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$\newcommand{\P}{\mathbb{P}}$$\newcommand{\E}{\mathbb{E}}$$X \sim N(0,1)$

$W$ is Rademacher distribution

$Y = WX$

https://en.wikipedia.org/wiki/Rademacher_distribution

In order to prove dependece of random variables I need to show, that $\P(X,Y) = \P(X)\P(Y)$ is violated.

I tried to find counterexample that would show, that the variables are dependent. First of all, I thought to show that $\E(XY) = \E(X)\E(Y)$ is not true. But in this case it is true $(\E(XY) = \E(X)\E(Y) = 0)$.

How can I prove, that variables are not independent

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  • $\begingroup$ Could you clarify that you mean $Y=WX$ and not $Y$ is the convolution of $W$ and $X$? Further to this, this seems like a self study question, so if it is could you add the self study tag? $\endgroup$ – Ed P Nov 13 '16 at 12:39
  • $\begingroup$ @StatsPlease edited $\endgroup$ – Daniel Yefimov Nov 13 '16 at 12:44
  • $\begingroup$ If you need more help just comment on my answer and I'll attempt to guide you. $\endgroup$ – Ed P Nov 13 '16 at 12:54
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What values can $Y$ take if $X=0$? What values can $Y$ take if $X=1$?

Are the conditional pdfs for $Y$ the same at the two values of $X$?

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  • $\begingroup$ If $X = 0$, than Y can take only $0$ value. If $X = 1$, than Y can take values $1$ or $-1$ with $P = 1/2$ for both values. And from the fact above, i think, we can conclude, that pdfs for $Y$ at different values of $X$ are different. $\endgroup$ – Daniel Yefimov Nov 13 '16 at 13:15
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    $\begingroup$ @DanielYefimov But can you follow up your conclusion that the conditional distributions are different to arrive at the conclusion that $X$ and $Y$ are dependent random variables? $\endgroup$ – Dilip Sarwate Nov 13 '16 at 18:22
  • $\begingroup$ @DilipSarwate So we concluded, that our conditional distributions are different. It means, when variable $X$ has different values, the output of $Y$ changes. So, change in $X$ provoke change in $Y$. So, $X$ and $Y$ are dependent. Am I right? $\endgroup$ – Daniel Yefimov Nov 13 '16 at 18:31
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    $\begingroup$ @DanielYefimov Perfect! $\endgroup$ – Dilip Sarwate Nov 13 '16 at 20:05
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I'll try and guide you to an answer.

We know:

$$X\sim N(0,1)$$

and

$$W=\begin{cases} \,\,\,\,\,1, & p=0.5\\ -1, & p=0.5\\ \end{cases}$$

We also know that $Y=WX$. So:

$$Y=\begin{cases} \,\,\,\,\,X, & p=0.5\\ -X, & p=0.5\\ \end{cases}$$

This shows that $Y$ is dependent on $X$.

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  • $\begingroup$ Yes, it is known. $\endgroup$ – Daniel Yefimov Nov 13 '16 at 12:54
  • $\begingroup$ This answer started out nicely but the conclusion (everything after the dotted line) is sheer nonsense. It is not true that $Y = X$; what is true is that they have the same distribution. -1 pending correction. $\endgroup$ – Dilip Sarwate Nov 13 '16 at 14:57
  • $\begingroup$ @DilipSarwate Apologies for my answer. I'm not sure where the mistake is (I'm guessing it's a technical one). As far as I can see, it's reasonable to say $Y=X\,\,\text{w.p. } 1$ and they are perfectly dependent. I'm obviously missing something, so your help would appreciated. $\endgroup$ – Ed P Nov 13 '16 at 23:43
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    $\begingroup$ @StatsPlease "$X$" is a function. It is a special kind of function that we call random variable, because its range is characterized by another function, the "distribution function" that $X$ "obeys". We never write $X=N(0,1)$ or for any other distribution, namely we never use the $=$ symbol, and this is because $X$ follows a (in this case) Normal distribution, it is not "equal" to a normal distribution. In most cases we do not have a clue about the functional form of $X$. $\endgroup$ – Alecos Papadopoulos Nov 14 '16 at 19:36

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