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If I have that $X_1, X_2, \ldots, X_n$ are iid Normal(0,1), then a standard result says that they are jointly or multivariate normal as well. The usual criteria for multivariate normal is if every linear combination of the random variables are normal as well. It certainly seems to be the case, as for any $a_1, a_2, \ldots, a_n \in \mathbb{R}$, $a_1X_1+ \ldots a_nX_n$ is normal as well, due to the fact that any linear combination of iid normals is normal. Is this a sufficient proof or is there something deeper I am missing? Thanks.

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    $\begingroup$ What's wrong with writing the joint density and deriving it is a multivariate normal density? $\endgroup$ – Xi'an Mar 5 '16 at 14:10
  • $\begingroup$ A frequently-observed hang-up on stats.SE is that multivariate normal distributions do not include independent normal random variables as a special case and one must jump through multiple hoops to prove that independent normal random variables can also be deemed to be enjoying a multivariate normal distribution. If independent standard normal random variables walk like a duck and quack like a duck, is the $p$-value small enough that we can call them a duck? Next theorem to be proved: Independent $N(\mu_i,\sigma_i^2)$ random variables are jointly normal a.k.a. multivariate normal as well. $\endgroup$ – Dilip Sarwate Mar 5 '16 at 14:43
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    $\begingroup$ Yes, that is a sufficient proof. This is also called the Cramér-Wald Theorem. $\endgroup$ – Greenparker Mar 5 '16 at 14:50
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    $\begingroup$ I'll be a pedant and point out that it's actually Wold and not Wald as in Abraham Wald. $\endgroup$ – dsaxton Mar 5 '16 at 15:12

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