# Suppose $\mathbf{X, Y}$ are independent random vectors. Are their components independent? [duplicate]

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Let $$\mathbf{X} = (X_1, \dots, X_p)^\top$$ and $$\mathbf{Y} = (Y_1, \dots, Y_p)^\top$$ be independent. Does it then follow that $$X_i$$ is independent with $$Y_j$$ i.e. cov$$(X_i, Y_j) = 0$$?

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• Since (obviously) the components are functions of the vectors, the duplicate says it all. Please note that independence and zero covariance are not equivalent: independence implies zero covariance if the covariance exists and zero covariance does not generally imply independence. – whuber Feb 12 at 20:25

## 2 Answers

If the two vectors are indepdendent, we have $$p(\textbf{X,Y})=p(\textbf{X})p(\textbf{Y})$$. Considering a specific pair $$X_i$$,$$Y_j$$, \begin{align}p(X_i,Y_j) &=\int_{X_k,k\neq i}\int_{Y_m,m\neq j}P(\textbf{X},\textbf{Y})\\ &=\int_{X_k,k\neq i}\int_{Y_m,m\neq j}P(\textbf{X})P(\textbf{Y})\\ &=\int_{X_k,k\neq i}P(\textbf{X})\int_{Y_m,m\neq j}P(\textbf{Y}) \\ &=P(X_i)P(Y_j)\end{align} So, they're independent, which means $$\operatorname{cov}(X_i,Y_j)=0$$. But, having $$\operatorname{cov}(X_i,Y_j)=0$$ doesn't mean that the two are independent, as you asked.

In addition to the answer by @gunes, here it is better to use the definitions directly. Two random variables (or vectors, as in this case) $$\mathbf{X}, \mathbf{Y}$$ are independent if all events determined by $$\mathbf{X}$$ are independent from all events determined by $$\mathbf{Y}^\dagger$$.

But an event determined by $$X_i$$ is certainly (indirectly) determined by $$\mathbf{X}$$. So the conclusion follows directly from the definition, without any need for integration or summation.

$$^\dagger$$ events determined by $$\mathbf{X}$$ means *member of the $$\sigma$$-algebra generated by $$\mathbf{X}$$.