# Mutual Independence in a Multivariate Normal with Identity Covariance

Consider a random vector $$X$$ which follows a multivariate nomal with zero means and Identity Covariance.

$$X\sim \mathcal{N}_n(\mathbf 0, \mathbf I)$$

We can say that the individual variables $$X_1, X_2, \cdots, X_n$$ are pair-wisely independent, since in the case of variable pairs in a multivariate normal, zero correlation implies independence (see ).

However, I am struggling to prove mutual independence of the individual variables. Note that pair-wise independence does not generally imply mutual independence, as the latter is a stronger condition (see these two links). In other words, I want to prove that not only any pair from $$X_1, X_2, \cdots, X_n$$ are independent but that any individual $$X_i$$ is independent of any intersection of the remaining variables.

I also suspect that mutual independence holds for any diagonal covariance matrix... 

 Robert V. Hogg, Joseph W. McKean, and Allen T. Craig. "Introduction to Mathematical Statistics, 7th Edition". In: Pearson, 2013, pp. 182-183. ISBN: 978-0-321-84943-4. • It seems the text you quoted already gives the answer. Or, maybe I'm not understanding what you're asking – user20160 Dec 10 '18 at 16:00
• @whuber, the text I quoted proves pair-wise independence of the individual variables, but not mutual independence, which is in general not equivalent. I modified the question to clarify this. – Daniel López Dec 10 '18 at 16:34
• Although I voted to reopen, I wish to note that a solution is immediately available through a generalization of this approach: a diagonal covariance matrix implies the moment generating functions partition into a product of $n$ univariate mgfs. – whuber Dec 10 '18 at 17:11

## 1 Answer

Here is a partial answer. Suppose $$D$$ is a diagonal matrix whose diagonal entries $$d_1,\ldots,d_n$$ are positive. Let $$\mathbf y' = (y_1,\ldots, y_n).$$ Then \begin{align} & \exp\left( \frac{-1}2 \mathbf y'D^{-1} \mathbf y \right) \\[8pt] = {} & \exp\left( \frac{-1} 2 \sum_{i=1}^n \frac{y_i^2}{d_i} \right) \\[8pt] = {} & \prod_{i=1}^n \exp\left( \frac{-1} 2 \frac{y_i^2}{d_i} \right). \end{align} This stops short of treating the case in which $$D = \operatorname{var}(Y) = \operatorname E(YY') \in \mathbb R^{n\times n}$$ is singular.

• Thanks for your answer. I also noted that the theorem I quoted (Theorem 3.5.2 from ) not only ensures pairwise independence among variables, but also independence of any variable with respect to any subset of the remaining ones, thus also ensuring mutual independence. This is because in the theorem $X_1$ and $X_2$ are not individual variables but arbitrary partitions of the variables in $X$. – Daniel López Dec 10 '18 at 19:32
• I would say that your "partial proof" is almost complete. The $X_i$ are $N(0,d_i^2)$ random variables and you have shown that the joint pdf of the $X_i$'s is the product of the marginal pdfs of the $X_i$'s. Ergo, the $X_i$ are mutually independent, not just pairwise independent. For the record, it is possible to have three pairwise independent normal random variables that are not mutually independent but the three don't have a multivariate normal density. See the last half of my answer to a question about pairwise vs mutual independence. – Dilip Sarwate Dec 10 '18 at 20:53