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Show that $U$ and $V$ are independent, where $U=X+Z$ and $V=X-Z$.

I´m given that $X\sim N(0,1)$, $Z\sim N(0,1)$ and X and Z are independent.

First I find U and V. This leads to $U\sim N(0,2)$ and $V\sim(0,2)$

From here I'm stuck, I know that $independence \implies Cov(V,U)=0$ but i'm not sure if $Cov(V,U)=0 \implies independence.$

Jean Marie
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Emil Dare
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  • This must be in your notes: if (U,V) is jointly gaussian with covariance zero then (U,V) is independent. Note "jointly", which is crucial since U gaussian + V gaussian + covariance zero does not imply independence. So, do you think your (U,V) is jointly gaussian? – Did Dec 30 '16 at 12:11
  • I have changed the Tag "statistics" into the tag "probability-distributions". It is not at all a statistical issue. – Jean Marie Dec 30 '16 at 13:34

1 Answers1

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The suggestion in the comment is correct. If you can show that U and V are jointly gaussian with covariance zero then you have their independence. There are details about this lemma online, for example this page.

In this case, as X and Z are independent it's easy to show that all linear combinations of U and V are gaussian RVs. From there you can show the independence.

You are also correct that zero covariance does not imply independence. For example, $A$ is a RV with $\mathbb{E}A = 0$ and $\mathbb{E}[A^2] = 0$, and let $B=A$. Then we have $Cov(AB) = 0$ but $A$ and $B$ are clearly dependent.

user403554
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  • The example to show that covariance zero does not imply independence is odd. What is the covariance of a single random variable already? Note also that if $E(A^2)=0$ then $A$ is independent of $A$. – Did Jan 02 '17 at 16:13