# Correlation coefficient confusion

1. Jan 28, 2006

### happyg1

Hi,
Here's my question:
Determine the correlation coefficient of the random variables X and Y if var(X)=4 and Var(Y)=2 and var(X+2Y)=15
I said let Z=X+2Y and then Var(Z)=E[Z^2]-E[Z]^2
Then I multiplied the thing out:
15=Var(z)=E[X^2]+4E[XY]+4E[Y^2]-E[X]-2E[Y]
I know that Var(X)=E[X^2]-E[X]^2 and var(Y)=E[y^2]-E[y]^2
I tried plugging in the mu and sigmas for the means and variances and I also know that E[X^2]=sigma^2+mu^2
I've been going in circles and getting nowhere for awhile. I know I'm missing something, but I don't know what.
the formula for the correlation coefficient, rho= cov[XY]/(sigmaX)(sigmaY)
doesn't get me anywhere, either. I know the sigmas but I am stuck. I have 3 pages of equations going in circles.
Any hint or pointer will be greatly appreciated.
Thanks,
CC

2. Jan 28, 2006

### happyg1

Sorry...I didn't mean to post it twice. My power went out and the computer blinked on and off and I hit submit again. Please delete one of these.
Sorry sorry sorry
CC

3. Jan 28, 2006

### 0rthodontist

I think the equation you want to use is var(X + Y) = var(X) + var(Y) + 2 cov(X, Y).

Last edited: Jan 28, 2006
4. Jan 28, 2006

### happyg1

Ok but how do I get the cov(X,Y) just from the variances I am given? I don't see how to get the means to plug in from what I was given.
CC

5. Jan 29, 2006

### 0rthodontist

You have all the information to use the formula directly except you have var(X + 2Y) instead of var(X + Y). You need to expand out var(X + 2Y). You don't need to know the means.

6. Jan 29, 2006

### happyg1

Ok,
I got this one at last...thank you! Where did that formula come from?
My book has dedicated only a half a page to the derivation and explanation of the correlation coefficient. My next few problems are all about the correlation coefficient. I am wondering if anyone knows a good internet source with more info on this topic. I have searched around, but I haven't found anything on my level.

My next question says X1, X2 and X3 are independent and rho12=.3, rho13=.5 and rho23=.2 and says that the variances are equal. It wants the correlation coefficient of Y=X1+X2 and Z=X2+X3. I don't know where to start. I'm really really confused and I feel like I ned to buy another book.
Any help will be appreciated.
CC

7. Jan 29, 2006

### 0rthodontist

I've never done anything with the correlation coefficient before, but if rho12 = the correlation coefficient between X1 and X2, then I don't understand your question. It seems like that should be 0 since the covariance between 2 independent variables is 0. Maybe you mean mu12? I don't understand the notation.

8. Jan 29, 2006

### happyg1

Hi again,
Let $$X_1 X_2 and X_3$$ be random variables with equal variances but with correlation coefficients $$\rho_12=.3,\rho_13=.5 and \rho_23=.2$$ Find the correlation coefficient of the linear functions $$Y=X_1+X_2 and Z=X_2+X_3$$.
The problem doesn't say they're independent.....then, as you pointed out, it makes no sense.

9. Jan 29, 2006

### 0rthodontist

All right (still assuming what rho12 means is cov(X1, X2)/sigma^2) what you need to do is expand out cov(X1 + X2, X2 + X3) in terms of variances and covariances, then divide by sigma^2. I found a nice formula in my book to do that but you can probably find it online.

Edit: also you'll need to find a constant to divide by since Sigma(X1 + X2) * Sigma(X2 + X3) in the denominator of the left hand side will work out to a constant times the variance.

Last edited: Jan 29, 2006
10. Jan 29, 2006

### happyg1

OK,
I worked the whole thing out like this:
cov(X1+X2,X2+X3)=cov(X1,X2)+cov(X1,X3)+cov(X2,X2)+cov(X2,X3)
the third term is zero, and I plugged in the $$\rho$$ values that I was given. This leads to:
$$\rho_{X1+X2,X2+X3}=\frac{.3\sigma^2+.5\sigma^2+.2\sigma^2}{\sigma_{X1+X2}\sigma_{X2+X3}}$$
I solved for the variances in the denominator by using
$$\sigma_{X1+X2}=Var(X1+X2)=Var(X1)+2cov(X1,X2)+var(X2)=\sigma^2+2(.3\sigma^2)+\sigma^2=2.6\sigma^2$$
and similarly for $$\sigma_{X2+X3}=2\sigma^2+.4\sigma^2$$
Since all of the variances of X1 X2 X3 are equal. I plug it into my equation for $$\rho_{X1+X2,X2+X3}$$
I got:
$$\rho_{X1+X2,X2+X3}=\frac{.3 \sigma^2+.5 \sigma^2+.2 \sigma^2}{\sqrt {(2.6\sigma^2)(2.4\sigma^2)}}$$
Which gives me:
$$\frac{\sigma^2}{\sigma^2(2.497999199...)}=.400320384$$
BOB says that the answer should be .801, so I'm off by a factor of 2 and i can't FIND IT!
Help
CC

Last edited: Jan 29, 2006
11. Jan 29, 2006

### 0rthodontist

cov(X2, X2) = ?

Last edited: Jan 29, 2006
12. Jan 29, 2006

### happyg1

ok,
cov(X2,X2)=var(X2)....which is another$$\sigma^2$$? and not zero?

13. Jan 29, 2006

### 0rthodontist

Yep, and everything else works out correctly (except for a notational quibble about sigma(X1 + X2) instead of sigma^2(X1 + X2))

Last edited: Jan 29, 2006
14. Jan 29, 2006

### happyg1

Wait wait
I thought that the denominator was supposed to be the standard deviations mulitplied together. Is that what you mean?

15. Jan 29, 2006

### 0rthodontist

Yes it should be and you took the square root correctly, but you wrote sigma(X1 + X2) = var(X1 + X2) and sigma(X2 + X3) = var(X2 + X3) at one point. Not that it mattered.

Last edited: Jan 29, 2006
16. Jan 29, 2006

### happyg1

And here's the next one:(It took me all day to get that last one, but dang it, I LEARNED something)
Find the variance of the sum of 10 random variables if each has variance 5 and if each pair has correlation coefficient .5.
Now, before I start messing up lots of sheets of paper, Is this just an expansion of the last one?....and does that mean that $$\rho_{X1,X2}=\rho_{X1,X3}=...\rho_{X1,X10}$$ and similarly for each pair?
Shove me in the right direction please.
THANK YOU SO MUCH for your pointers on the last one. This stuff is really hard for me.
And I see what you mean about my error.
CC

17. Jan 29, 2006

### 0rthodontist

Yes, I guess that's what it would mean. But it may be easier to find the covariance between each pair instead of the correlation coefficient before you start expanding. It's not quite the same as the last since you're expanding a variance instead of a covariance, but since var(X) = cov(X, X) you can use the same expansion formula.

18. Jan 30, 2006

### happyg1

aall right,
Here's what I am doing:
I am using this formula:
$$Var\left(\sum_{i=1}^{10} X_i\right)=\sum_{i=1}^{10}Var(X_i)+2\sum\sum_{i<j}Cov(X_i X_j)$$
and I know from what I am given that:
$$\rho_{i,j}=.5=\frac{Cov(X_i,X_j)}{\sigma_{i,j}=\sqrt 5}=>(.5 \sqrt 5)=Cov(X_i,X_j)$$
So ultimately I got:
$$((10(5)) + .5 \sqrt 5 (9+8+7+6+5+4+3+2+1))=95$$
Is that remotely correct? It seems to be quite large. I don't know what I'm supposed to get, i.e. no BOB answer his time.
Any guidance or hints will be appreciated.
Thanks,
CC

Last edited: Jan 30, 2006
19. Jan 30, 2006

### 0rthodontist

The 10(5) is right but I don't know what you're doing in the second part. First thing, remember that the definition of the correlation coefficient is the covariance of the two variables divided by the _product_ of the two standard deviations. I don't see how you're summing 9 + 8 + ... + 1 and you forgot the factor of 2.

You're using a more complicated formula than you need to. That's the one my book gives, but if you use the generalized way of expanding the covariance it is easier to understand. Summarizing that again, an expression like var(X1 + X2 + ... + Xn) = cov(X1 + X2 + ... + Xn, X1 + X2 + ... + Xn) is expanded similar to if you were multiplying out (X1 + X2 + ... + Xn)(X1 + X2 + ... + Xn). For each term on the left, you take the covariance of that term and each term on the right, just as you do when expanding the product of two polynomials, except with covariance instead of multiplication.

20. Jan 30, 2006

### happyg1

Ok,
I think I understand what you're saying, and I do see that I forgot my 2.
Here's how I got the sum(1+2+....+9):
$$Var\left(\sum_{i=1}^{10} X_i\right)=\sum_{i=1}^{10}Var(X_i)+2\sum\sum_{i<j}Cov(X_i X_j)$$
gives:
$$10(5)+2(cov(X_1X_2)+Cov(X_1 X_3)+Cov(X_1 X_4)+...+Cov(X_1 X_{10})$$ nine terms, all $$\sqrt 5(.5)$$
continuing on to i=2,
$$Cov(X_2 X_3)+Cov(X_2 X_4)+...+Cov(X_2,X_10)=>$$8 terms, all $$\sqrt 5(.5)$$
Continuing, when i=3, there are 7 terms, all $$\sqrt 5(.5)$$
and so on until i=9 where there's only one term.
factor out the common $$\sqrt 5(.5)$$
That's how I got the sum, which is 45, the total number of terms in my expansion there.
My revised total is 150.62, which still seems wrong.
tell me what you think.
CC

Last edited: Jan 30, 2006