# Independence of a random vector [closed]

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?

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## closed as unclear what you're asking by Yes, Etienne, voldemort, Aaron Maroja, sazJan 19 at 18:54

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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

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.

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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

My undestanding of the question is to give an example of four random variable where the above occurs. Here is one: Let A and B be two independent random variables, the distribution actually does not matter much, you can mostly choose one you like, see below. (Note that the components of a product probability space are always independent. So if you have chosen two distributions you like, build the product space and you get independent variable with the given distribution.) Then take X1 := A, X2 := A, Y1 := B, Y2 := B. Apart from exceptional cases X1 and X2 are correlated. It is a good exercise to find out which these exceptional cases are. To nevertheless answer the question more concretely:

Take A and B as independent Bernoulli distributions, i.e. P(A = i, B = j) = p^i (1-p)^(1-i) q ^j (1-q)^(1-j) (i,j \in {0,1}; p,q \in (0,1)) (q=p is allowed, e.g. p= q = 1/2). Then cov(X1, X2) = cov (A, A) = E( (A-E(A))^2)= E(A^2) - E(A)^2 = E(A) - E(A)^2 = p - p^2 = p (1-p) <> 0 since p <>0 and p<>1.

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