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careless25

Real Member

Registered: 2008-07-24

Posts: 560

Correlation Coefficient

This was a question on my exam and I am not sure I did it correctly. Can someone show me the steps to derive it?

Given X1, X2, X3, X4 are random uncorrelated variables with variance of 1 and mean of 0. Find the correlation coefficient for (X1 + X2 and X2 + X3).

I got the answer as -1/2. is this correct?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,627

Re: Correlation Coefficient

hi careless25,

I'm well out of my comfort zone with this one so what follows may be rubbish.

Intuitively, I feel it ought to be zero, as the variables are all uncorrelated.

So I thought I'd better check on the formulas:

http://www.math.uiuc.edu/~hildebr/461/variance.pdf

So we have V[X1] = V[X2] = V[X3] = V[X4] = 1

So V[X1+X2] = V[X3+X4] = 1 + 1 = 2

and V[X1+X2+X3+X4] = 4

So Cov[X1+X2.X3+X4] = 0.5(V[X1+X2+X3+X4] - V[X1+X2] - V[X3+X4]) = 0.5(4-2-2) = 0

P[X1+X2,X3+X4] = Cov[X1+X2.X3+X4] /(ROOT(V[X1+X2]*V[X3+X4])= 0/2 = 0

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

careless25

Real Member

Registered: 2008-07-24

Posts: 560

Re: Correlation Coefficient

Hi Bob Bundy,

I am asking for the Correlation coefficient of X1 + X2 and X2+X3. Not X1+X2 and X3+X4.

I agree with your answer if it was that .

Bob

Administrator

Registered: 2010-06-20

Posts: 10,627

Re: Correlation Coefficient

Oh sorry.  I'll recalculate. 

but it'll have to be a little later as I've got to log off now.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Bob

Administrator

Registered: 2010-06-20

Posts: 10,627

Re: Correlation Coefficient

revised version 1:

V[X1] = V[X2] = V[X3] = V[X4] = 1

So V[X1+X2] = V[X2+X3] = 1 + 1 = 2

and V[X1+X2+X2+X3] = 4

So Cov[X1+X2.X2+X3] = 0.5(V[X1+X2+X2+X3] - V[X1+X2] - V[X2+X3]) = 0.5(4-2-2) = 0

P[X1+X2,X3+X3] = Cov[X1+X2.X2+X3] /(ROOT(V[X1+X2]*V[X2+X3])= 0/2 = 0

revised version 2:

and V[X1+X2+X2+X3] = V[X1+2X2+X3] = 1 + 4 + 1 = 6

So Cov[X1+X2.X2+X3] = 0.5(V[X1+X2+X2+X3] - V[X1+X2] - V[X2+X3]) = 0.5(6-2-2) = 1

P[X1+X2,X2+X3] = Cov[X1+X2.X2+X3] /(ROOT(V[X1+X2]*V[X2+X3])= 1/2

This feels better to me.  The X1+X2 and X2 + X3 variables have a 'bias' towards X2 so I'd expect them to correlate positively.

I'm going to try to construct a simulation to sort this out.  It may take a while ........................

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Bob

Administrator

Registered: 2010-06-20

Posts: 10,627

Re: Correlation Coefficient

Here's how I constructed the simulation:

(1)  I used MS Excel.

(2)  I used the function RAND() to generate numbers in the range 0 to 1.

(3) I used the function NORMSINV() to convert the number from step (2) into an x value from a Normal distribution, with mean=0, variance=1.

(4)  I copied the formulas into 200 rows.

(5)  To check that these values were as required I computed the mean and variance of the numbers.

(6) I repeated this for three more columns.

(7) In two more columns I computed X1+X2 and X2+X3.

(8) I made a scatter graph for these two columns and also computed the correlation coefficient.

This screen shot shows the last few rows, and the graph.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

careless25

Real Member

Registered: 2008-07-24

Posts: 560

Re: Correlation Coefficient

I must have messed up some of my signs somewhere.

Thanks Bob!

EDIT:

Actually how is V[X1+X2+X2+X3] = 6?? shouldn't it be 4 as V[X1 + 2*X2 + X3] = V[X1] + 2*V[X2] + V[X3] = 1+2+1?

Last edited by careless25 (2014-08-08 05:46:29)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,627

Re: Correlation Coefficient

VAR[cX] = c^2.VAR[X]

It happens because you square the differences from the mean.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob