Multiplying two dependent normal distributions (statistics)

[mb12777]

So I've recently come across a statistics problem that interested me, and it's not covered at the level I'm currently being taught at (my teacher still wants me to have a go at it). I'm guessing it'll be easiest if I just post the question and where I'm up to.

Q: X and Y are independent random variables with variances 4 and 12 respectively. Find the correlation coefficient between 2X+Y and 5X-Y.

Where I'm up to so far

X~N(x, 4) and Y~N(y, 12) (I'll use x and y as the means, I don't know how to get the actual symbols)

So call 2X+Y A, and 5X-Y B, then A~N(2x+y, 28) and B~N(5x-y, 112)

I'll just say that there were n numbers taken from each distribution

Then I got S_aa=28n, and S_bb=112n, so the denominator for the correlation coefficient should be the square root of 28n*112n, which comes out as 56n. So I already know the denominator is 56n

(I'll also use {Sum} in place of a summation sign, again I don't know how to get those on here)

Where I'm stuck is at finding the numerator. I expanded out S_ab={Sum}(a_i-a)(b_i- to get this...

S_ab= {Sum}a_ib_i - b{Sum}a_i - a{Sum}b_i +ab

But {Sum}a_i = a*n, and similar for {Sum}b_i

So then S_ab = ab - 2abn + {Sum}a_ib_i

=> S_ab = ab(1-2n) + {Sum}a_ib_i

Now, at this point I'm thinking I could just find the mean of A*B, and multiply this by n to get {Sum}a_ib_i. But I have no idea how I could find a distribution for A*B, all I know is I can't simply multiply the means of each distribution.

I hope this made sense, and any help would really be appreciated.

Edited March 29, 2015 by mb12777

[mb12777]

Bump it up... I'm still clueless on this, I've tried googling it as well. Can anyone help at all?

[Linear]

Of all of the things people ask me, stats must be up there with the most asked topic.

I can't do it. I'm ashamed of it.

I thought i'd let you know - anyone who can get through stats without turning into a vegetable gets top marks from me.

[mb12777]

Of all of the things people ask me, stats must be up there with the most asked topic.

I can't do it. I'm ashamed of it.

I thought i'd let you know - anyone who can get through stats without turning into a vegetable gets top marks from me.

Haha, well thanks for a reply. It does get pretty confusing at times, and plain boring at other times.

[impwarhamer]

goddamit I have enough stats at school!