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I need to state and prove a general sufficient condition on(a,b,c) for independence of two random Variables. We have that $a,b$ and $c$ are real numbers and the random variables are below:

$$ Y_1=aZ_1+bZ_2+cZ_3 $$ $$ Y_2=aZ_2+bZ_3+cZ_4 $$

Where $Z_i$ are iid from $Z$. Here, I need to prove this result regardless of the probability distribution $Z$. I am also given that

$$ Cov[Y_1,Y_2]=(ab+bc)Var[Z] $$

Thanks for the help in advance,

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  • $\begingroup$ Is this homework? If so, show some work first. $\endgroup$ – user76844 Jan 12 '15 at 15:36
  • $\begingroup$ Hello, This is a part of a qustion. I have managed to derive the Covariance, show a counter example with covariance=0 and dependent, but couldn't figure this out. I tried it from first principles. $\endgroup$ – Math525 Jan 12 '15 at 15:43
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Hint 1: What if $a=b=c=k$ this implies $Y_1=W+kZ_1$ and $Y_2=W+kZ_4$, $Z_1$ is independent of $Z_4$ but we still have the common $W$.

Hint 2: $X \text{ ind } Y \implies Cov(X,Y)=0$, therefore $(a=-c) \text{ or } b=0$. Note that at this point, this is just a necessary conditions.

Lets see if $b=0$ is sufficient. $b=0 \implies Y_1=aZ_1+cZ_3$ and $Y_2=aZ_2+cZ_4$, but then each $Y$ is a sum of separate, iid random variables. Hence, $Y_1,Y_2$ they are independent regardless of the distribution of $Z$.

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  • $\begingroup$ Thanks for the reply. If $a=b=c=k \Rightarrow $ independence, and independence $ \Rightarrow a=-c$, then does this mean that $k=0$? Thank you. $\endgroup$ – Math525 Jan 12 '15 at 15:52
  • $\begingroup$ @Math525 see my revised post...$b=0$ is sufficient. $\endgroup$ – user76844 Jan 12 '15 at 15:59
  • $\begingroup$ (i) Is there a nice proof of that last statement. I.e, because $Y_1$ and $Y_2$ are sums of different independent random variables, they will independent? (ii)Also, why did the first Hint break down? Why was that not enough?\\ (iii)Surely your final statement says this is true for any a and c, but the hint 2 has the necessary statement that If they are independent, then $a=-c$? Thanks $\endgroup$ – Math525 Jan 12 '15 at 16:07
  • $\begingroup$ @Math525 (i) There are various levels of sophistication one could prove this with. Here's a simple one: $P(Y_1=x|Y_2=w)=P(aZ_1+cZ_3 = x)$ since $Z_2,Z_4$ do not figure into the value of $Y_1$, yet this equation is the same as $P(Y_1=x)$ hence they are independent. (ii) It fails because the covariance is positive. This can also be seen because they share a common variance of $Var(W)$. (iii) I've clarified the condition...I assumed that $b\neq 0$ in the first part of the or statement. $\endgroup$ – user76844 Jan 12 '15 at 16:17
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    $\begingroup$ I think I see it now. So you see spotted a necessary condition, and tried to do a proof with the $b=0$ statement which worked nicely, so you chose that one? Thank you, it makes sense now. $\endgroup$ – Math525 Jan 12 '15 at 16:22
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As Eupraxis1981 showed, $b=0$ is a sufficient condition. If $b\ne 0$, we can set $a=c=0$, and we have another solution. What remains it to prove more strictly that these conditions:

  • are sufficient. Namely, if $Y_1 = g_1( \{X_i\ : i\in A\})$ and $Y_2 =g_2( \{X_j\ : j\in B\})$ where the sets of indexes $A,B$ are disjoint, then $Y_1,Y_2$ are independent

  • are necessary. For example, if $b\ne0$ and $a = -c \ne 0$ then, for some distribution , $Y_1$ and $Y_2$ are not independent

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  • $\begingroup$ Are you saying that Hint 1 does not provide a counterexample to bullet 2? $\endgroup$ – user76844 Jan 12 '15 at 16:42
  • $\begingroup$ Why is it that just saying $b=0$ is not enough. Eupraxis1981 showed that $b=0$ leads to Y_1 and Y_2 being independent $\forall a,b \in \Re$. Is this not what the question required? $\endgroup$ – Math525 Jan 12 '15 at 16:49
  • $\begingroup$ @Math525 No that's not what he's saying. He agreed that b=0 is sufficient. He's giving more rigorous statements that would need to be shown if $b\neq 0$ $\endgroup$ – user76844 Jan 12 '15 at 16:57
  • $\begingroup$ I understand. Thank you for your help throughout. Really helpful $\endgroup$ – Math525 Jan 12 '15 at 17:10
  • $\begingroup$ $b=0$ is enough, but it's just not necessary (and in that sense might not be a "general sufficient condition" - though the expression is rather curious). I've given another solution (also sufficient), with $a=c=0$ and $b\ne 0$ $\endgroup$ – leonbloy Jan 12 '15 at 18:47

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