables exists, then $$\mbox{E}X=\int_\mathbb{R}tdF(t)$$ and $$\mbox{Var}X=\mbox{E}X^2-(\mbox{EX})^2=\int_\mathbb{R}t^2dF(t)-(\mbox{E}X)^2.$$ If $F(t)$ is differentiable then you can write $\int_\mathbb{R}tdF(t)$ as $\int_\mathbb{R}tf(t)dt$ where $f(t)$ is a density function of $X$ and $Y$. So if iid (independent and identically distributed) variables are defined with such distribution functions, then there is no difference in means and variances. But it might happen that your variable does not have variance or mean. One example is Cauchy random variable, then you can't say anything about the mean or variance, because those characteristics simply do not exist.

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The abbreviation iid usually refers to independent and identically distributed" (in linear modelling, error distributions need to be iid)

  • NOT identical and independently distributed (as Daniel's question mentions).

Two variables are "independent" if they are not correlated

Two variables are identically-distributed if they have the same frequency distribution (the same skew/kurtosis/variance etc.)

Two variables are identical if they are exactly the same. This would mean that these variables would be perfectly correlated. Therefore, by definition, they cannot be "independently distributed".

In other words, " identical and independently distributed" is a contradiction in terms.

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