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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.

ANSWER:

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.

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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.


Addendum

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.

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  • $\begingroup$ may I ask why X and ZX being independent would imply that X^2 and (ZX)^2 are independent too? and how E(X^2 (ZX)^2) = E(X^4)? $\endgroup$
    – mormey
    Commented Nov 24, 2023 at 21:14
  • $\begingroup$ (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. $\endgroup$
    – whuber
    Commented Nov 24, 2023 at 22:18
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    $\begingroup$ @whuber Make sense, now I understand how its "ANSWER" works... I will update my answer. $\endgroup$
    – Zhanxiong
    Commented Nov 24, 2023 at 23:16

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