0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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
0
$\begingroup$

$(1 + T)X$ 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*}

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.