# Pairwise independence implies independence for normal random variables

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!

• "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
– Did
Apr 19, 2016 at 21:05
• Thanks. I will correct my question. Apr 19, 2016 at 21:06
• It would also suffice to show that they are uncorrelated. Does the correction/hint of Did help you? Apr 19, 2016 at 21:42
• 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. Apr 19, 2016 at 21:49

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.