# Using linear combination of multivariate normal $X$

I have a question and its answer but I don't understand it. would you please explain it to me?

QUESTION: Let $$X$$ follow a $$N(0, 1)$$ and $$Z$$ a random variable independent of $$X$$ following a $$U(-1, 1)$$. It can be proved that $$ZX \sim N(0, 1)$$ . Prove that $$(X, ZX)$$ is not a Gaussian vector.

It is not a Gaussian vector because there exists a vector $$a$$, such that $$a^{\prime}[X, ZX]^{\prime}$$ is not Gaussian, specifically $$a = [1, 1]'$$,

$$P (X + ZX = 0) = P (Z = - 1) = 1/2;$$

then it is not a full continuous variable, it cannot be normal. Then, $$[X, ZX]^{\prime}$$ is not a Gaussian vector.

Clarification. "$$U(-1, 1)$$" in your question should be understood as $$P(Z = \pm 1) = \frac{1}{2}$$, instead of the continuous uniform random variable (but please note that the notation "$$U(-1, 1)$$" is typically reserved for the continuous uniform random variable whose range is $$(-1, 1)$$, so do not use it for the Rademacher variable). This interpretation is based on the fact that $$ZX$$ cannot be $$N(0, 1)$$ if $$Z$$ is the continuous $$U(-1, 1)$$ random variable, see Addendum for the proof.

Given that, your provided answer works: for $$X + ZX = 0$$ implies $$X = 0$$ or $$Z = -1$$. On the other hand $$P(X = 0) = 0$$, which implies that \begin{align*} P(X + ZX = 0) &= P(X = 0) + P(Z = -1) - P(X = 0, Z = -1) \\ &= 0 + P(Z = -1) - 0 = \frac{1}{2}. \end{align*}

Here is another way of showing $$(X, ZX)$$ is non-Gaussian using proof by contradiction. If $$(X, ZX)$$ was Gaussian, then because $$X$$ and $$ZX$$ are uncorrelated (as a result of $$\operatorname{Cov}(X, ZX) = E[X^2Z] = E[X^2]E[Z] = 0$$), they must be independent (this is due to that for a Gaussian vector, uncorrelation between components is equivalent to their independence). This would imply that $$X^2$$ and $$(ZX)^2$$ are independent too, hence $$E[X^2 (ZX)^2] = E[X^2]E[(ZX)^2].$$

However, $$E[X^2(ZX)^2] = E[X^4]E[Z^2] = 3 \times 1 = 3$$, while $$E[X^2]E[(ZX)^2] = 1 \times 1 = 1$$, contradiction.

Note that if $$Z \sim U(-1, 1)$$ (continuous), then $$E(Z^2) = \frac{1}{3}$$, this implies that $$\operatorname{Var}(ZX) = E(Z^2X^2) = E[Z^2]E[X^2] = \frac{1}{3} \neq 1,$$ hence $$ZX$$ is not $$N(0, 1)$$.
In fact, $$ZX$$ cannot be normally distributed: suppose $$ZX \sim N(0, \frac{1}{3})$$ (for its variance has to be $$\frac{1}{3}$$, as shown in the preceding paragraph), then on one hand, based on $$\sqrt{3}ZX \sim N(0, 1)$$, we have $$E[(ZX)^4] = \frac{1}{9}E[(\sqrt{3}ZX)^4)] = \frac{3}{9} = \frac{1}{3}$$. On the other hand, by the independence of $$Z^4$$ and $$X^4$$ and $$E[Z^4] = \int_{-1}^1z^4\frac{1}{2}dz = \int_0^1z^4dz = \frac{1}{5}$$, we have $$E[(ZX)^4] = E[Z^4]E[X^4] = \frac{3}{5}$$. This contradiction shows that $$ZX \not\sim N(0, \frac{1}{3})$$ either.
• (1) Although the events in your first example are not equivalent, their probabilities are indeed both equal. (2) The context of the question makes it clear that "$U(-1,1)$" means a Rademacher variable, not a uniform variable, for otherwise the product $ZX$ would not be Normal. This helps us see in what sense the answer is correct.