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Suppose I have a random vector $\boldsymbol{Z}$, if I can prove that for $\forall \boldsymbol{\lambda} \neq \boldsymbol{0}$ where $\boldsymbol{\lambda}$ is a fixed vector, not a random vector,

$\boldsymbol{\lambda}^{\text{T}}\boldsymbol{Z} \sim N(0, \boldsymbol{\lambda}^{\text{T}}\boldsymbol{C}\boldsymbol{\lambda})$

where $\boldsymbol{C}$ is a positive definite matrix. Can I establish accordingly, that

$\boldsymbol{Z} \sim N(\boldsymbol{0}, \boldsymbol{C})$

?

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2 Answers 2

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Just to slightly expand on the existing answer: you can see that it is indeed true as an immediate consequence of Bochner's theorem. Also, Gaussian measures on infinite-dimensional spaces are actually defined in this way. (A measure is Gaussian if and only if its push-forwards under continuous linear functionals are all Gaussian.)

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Yes assuming $C$ is positive semi definite. This is an alternative definition of the multivariate normal distribution.

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  • $\begingroup$ That $C$ is positve semi definite follows from the assumption. One has to allow variance $0$ for a normal distribution (which is then a Dirac measure). $\endgroup$
    – Jochen Wengenroth
    Commented Mar 19, 2015 at 9:56
  • $\begingroup$ I forgot to mention that $\boldsymbol{C}$ is positive definite. Thanks for the answer! $\endgroup$
    – Y.X
    Commented Mar 19, 2015 at 10:16
  • $\begingroup$ Also, is there any reference or proof for this? $\endgroup$
    – Y.X
    Commented Mar 19, 2015 at 12:20
  • $\begingroup$ @YuchenXie off the top of my head I don't know a link. But one technique is to write down the moment generating function and use the uniqueness theorem of moment generating functions. $\endgroup$
    – Kian
    Commented Mar 19, 2015 at 12:23
  • $\begingroup$ @fushsialatitude Thanks you very much for the hint, I will look for further information based on it. $\endgroup$
    – Y.X
    Commented Mar 19, 2015 at 13:26

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