The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » Courses » ap-stat

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: exponential regression
Replies: 4   Last Post: Jan 5, 1997 9:57 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]

Posts: 119
Registered: 12/6/04
exponential regression
Posted: Jan 2, 1997 3:39 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

The game Dead Dice, where one begins with a fairly large number of dice and
rolls 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
exponential decay. If you start with 72 dice, the theoretical decay
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-blnXi)^2. Then it calculates A=e^a and B=e^b for the equation

It isn't surprising then 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 doing it this way 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 = 0. 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)).



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:


(I used X in place of B but if you differentiate wrt B as nDeriv allows you
will get the correct answer.)

I stored this in Y1 and used 5/6 as my initial seed value and iterated.
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

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.

P.S. Thanks to Al Coons for proofreading this.

Doug Kuhlmann (508)-749-4242
Phillips Academy
Andover, MA 01810

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.