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Topic: exp reg again
Replies: 0

exp reg again
Posted: Jan 4, 1997 1:50 PM

My apologies to you if you have already seen this but apparently there were
transmission problems the first time I posted it. If you have already seen
this, hit the delete key now. Happy New Year to all.

------

Fitting an Exponential Regression Curve "Exactly"

The game Dead Dice, where you begin with a fairly large number of dice and
roll them all, removing the sixes, and then rolling again and removing the
sixes until you are left with just one die, is a great way to demonstrate
equation is Y=72*(5/6)^X, where Y is the number of dice after X throws. I
wrote a short program to simulate this. (Although 72 dice is not all that
many--I recommend it as a class activity and give credit to Martha Lowther
of the Tatnall School for showing it to me.)

I generated the data and fit an exponential regression equation (with
"correlation" r = -.9717). For my data I got an equation of
Y=86.9*(.8086)^X. Hmm, that 86.9 looked big and .8086 was not terribly
close to 5/6. So I looked at the residuals using both the theoretical
equation Y=72*(5/6)^X and the exponential regression equation. For the
regression equation the sum of the squares of the residuals was 490.88 and
for the theoretical equation the sum of squares was 50.97 - a change by a
factor of almost ten!

Of course the exponential regression is not fit to the actual data, but
instead is a linear regression fit to (X,lnY) and then transformed back to
exponential form. This technique does not minimize the sum of squares of
the residuals, sum(Yi-A*B^Xi)^2 , but instead minimizes the sum of the
residuals of the log of the data, i.e. it finds a and b that minimizes
sum(lnYi-a-bXi)^2. Then it calculates A=e^a and B=e^b for the equation
Y=A*B^X.

It isn't surprising that this A and B do not minimize the sum of the squares
of the residuals for the exponential equation; they weren't calculated that
way. Explanations for using this method often include (1) it's easy and
doable, even by "hand" and (2) minimizing the sum of squares directly leads
to equations which are not easily solvable - insoluble in closed form. So I
looked at the problem of minimizing the sum of squares (SS from now on) for
the exponential equation.

To minimize SS=sum(Yi-A*B^Xi)^2 we need to take the partial derivatives of
SS wrt A and B and set them equal to zero. If we do so and simplify,
remembering that in this context A and B are the variables and the Xi's and
Yi's are constants, we get these two equations:

sum(Yi*B^Xi) = Asum(B^(2Xi))

sum(Xi*Yi*B^Xi) = Asum(Xi*B^(2Xi))
Eliminate A and get

sum(Yi*B^Xi)/sum(B^(2Xi)) = sum(Xi*Yi*B^Xi)/sum(Xi*B^(2Xi)).

Or

sum(Yi*B^Xi)/sum(B^(2Xi))-sum(Xi*Yi*B^Xi)/sum(Xi*B^(2Xi))=0.

This looks difficult but we can solve it numerically using Newton's method
(actually an approximation to Newton's method using nDeriv instead of the
actual derivative). If the Xi's are in L1 and the Yi's are in L2, the
equation for which we must find a zero is:

sum(L2*B^L1)/sum(B^(2L1))-sum(L1*L2*B^L1)/sum(L1*B^(2L1)).

I stored this in Y1 and used 5/6 as my initial seed value and iterated. (I
used X in place of B but if you differentiate wrt B as nDeriv allows you
will get the correct answer.) Warning: If you try this make sure your
batteries are charged. Each iteration can take over a minute of calculation
time. This might not sound like a lot, but you'll be surprised.

After determining the value for B I calculated A=sum(Yi*B^Xi)/sum(B^(2Xi)).
For the data I was looking at A=71.32699599 and B=.8387668059. Using the
equation Y=A*B^X with these values, I then calculated the residuals and
found the SS. It was 45.218, smaller then either of the other two.

My final conclusion is that with the power of even a small computer like the
TI-82/83 we can solve a problem considered impossible in the past. If our
criterion for the "best fitting" equation is one that minimizes the sum of
squares of the residuals, why accept a poor approximation? (Time isn't
everything.)

The above technique can be applied for PwrReg--fitting and equation of the
form Y=A*X^B. It can even be applied to equations of the form Y=A*X^B+C, as
in Newton's Law of Cooling experiments. This post is too long already so I
leave it to you to do the calculations.

--
Doug Kuhlmann (508)-749-4242