^
^
Y = ln f = ln (a RRbRec) = ln a + b ln RR + c ln Re
= β0+ β1X1+ β2X2
(A-5)
where β0 = ln a
β1 = b
β2 = c
X1 = ln RR
X2 = ln Re.
Now we can restate the problem in a linear form
(
)
(
)
2
2
n
n
^
= ∑ Yi - β0 - β1X1,i - β2X2,i .
min ∑ Yi - Yi
(A-6)
i =1
i =1
The summation index i has now been added. The summation occurs over the total
number of observations n. To find the minimum for this expression with respect to
respect to each of the parameters and set the result to zero in each case
(
)
n
/ β0 = ∑ Yi - β0 - β1X1,i - β2X2,i = 0
(A-7)
i =1
(
)
n
/ β1 = ∑ X1,i Yi - β0 - β1X1,i - β2X2,i = 0
i =1
(A-8)
(
)
n
/ β2 = ∑ X2,i Yi - β0 - β1X1,i - β2X2,i = 0 .
i =1
(A-9)
We now have three linear equations in the three unknown parameters by rear-
ranging as follows
n
n
n
β0 + β1 ∑ X1,i +β2 ∑ X2,i = ∑ Yi
(A-10)
i =1
i =1
i =1
n
n
n
n
β0 ∑ X1,i + β1 ∑ (X1,i )2 +β2 ∑ X2,i X1,i = ∑ Yi X1,i
i =1
i =1
i =1
i =1
(A-11)
n
n
n
n
β0 ∑ X2,i + β1 ∑ (X1,i X2,i ) +β2 ∑ (X2,i )2 = ∑ YiX2,i .
i =1
i =1
i =1
i =1
(A-12)
These equations can be written in matrix form as
A11A12 A13 β0 C1
A21A22 A23 β1 = C2
(A-13)
A31A32 A33 β2 C3
where A11 = n
n
A12 = A21 = ∑ X1,i
i =1
n
A13 = A31 = ∑ X2,i
i =1
n
(X1,i )2
A22 = ∑
i =1
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