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The Econonometrics book by Hansen mentions this property of a multivariate normal distribution: If $X∼N(\mu,\Sigma)$ then $AX+b∼N(Aμ+b,AΣA^\prime)$. It goes to say that a simple implication of this that each individual $X_j$ in $X$ is univariate normal. Could someone please explain how is this implied from the linearity property? Also is the converse true? If each $X_j$ in a $k$-vector random variable $X$ is normally distributed, is $X$ normally distributed?

StatsStudent
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  • X∼N(μ,Σ) is enough for saying Xj is univariate normal – gunes Aug 25 '20 at 14:28
  • Each $X_j$ is (in an obvious, trivial way) a linear transformation of $X,$ *qed.* The converse is false: this is discussed in several nice threads, including https://stats.stackexchange.com/questions/30159. For a compendium of related results and characterizations of Normal distributions, see https://stats.stackexchange.com/questions/4364. – whuber Aug 25 '20 at 14:32
  • The reverse is not true; marginal normality is not enough to assure joint normality: https://stats.stackexchange.com/a/189633/247274 – Dave Aug 25 '20 at 14:32

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