According to Rohatgi & Saleh [1], random variables X and Y are said to be identically distributed if they have the same distribution function, i.e. $F_X$(x) = $F_Y$(y).

Moreover, the distribution function of a normal random variable is given by

$F_X$(x) = $\frac{1}{σ\sqrt{2π}}\int_{-\infty}^xe^{-(x'-μ)^2/2σ^2}dx'$ [2]

i.e. it is a function of only μ and σ². Therefore, I would assume that the statements "normally distributed with the same unconditional mean μ and unconditional variance σ²" and "identically normally distributed" are equivalent.

Yet, wouldn't the following setting be a counterexample to above conclusion? Let $X_1$, ..., $X_n$ and $Y_1$, ..., $Y_n$ be normally distributed random variables with same unconditional mean and unconditional variance. According to above statement, they are also identically distributed. However, wouldn't it still be possible that there is conditional heteroscedasticity, i.e. the conditional variance of ${Y_i}$ given ${X_i}$, changes with i? And wouldn't conditional heteroscedasticity contradict the statement that $X_1$, ..., $X_n$ and $Y_1$, ..., $Y_n$ are identically distributed?


[1] https://books.google.ch/books?id=IMbVyKoZRh8C&pg=PA123&lpg=PA123&dq=definition+identical+distributed&source=bl&ots=nFoTq7EOER&sig=QMh_gZKiuDn_HX7Q7OsfuTymApA&hl=en&sa=X&ei=L0RyVaesIInwUIqrgbAP&redir_esc=y#v=onepage&q=definition%20identical%20distributed&f=false
[2] Equation 9 on http://mathworld.wolfram.com/NormalDistribution.html

up vote 4 down vote accepted

If you consider the univariate distribution of $X$ alone, a univariate mean $\mu_X$ and a univariate variance $\sigma^2_X$ fully characterizes the distribution, if $X$ is normal. Then saying that a bunch of $X$'s are i.i.d. normal or saying that they are normal with the same mean and the same variance is equivalent. That can be applied for $Y$, too.

Now if you consider bivariate distribution of a vector $(X,Y)$ then the univariate distributions characterized by normality and by $\mu_X, \mu_Y, \sigma^2_X, \sigma^2_Y$ are not enough to fully specify it, if $X$, $Y$ are univariately normal. The bivariate distribution of two univariate normals need not even be bivariate normal (pick a different copula, and you will get a different bivariate distribution). Thus the statement

According to above statement, they are also identically distributed.

should be considered carefully.

However, wouldn't it still be possible that there is conditional heteroscedasticity, i.e. the conditional variance of $Y_i$ given $X_i$, changes with $i$?

I just showed that there may be unconditional heteroskedasticity. Also, there may be conditional heteroskedasticity. Suppose for a while that $(X,Y)$ are i.i.d. bivariate normal so that there is no unconditional heteroskedasticity. However, conditional heteroskedasticity is easily possible. $\operatorname{Var}(Y|X=1)$ will not generally be equal to $\operatorname{Var}(Y|X=5)$.

However, conditional distributional properties do not contradict i.i.d.'ness, because when defining i.i.d.'ness we use the unconditional distribution function.

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