Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

closed as unclear what you're asking by Fundamental, Etienne, voldemort, Aaron Maroja, saz Jan 19 at 18:54

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

    
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
2  
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

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer
    
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.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.