1

Let $X$ be a random variable with support $\mathcal{X}$. Let $r:\mathcal{X}\to\{0,1\}$.

Assume the random variables in the sequence $(\epsilon_i)_{i=1}^N$ are i.i.d. conditional on $X$.

Does this imply that the random variables in the sequence $(\epsilon_i)_{i=1}^N$ are i.i.d. conditional on $r(X)=1$?

Star
  • 222
  • Hint: Define $r(x)=1$ for all $x \in \mathcal{X}$. – Michael Jun 12 '19 at 03:29
  • Thanks, I'm still confused. Take $N=2$. The assumption (focusing on the independence part just for simplicity) tells me that $f_{\epsilon_1,\epsilon_2|X=x}=f_{\epsilon_1|X=x}\times f_{\epsilon_2|X=x}$ $\forall x\in \mathcal{X}$. What I want to show is that this implies $f_{\epsilon_1,\epsilon_2|{x\in \mathcal{X}: r(x)=1}}=f_{\epsilon_1|{x\in \mathcal{X}: r(x)=1}}\times f_{\epsilon_2|{x\in \mathcal{X}: r(x)=1}}$ – Star Jun 12 '19 at 07:56
  • Maybe my implication is stronger than the assumption? I don't know, any further help is appreciated. – Star Jun 12 '19 at 07:57
  • Now if you use my hint in what you are trying to show, what is it equivalent to showing? – Michael Jun 12 '19 at 09:53
  • Under you condition on $r$, I have to show that $f_{\epsilon_1,\epsilon_2| \mathcal{X}}= f_{\epsilon_1| \mathcal{X}}\times f_{\epsilon_2| \mathcal{X}}$. Which looks like I'm not conditioning? – Star Jun 12 '19 at 09:58
  • I don't know what your notation means, but it looks not quite correct. Can you get rid of $\epsilon_1,\epsilon_2$ and use standard $Y_1,Y_2$? So the marginal for $Y_1$ is $f_{Y_1}(y_1)$ and conditional on $X$ is $f_{Y_1|X}(y_1|x)$. This will allow you to use standard "capital letters" for random variables and also small case. I don't know what "small case $\epsilon$" is. – Michael Jun 12 '19 at 10:00
  • OK. I have to show that $\forall (y_1, y_2)$ $f_{Y_1,Y_2|r(X)}(y_1, y_2| 1)= f_{Y_1|r(X)}(y_1| 1)\times f_{Y_2|r(X)}(y_2| 1)$. Under you condition on $r$, $f_{Y_1,Y_2|r(X)}(y_1, y_2| 1)=f_{Y_1,Y_2}(y_1, y_2)$, $f_{Y_1|r(X)}(y_1|1)=f_{Y_1}(y_1)$, $f_{Y_2|r(X)}(y_2|1)=f_{Y_2}(y_2)$ $\forall (y_1, y_2)$. Hence, I have to show that $f_{Y_1,Y_2}(y_1, y_2)=f_{Y_1}(y_1)\times f_{Y_2}(y_2)$ $\forall (y_1, y_2)$, which is true under my assumption. – Star Jun 12 '19 at 10:07
  • Is this correct? – Star Jun 12 '19 at 10:07
  • Are you sure your conclusion is true? You are saying that conditional independence always implies unconditional independence. – Michael Jun 12 '19 at 10:08
  • 1
    I see, conditional independence does not imply independence (https://math.stackexchange.com/questions/22407/independence-and-conditional-independence-between-random-variables). Therefore, the answer to my question is NO. Correct? – Star Jun 12 '19 at 10:14
  • Correct. If it works for all functions $r$, it must work for the always-1 function. A complete answer would give some simple counter-example where you have $Y_1,Y_2$ conditionally independent given $X$, but not independent. – Michael Jun 12 '19 at 10:15

0 Answers0