I am coming back to a well-known example: let $X$ follow $N(0,1)$ and $T$ follow a Rademacher distribution $(p(T=1)=p(T=-1)=1/2)$. Then, it can be easily demonstrated that $TX$ follows also $N(0,1)$.

My problem is about the demonstration that the random vector $(X,TX)$ is not gaussian.

Usually, the demonstration shows that the linear combination $X+TX$ doesn't follow a normal distribution (thus implying the desired result that $(X,TX)$ is not gaussian...).

This demonstration is usually explained by setting: $p(X+TX=0)=p((1+T)X=0)=p(1+T=0)=p(T=-1)=1/2$

which shows a concentration of probability in $0$ and contradicts the fact that $X+TX$, as normal, should be continuous.

This is perfectly clear... as far as the above demonstration is perfectly understood... But the problem is that I don't understand why the equality $p((1+T)X=0)=p(1+T=0)$ holds!

I know it must be obvious,but I'm just stuck on it.

  • $\begingroup$ The rules of arithmetic tell us $(1+T)X=0$ if and only if either $1+T=0$ or $X=0.$ $\endgroup$ – whuber Oct 9 '19 at 13:26

$(1 + T)X$ is zero if $(1 - T) = 0$ or if $X = 0$. Since the probability that $X=0$ is zero, $Pr((1 + T)X = 0) = Pr(1 + T = 0).$

By conditioning on the event that $X=0$, you can see that

\begin{align*} Pr((1 + T)X = 0) &= Pr((1 + T)X = 0 \vert X=0)Pr(X=0)\\ &+Pr((1 + T)X = 0 \vert X\neq 0)Pr(X\neq 0)\\ &= Pr((1 + T)X = 0 \vert X=0)\cdot 0 + Pr((1 + T)X= 0 \vert X\neq 0)\cdot 1 \\ &=Pr((1 + T)=0) \end{align*}

  • $\begingroup$ Do you mean we just have to write: $Pr((1+T)X=0)=Pr(1+T=0 \,or\, X=0)=Pr(1+T=0)+Pr(X=0)+Pr(1+T=0 \,and\, X=0)=Pr(1+T=0)+Pr(X=0)Pr(1+T=0 \,knowing\, \,that\, X=0)=Pr(1+T=0)$ $\endgroup$ – Andrew Oct 9 '19 at 10:16
  • $\begingroup$ Essentially yes, if you want something that resembles a proof. An easier technique is to condition on the event that $X=0,$ so you get \begin{align*} Pr((1 + T)X = 0) = Pr((1 + T)X &= 0 \vert X=0)Pr(X=0) +Pr((1 + T)X = 0 \vert X\neq 0)Pr(X\neq 0)\\ &= Pr((1 + T)X = 0 \vert X=0)\cdot 0 + Pr((1 + T)X= 0 \vert X\neq 0)\cdot 1 \\ &=Pr((1 + T)=0) \end{align*} $\endgroup$ – Simon Boge Brant Oct 9 '19 at 11:14
  • $\begingroup$ Thanks, it's clear. Sorry for the mistake in my comment; it is $-Pr(1+T=0 \,and\, X=0)$... $\endgroup$ – Andrew Oct 9 '19 at 11:24

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