# Why Normal + Normal x Rademacher is not Normal?

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.

• The rules of arithmetic tell us $(1+T)X=0$ if and only if either $1+T=0$ or $X=0.$ – 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*}
• Do you mean we just have to write: $Pr((1+T)X=0)=Pr(1+T=0 \,or\, X=0)=Pr(1+T=0)+Pr(X=0)+Pr(1+T=0 \,and\, X=0)=Pr(1+T=0)+Pr(X=0)Pr(1+T=0 \,knowing\, \,that\, X=0)=Pr(1+T=0)$ – Andrew Oct 9 '19 at 10:16
• 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*} – Simon Boge Brant Oct 9 '19 at 11:14
• Thanks, it's clear. Sorry for the mistake in my comment; it is $-Pr(1+T=0 \,and\, X=0)$... – Andrew Oct 9 '19 at 11:24