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Linear (?) Model Question

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elodie....@gmail.com

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Nov 12, 2007, 1:07:18 PM11/12/07
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Consider the model

Yi=1+beta*Xi+beta^2*Xi^2+epsilon_i
withi=1...n

where the epsilon_i are uncorrelated random vars with mean 0 and var
sigma^2.

Xi....Xn are known constants. beta and sigma^2 are unknown parameters.

Question 1:
Is this a linear model?
I would say that it is not. That model is not linear in the
parameters, because of beta^2.

Question 2:
Indicate how to find the least squares estimator (LSE) of beta. Is
this a straightforward estimator? Does one need an algorithm to obtain
the estimator?

I worked out the (X'X)^-1*X'Y estimator. It looks like I am getting
two equations of the form
beta=f1(sumXi, other parameters)
beta^2=f2(sumXi, other parameters)

This will require numerical computation, right?

Does anybody have a pointer to some material on this model.

Thanks for your help.

Jack Tomsky

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Nov 12, 2007, 1:45:32 PM11/12/07
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1. It is not a linear model, as you say, because a linear model means linearity in the parameters.

2. To get the least-squares estimate for beta, first let Zi = Yi - 1. Then the sum of squares is

SS = Sum[Zi - beta*Xi - (beta*Xi)^2]^2

After taking derivatives, d(SS)/d(beta), and setting it equal to zero, you end up with a cubic equation for beta.

There is a closed-form expression for the three roots. Either all three betas are real or one of them is real and the other two are complex conjugates.

In a practical sense, you could probably solve this numerically. One way is to use Excel's Solver to minimize the sum of squares SS.

Jack

Ray Koopman

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Nov 12, 2007, 2:30:17 PM11/12/07
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If the gradient has three real roots then SS has two local minima,
at the outer roots; the middle root gives a local max.
If you look for *a* solution, you may get the wrong one.
If you turn a minimizer loose, it may return the wrong min.

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