# Covariance of linear combination of normal distribution is zero implies independence?

Let's say we have i.i.d. normal distributions $$Z_1, ..., Z_n$$.

For some nonzero constants $$a_1, \cdots, a_n$$ and $$b_1, \cdots, b_n$$, let

$$X = a_{1}Z_{1} + \cdots + a_{n}Z_{n}\\ Y = b_{1}Z_{1} + \cdots + b_{n}Z_{n}$$

If $$X$$, $$Y$$have covariance zero, are $$X$$ and $$Y$$ necessarily independent?

• I'm pretty sure this is effectively a duplicate. This one addresses the central issue (that (X,Y) will be bivariate normal), from which the answer to your question follows (if that's not obvious see here), but I think there's another one which even more directly addresses this.question Commented Mar 3, 2017 at 8:26
• Yes, $X$ and $Y$ will have a bivariate normal distribution, therefore zero covariance implies independence. Commented Mar 5, 2017 at 3:57

As gammer pointed out in a comment, $$X$$ and $$Y$$ enjoy a bivariate normal distribution and so the assumption that $$X$$ and $$Y$$ are uncorrelated suffices to show that $$X$$ and $$Y$$ are indeed independent (normal) random variables. However, arriving at the point that $$X$$ and $$Y$$ are bivariate normal requires just a tad more work.

A standard definition of joint normality of $$n$$ random variables $$Z_1, Z_2, \ldots, Z_n$$ is

$$n$$ random variables $$Z_1, Z_2, \ldots, Z_n$$ are said to be jointly normal if for all choices of real numbers $$a_1, a_2, \ldots, a_n$$, $$X=\sum_{i=1}^n a_iZ_i$$ is a normal random variable.

Note that $$Z_1, Z_2, \ldots, Z_n$$ enjoying the property of being jointly normal implies that $$Z_1, Z_2, \ldots, Z_n$$ themselves individually are normal random variables. So, in the OP's problem where $$Z_1, Z_2, \ldots, Z_n$$ are i.i.d. normal, (and hence jointly normal), $$X=\sum_{i=1}^n a_iZ_i$$ and $$Y=\sum_{i=1}^n b_iZ_i$$ certainly are normal random variables. Even more, $$X$$ and $$Y$$ are jointly normal random variables because every linear combination $$\alpha X + \beta Y$$ of $$X$$and $$Y$$ is a normal random variable. Note that \begin{align} W &= \alpha X + \beta Y\\ &= \alpha\sum_{i=1}^n a_iZ_i + \beta\sum_{i=1}^n b_iZ_i\\ &= \sum_{i=1}^n (\alpha a_i + \beta b_i)Z_i\\ &= \sum_{i=1}^n cZ_i \end{align} and so $$W,$$ being a linear combination of the jointly normal $$Z_i$$ is normal by definition. So, $$X$$ and $$Y$$ are themselves jointly normal, and since they have zero covariance, they are independent too.

Jointly normal random variables that are uncorrelated (a.k.a. have zero covariance) are independent normal random variables.

Normal random variables that are uncorrelated (a.k.a. have zero covariance) are not necessarily independent.

A common example of normal uncorrelated $$X$$ and $$Y$$ that are not independent is $$X$$ being a standard normal random variable and $$Y = BX$$ where $$B$$ is a discrete random variable (independent of $$X$$) that takes on values $$\pm 1$$ with equal probability $$\frac 12.$$. So, a little extra work is needed because we need to supplement the fact that $$X$$ and $$Y$$ are normal random variables (which is immediate from the definition of joint normality of the $$Z_i$$) with the additional fact that $$X$$ and $$Y$$ are also jointly normal in order to claim that since $$X$$ and $$Y$$ are assumed to be uncorrelated, they are also independent.

This must be a duplicate, but impossible to find. Since $$Z_a, Z_2, \dotsc, Z_n$$ are iid normal, the vector $$Z$$ have a multinormal distribution (with a diagonal covariance matrix.) Now, a linear combination of the components of a multivariate normal distribution is itself multivariate normal, as proved at Proof for linear combination of multivariate normal X? and elsewhere.

So your vector $$(X,Y)$$ is bivariate normal, and a bivarite normal with correlation zero have independent components. So the answer is yes.