# Independence of a random vector

Let $X=(X_1,X_2)$ and $Y=(Y_1,Y_2)$. Suppose $X$ and $Y$ are independent, although $X_1$ and $X_2$, $Y_1$ and $Y_2$ are correlated. How can we define these relationships?

-
Is it $Y=(Y_1,Y_2)$ ? Otherwise, I am afraid $Y$ is defined in terms of itself. –  user21436 Jan 13 '12 at 2:58
@KannappanSampath: Yes, looks like a simple typo to me. I fxed it. –  Nate Eldredge Jan 13 '12 at 2:59
I don't understand what you're asking by "How can we define these relationships?" Can you elaborate? –  Nate Eldredge Jan 13 '12 at 3:00
By independence of $X$ and $Y$, $$f_{X_1,X_2,Y_1,Y_2}(x_1,x_2,y_1,y_2) = f_{X_1,X_2}(x_1,x_2)f_{Y_1,Y_2}(y_1,y_2) ~ \forall x_1,x_2,y_1,y_2$$ and by non-independence of $X_1,X_2$ and similarly $Y_1,Y_2$, $$f_{X_1,X_2}(x_1,x_2) \neq f_{X_1}(x_1)f_{X_2}(x_2), ~~ \text{and}~~ f_{Y_1,Y_2}(y_1,y_2) \neq f_{Y_1}(y_1)f_{Y_2}(y_2)$$ –  Dilip Sarwate Jan 13 '12 at 3:07

## 1 Answer

Let $X=(X_1,X_2)$ and $Y=(Y_1,Y_2)$ be two random vectors. If $X$ and $Y$ are independent, it means, $$P(X=(x_1,x_2);Y=(y_1,y_2))=P(X=(x_1,x_2)) \cdot P(Y=(y_1,y_2))$$ $$=P(X_1=x_1;X_2=x_2) \cdot P(Y_1=y_1; Y_2=y_2)$$

Now, sice $X_1$ and $X_2$ are correlated, they are not independent. That is, since, $$\rho(X,Y)=\dfrac{Cov(X,Y)}{\sqrt{Var X \cdot Var Y}}$$ $Cov(X_1,X_2)\neq0 \implies~X_1$ and $X_2$ are not independent. This in turn means, $$P(X_1=x_1;X_2=x_2) \neq P(X_1=x_1) \cdot P(X_2=x_2)$$

The same goes for $Y_1$ and $Y_2$ as well.

The following remarks are in order:

1] Some interesting properties of covariance, like its invariance under translation; its connection with Variance are described here.

2] Random Variables may be uncorrelated but not independent. But, if they are correlated, they can never be independent.

-
I think the implication $Cov(X_1,X_2)\neq0 \implies~X_1$ and $X_2$ are not independent is not true is general,only the converse is true. –  Mathematics Jan 13 '12 at 4:11
Statement 1, p: $X$ and $Y$ are independent. Statement 2, q: $Cov(X,Y)=0$ We Know, p $\implies$ q. So, ~q $\implies$ ~p –  user21436 Jan 13 '12 at 4:13
This answer applies to discrete random variables. –  Did Jan 13 '12 at 7:20
@Mathematics It is not true that $X_1$ and $X_2$ are not independent implies that $X_1$ and $X_2$ have nonzero correlation. Dependent random variables can be uncorrelated. For example, a standard normal $X$ and its square $Y = X^2$ are dependent but uncorrelated since $E[X^3] = 0$. –  Dilip Sarwate Jan 13 '12 at 11:06