Question

If independent random variable X and Y have variances 36 and 16 resp. What should be the correlation coefficient between (X+Y) and (X-Y) ?

Answer 1

@amistre64 any idea?

Answer 2

well, the respect variances would just be the addition or subtraction of the parts right? we aint go not -1Y or 4X stuff to get in the way

Answer 3

how does variance relate to a correlation coeff?

Answer 4

as per my knowledge correlation coeff is given by
\[r(x,y)=\frac{ Covariance of x and y }{ \sigma x*\sigma y }\]

Answer 5

here we have to fine for x+y and x-y instead of x and y

Answer 6

since X and Y are independant; VAR(X+Y) = VAR(X) + VAR(Y)

Answer 7

but all the covariance stuff i see has to do with the expected value E(XY) = E((X-Xu)(Y-Yu))

Answer 8

yeah that's right variance of x+y=52 and variance of x-y=20

Answer 9

yeah that's where I am stuck how to find covariance if only variances are known

Answer 10

i read that the covar of independant events is zero ....

Answer 11

but this is the correlation between x-y and x+y .... not between x and y

Answer 12

yep x and y are independent but x+y and x-y are dependent hence there should be covariance between them

Answer 13

so at least the under part is known :) sqrt(52*20)

Answer 14

yeah

Answer 15

you dont have parts of this buried in another problem do you? sometimes they say, use this from that in here to do stuff with

Answer 16

nope nothing like that. I have posted the question as appeared in the university question paper. :(

Answer 17

i recall reading something about this a few months back ... but cant quite find the site i saw it on

Answer 18

okay that's fine just let me know if you think of anything.

Answer 19

one thing i have rattling around up in my head is:

if we assume X to have a mean of 0; then mean X+Y = Yu
mean X-Y = -Yu

not too sure how to use that tho

Answer 20

why would it have mean zero?

Answer 21

dunno, its just something in my head .....

lets say we have a set of number a,b,c,....,k whose mean is m, and var is q

an equivalent set of a-m,b-m,c-m,...,k-m will have a mean of 0, and retain the same var of q

Answer 22

the relationship between X and Y is the same regardless if we move at all such that X has a mean of 0

Answer 23

without knowing a mean for the Y, that simply leaves Y as some unknown

Answer 24

okay lets assume but still how we can find E(XY) = E((X-Xu)(Y-Yu))

Answer 25

i believe we want \[E(X-Y)(X+Y)\]
\[E([X-Y]+Yu)([X+Y]-Yu)\]

Answer 26

not sure how this would help tho

Answer 27

yeah that's exactly what we want....

Answer 28

can we construct that from using standard deviations as random X and Y values?

Answer 29

ahh don't think so. can't remember any formula for such condition.

Answer 30

say:

X = {-18, -12, -6,0, 6, 12, 18}

Y = {-12,-8,-4, 0,4,8,12}+Yu

Answer 31

okay then?

Answer 32

then use those to determine X+Y and X-Y

and run those values thru Sxy Sxx and Syy

Answer 33

in other words, start from scratch to see if we can get to a viable result

Excel might help

Answer 34

okay I'll do it later now I got to go..

Answer 35

well, heres what i got wroted up, might want to chk it for mistakes tho :)

Answer 36

after a got home last night, i realized why i was centering X at zero, and for that matter Y as well.

Notice that the operation: X-Xu does this by default. take Xu - Xu, = 0, so Xu is shifted to 0, and all the X values about Xu are shifted center around 0 as well

X-Xu is X "centered at 0"

Y-Yu is Y "centered at 0"

Answer 37

if we take Xu and Yu to be some fixed interval apart and move the whole shebang to X = 0, we retain that same relationship regardless of where they actually started at.

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