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Example text

Assume that 8, is not a stationary point of SSE(8); that is, ( d / a 8 ) S S E ( 8 , ) # 0. Set forth regularity conditions (Taylor's theorem) such that SSE(B,+ h ( 8 , = SSE(8,) Let F, = F( d,), 8, = y - @,>I + h [ aaS S E ( @,)]'( - f(B,), to There must be a A* such that 8, - 6,) -k o( A'). ). Thus s s ~8,[ + A( 8, 4. - e,)] iSSE( 8,) for all X with 0 < X c A*. ) Supply the missing details in the proof of the following result. THEOREM. Let Conditions: There is a convex, bounded subset to S such that: S of RP and 8, interior 1.

And we shall show it directly below. Put A' (0,1,0, - 1). Then exists, =i Since it is zero for every n, X'[lim,,,,Q,(8)~el~o]h = 0 by continuity of A'AX in A . Since it is an age distribution, there is some (possibly unknown) maximum attained age c that is biologically possible. Then for any continuous function g ( x ) we must have l:}g(x)lp3(x) dx < 00, so that by Kolmogorov's strong law of large numbers (Tucker, 1967) 22 UNIVARIATE NONLINEAR REGRESSION Applying these facts to the treatment group, we have Applying them to the control group, we have Then Suppose we let Fiz(x,,x,) be the distribution function corresponding to the discrete density and we let F3(x,) be the distribution function corresponding to p 3 ( x ) .

Golub and Pereyra (1973) obtain an analytic expression for ( a / a p ’ ) / ( p ) and present an algorithm exploiting it that is probably the best of its genre. Marquardt’s algorithm is similar to the Gauss-Newton method in the use of the sum of squares SSE,(8) to approximate SSE(0). The difference between the two methods is that Marquardt’s algorithm uses a ridge regression improvement of the approximating surface instead of the minimizing value 6,. For all 6 sufficiently large 19, is an improvement over 8, [SSQ8B,) is smaller than SSE(fl,)] under appropriate conditions (Marquardt, 1963).

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