Consider the random vectors $Z_1, Z_2, Z_3, Z_4$ and assume they are i.i.d.

Moreover, $Z_i\equiv (Y_i, X_i)$, where $Y_i$ $X_i$ are scalar random variables, and $\epsilon_i\equiv g(Y_i, X_i)$ for $i=1,2,3,4$ for some function $g$

I want to show that $\epsilon_i \perp X_i $ $\forall i$ implies $\epsilon_i \perp(X_1, X_2, X_3, X_4)$ $\forall i$

Could you help me? ($\perp$ denotes independence)

My attempt (correct?)

  • Let $i=1$

  • $Z_1, Z_2, Z_3, Z_4$ i.i.d. and $\epsilon_1\equiv f(Y_1, X_1)$ $\Rightarrow$ $(\epsilon_1, X_1) \perp (X_2, X_3, X_4) $

  • By assumption $\epsilon_1 \perp X_1$

  • $Z_1, Z_2, Z_3, Z_4$ i.i.d.$\Rightarrow$ $X_1 \perp X_2, X_3, X_4$

  • Hence, $$ f_{\epsilon_1, X_1, X_2, X_3, X_4}=f_{\epsilon_1, X_1} f_{X_2, X_3, X_4}=f_{\epsilon_1} f_{X_1} f_{X_2, X_3, X_4}= f_{\epsilon_1}f_{X_1,X_2, X_3, X_4} $$

  • repeat for $i=2,3,4$


(2017.09.16) Is it me or is the plague of silent revenge downvotes spreading? Yet one more...

You might recognize that your question is a special case of the following:

Let $(U,V,W)$ denote random variables such that $U$ is independent of $V$, and $(U,V)$ is independent of $W$, then $U$ is independent of $(V,W)$.

This seems direct, for example computing the characteristic function $$\varphi_{U,V,W}(u,v,w)=E[\exp(iuU+ivV+iwW)]$$

  • $\begingroup$ +1. Downvoting this accepted answer is certainly ridiculous. $\endgroup$ – user9464 Sep 23 '17 at 12:58
  • $\begingroup$ @Jack Thanks. I guess we have to live with such erratic behaviours. :-) $\endgroup$ – Did Sep 24 '17 at 6:58

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