here is one VERY simple example
the COV is the AVERAGE of the PRODUCTS of the deviations around the means
... of two variables
if the cov is + .. there is a + relationship between X and Y, if it is -,
there is a - relationsip between X and Y
X Y devX devY (devX)(devY)
10 20 1-2 - 2
9 24 0 2 0
8 22 -10 0
the sum of the products of the deviations around the two means is -2
the average is -2 / 3 = -.67 = the covariance
now, some books will divide by n-1 or 2 in this case ... which would = -2/2
= -1 for the covariance
here is the minitab output ... note, mtb and most software packages will
divide by n-1
Y - *
-
-
22.5+
- *
-
-
-
20.0+ *
-
+-+-+-+-+-+--X
8.00 8.40 8.80 9.20 9.60 10.00
MTB prin c1-c5
Data Display
Row X Y devX devY product
1 10 20 1 -2 -2
2 9 24 0 20
3 8 22 -1 00
MTB sum c5
Sum of product
Sum of product = -2.
MTB cova c1 c2
Covariances: X, Y
XY
X 1.0
Y-1.0 4.0
MTB NOTE ... THE INTERSECTION OF THE X AND Y OUTPUT ABOVE ... -1 ... IS
THE COVARIANCE IN THIS PROBLEM
by the way, the CORRELATION is
MTB corr c1 c2
Correlations: X, Y
Pearson correlation of X and Y = -0.500
At 06:06 PM 2/2/02 +, Maja wrote:
Hi everyone,
we just leanred the cov, and I'm trying to apply the formula
cov(X,Y)=E(XY)-E(X)E(Y) to the following question
f(x,y)=(0.6)^x(0.4)^1-x(0.3)^y(0.52)^1-y(2)^xy where poss. values for X and
Y are x=0,1 and y=0,1.
The prof. never gave us any examples for cov and neither does the text book.
Now I don't know what values to plug in for X and what values to plug in for
Y.
Could someone PLEASE explain to me how am I supposed to know what vales to
plug into the equation???
TNX
Maja
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