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I'm reading a book on Brownian Motion. In the proof of the existence of such random function (Wiener, 1923), the following is stated:

Indeed, all increments $B(d)-B(d-2^{-n})$, for $d\in \mathcal{D}_n\setminus \{0\}$, are independent. To see this it suffices to show that they are pairwise independent. as the vector of these increments is Gaussian.

The last part of this quote is the claim that pairwise independent normal variables from a Gaussian family are independent. Could anyone provide/direct me to a proof of this claim?

Thanks!

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  • $\begingroup$ "The last part of this quote is the claim that pairwise independent normal variables are independent" No, the claim is that pairwise independent normal variables from a gaussian family are independent $\endgroup$ – Did Apr 19 '16 at 21:05
  • $\begingroup$ Thanks. I will correct my question. $\endgroup$ – EZLearner Apr 19 '16 at 21:06
  • $\begingroup$ It would also suffice to show that they are uncorrelated. Does the correction/hint of Did help you? $\endgroup$ – user190080 Apr 19 '16 at 21:42
  • $\begingroup$ Not so much. I have edited my question as a result, but it didn't answer my question. The idea behind such a proof will also suffice. I am trying to understand the book all the while I am refreshing my knowledge in probability theory, so I make sure I understand every claim. $\endgroup$ – EZLearner Apr 19 '16 at 21:49
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Actually it is rather straight forward:

Assume you have a gaussian vector (so the joint distribution follows a gaussian distribution) $$ \mathbf X=(X_1,X_2,\ldots,X_n)\sim N(\mathbf \mu,\Sigma) $$ where $\Sigma\in M^{n\times n}(\mathbb R)$ describes the covariance matrix. The density function looks like $$ f_{\mathbf X}(x_1,\ldots,x_n) = \frac{1}{\sqrt{(2\pi)^{n}\lvert\Sigma\rvert}} \exp\left(-\frac{1}{2}({\mathbf X}-{\mathbf\mu})^\mathrm{T}{\Sigma}^{-1}({\mathbf X}-{\mathbf\mu}) \right), $$ Now, we have independence (mutually) iff the density function factorizes. Since the random variables are pairwise independent (uncorrelated suffices), the covariance matrix is a diagonal matrix $$ \Sigma=\mathrm{diag}(\sigma_1,\ldots,\sigma_n) \text{ and }\mathbf \mu=(m_1,\ldots,m_n) $$ since the above holds, the density indeed factorizes and we get $$ f_{\mathbf X}(x_1,\ldots,x_n)=\prod_{i=1}^n\frac{1}{\sqrt{2\pi{\sigma_i}^2}}\exp\bigg(-\frac{{(x_i-m_i)}^2}{2{\sigma_i}^2}\bigg) $$ and we have mutually independent gaussians.

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  • $\begingroup$ In which do specific properties of normal distribution make the difference in making your result true? I am a bit confused, sorry. It seems to me that your result holds true for every possible vector of random variables (that is, of random variables following every possible distribution, not necessarily Normal) @user190080 $\endgroup$ – Strictly_increasing Sep 15 at 18:31
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    $\begingroup$ @Strictly_increasing but you are already convinced, that pairwise independence does not necessarily imply mutually independence? The joint normal distribution factorizes if the covariance matrix is diagonal (e.g. if we have pairwise independence), and this distributional fact is special to the normal distribution. $\endgroup$ – user190080 Sep 16 at 9:14
  • $\begingroup$ Yeah, I was asking about that since I already knew that pairwise independence does not necessarily imply mutual independence, generally speaking. Could you please give me some reference to understand properly why that distributional fact is specific to the normal distribution? @user190080 $\endgroup$ – Strictly_increasing Sep 16 at 11:16
  • $\begingroup$ Could I convince myself that that fact is special to the normal distribution by reasoning on normal distribution characteristic function? @user190080 $\endgroup$ – Strictly_increasing Sep 17 at 9:45
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    $\begingroup$ @Strictly_increasing actually this might be a good question to be asked separately here...it comes down to the structure of the joint distribution (and the marginals ofc too) and how it factorizes when the covariance structure is diagonal. I don't know any distribution which behaves in a similar way. To study the CF of a normal will definitely help to to spot immediately that the distribution is stable but you don't need this here. $\endgroup$ – user190080 Sep 17 at 10:01

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