R help - Aug 2006 - solving non-linear system of equations

Didn't get any useful response to the following question. Trying again.
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I can't seem to get computationally stable estimates for the following
system:

Y=a+bX+cX^2+dX^3, where X~N(0,1). (Y is expressed as a linear combination
of the first three powers of a standard normal variable.) Assuming that
E(Y)=0 and Var(Y)=1, one can obtain the following equations after tedious
algebraic calculations:

1) b^2+6bd+2c^2+15d^2=1
2) 2c(b^2+24bd+105d^2+2)=E(Y^3)
3) 24[bd+c^2(1+b^2+28bd)+d^2(12+48bd+141c^2+225d^2)]=E(Y^4)-3

Obviously, a=-c. Suppose that distributional form of Y is given so we know
E(Y^3) and E(Y^4). In other words, we have access to the third and fourth
raw moments. How do we solve for these four coefficients? I reduced the
number of unknowns/equations to two, and subsequently used a grid
approach. It works well when I am close to the center of the support, but
fails at the tails. Any ideas?

Hakan Demirtas

Are you saying you did not receive my reply stamped "8/10/2006 2:12 
AM" in my Sent folder, or that my reply was not useful?  In case the 
former is correct, my comments were as follows:

"Have you tried writing a function to compute SS = sum of squares 
deviations between the the left and right hand sides of your three 
equations, then using 'optim'?  See also Venables and Ripley (2002) 
Modern Applied Statistics with S, 4th ed. (Springer).

p.s.  I don't see how it's obvious that 'a=-c'."

	  If you think these comments are not useful, I would appreciate the 
courtesy of a reply.  I have solved many superficially similar problems 
in this way, and I would like to know why this would not work for your 
case.

       hope this helps.
       Spencer Graves	

HAKAN DEMIRTAS wrote:
> Didn't get any useful response to the following question. Trying again.
>
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> I can't seem to get computationally stable estimates for the following
> system:
> 
> Y=a+bX+cX^2+dX^3, where X~N(0,1). (Y is expressed as a linear combination
> of the first three powers of a standard normal variable.) Assuming that
> E(Y)=0 and Var(Y)=1, one can obtain the following equations after tedious
> algebraic calculations:
> 
> 1) b^2+6bd+2c^2+15d^2=1
> 2) 2c(b^2+24bd+105d^2+2)=E(Y^3)
> 3) 24[bd+c^2(1+b^2+28bd)+d^2(12+48bd+141c^2+225d^2)]=E(Y^4)-3
> 
> Obviously, a=-c. Suppose that distributional form of Y is given so we know
> E(Y^3) and E(Y^4). In other words, we have access to the third and fourth
> raw moments. How do we solve for these four coefficients? I reduced the
> number of unknowns/equations to two, and subsequently used a grid
> approach. It works well when I am close to the center of the support, but
> fails at the tails. Any ideas?
> 
> Hakan Demirtas
> 
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