1. ## correlation coefficient help

Hey, im a first year in leeds uni doing maths and there are some things im starting to get quiet stressed over and they arnt even hard so you'll probably see quite a few of my posts pop up with my exams coming in a few days.

The question im stuck on is:

Let X1 , X2 and X3 be independent random variables with zero mean and variance 1.
Find the correlation coefficient between Y1 = 2(X1) + (X2) and Y2 = (X2) ¡ 2(X3) .

I really dont know where to start and its quite conserning because i dont think its even hard .

I beleive that cov(x,y) = E(XY) -E(X)E(Y) but with a zero mean of the variables im unsure how this works. Thanks for all help.

2. Originally Posted by dhowlett
Hey, im a first year in leeds uni doing maths and there are some things im starting to get quiet stressed over and they arnt even hard so you'll probably see quite a few of my posts pop up with my exams coming in a few days.

The question im stuck on is:

Let X1 , X2 and X3 be independent random variables with zero mean and variance 1.
Find the correlation coefficient between Y1 = 2(X1) + (X2) and Y2 = (X2) ¡ 2(X3) .

I really dont know where to start and its quite conserning because i dont think its even hard .

I beleive that cov(x,y) = E(XY) -E(X)E(Y) but with a zero mean of the variables im unsure how this works. Thanks for all help.
You want to find the population correlation coefficient $\rho_{Y_1,Y_2}$

In you book,
$\rho{X,Y}=\frac{cov(X,Y)}{\sigma_X \sigma_Y}=\frac{E[(X-\mu_X)(Y-\mu_Y)]}{\sigma_X \sigma_Y}$

You can simply substitute $Y_1$ into $X,$ and $Y_2$ into $Y$.

Since $\sigma_{Y_1}=1, \sigma_{Y_2}=1, \mu_{Y1}=0, \mu_{Y2}=0$,

The result becomes $\rho_{Y_1,Y_2}= cov(Y_1,Y_2)=E[Y_1Y_2]$

3. Thankyou. Though i am confused. The question says the random variables are independent, doesn't that mean the correlation coefficient is 0? the answer to the question is 0.2.

E(xy) = E(x)E(y) doesn't it when they are independent?

This gives E(y1) = 2E(x1) + E(x2) and E(y2) = E(x2) - 2E(x3) but the question states the mean of all these random variables is 0 thus that gives 0 for both of them. But the answer is 0.2 :S.

4. Originally Posted by dhowlett
Thankyou. Though i am confused. The question says the random variables are independent, doesn't that mean the correlation coefficient is 0? the answer to the question is 0.2.

E(xy) = E(x)E(y) doesn't it when they are independent?

This gives E(y1) = 2E(x1) + E(x2) and E(y2) = E(x2) - 2E(x3) but the question states the mean of all these random variables is 0 thus that gives 0 for both of them. But the answer is 0.2 :S.
Are you looking for the linear correlation or the population correlation?

5. i dont know unfortunetly, i copyed out the whole question. Im really getting lost on these "introduction to probability" exams.

6. Originally Posted by dhowlett
i dont know unfortunetly, i copyed out the whole question. Im really getting lost on these "introduction to probability" exams.
I think you are looking for the population correlation coefficient, $\rho_{Y_1,Y_2}$

When $Y_1$ and $Y_2$ are dependent random variables, after algebraic reduction, it became $\rho_{Y_1,Y_2}=Cov(Y_1,Y_2)=E[XY].$

In the case where $Y_1$ and $Y_2$ are independent, $E[XY]$ becomes $E[X]E[Y].$

The linear correlation coefficient is $r=\frac{\Sigma Y_1Y_2}{\sqrt{(\Sigma Y_1)^2(\Sigma Y_2)^2}}$. If this does not look familiar, the population correlation coefficient must be what you are looking for.

7. Originally Posted by dhowlett
Thankyou. Though i am confused. The question says the random variables are independent, doesn't that mean the correlation coefficient is 0? the answer to the question is 0.2.

If the two random variables are independent, the coveriance =0. It follows that the population correlation coefficient must equal to zero. It cannot be anything but zero; however, it's not true for the linear regression coefficient of correlation.

For independent variable, the linear correlation need not be zero. 0.2 can be true.

I have a hunch that you are to find the linear correlation coefficient.