dave fournier davef at otter-rsch.com
Tue Nov 16 04:21:47 PST 2010

```Chris Gast wrote:

I think this is discussed in the paper Hans and I wrote.
anyway since  uhat(x) is the value of uhat which maximizes L(x,u)
It follows that

L_u(x,uhat(x))=0  for all x so differentiating wrt x you get

L_xu + L_uu uhat'(x)=0   so that

uhat'(x) = - L_uu^{-1} * L_xu

If there are n x's and m u's then uhat'(x) is an n x m matrix.

> Thanks again, Dave.  Could you clarify what the general entry of the
> dx_i/du_j, but I think these would all be 1's or 0's (since, for
> instance , Shat_1 = shat + uhat_1).  It is likely that I am
> misunderstanding something.
>
>
> Thanks,
>
> Chris
>
>
>
>
> -----------------------------
> Chris Gast
> cmgast at gmail.com <mailto:cmgast at gmail.com>
>
>
> On Mon, Nov 15, 2010 at 7:00 PM, dave fournier <davef at otter-rsch.com
> <mailto:davef at otter-rsch.com>> wrote:
>
>     OK,
>
>     there are inly ar efew thing to play with.  for notation let
>
>     F(x,u) be the likelihood for x and u.
>
>     let
>
>       L(x) = int F(x,u) du
>
>     be what we get after integrating out u either exactly or by the
>     laplace approx.
>
>     and let xhat uhat(xhat) be the miximizing values.
>
>     We have   the Hessians L_xx(xhat), and F_uu(xhat,uhat(xhat)
>     Then the covariance matrix for x,u is assumed to be
>
>         L_xx^{-1}                        L_xx^{-1} * uhat'(x)
>
>          uhat'(xhat)*L_xx^{-1}    F_uu^{-1} +
>     uhat'(xhat)*L_xx^{-1}*uhat'(x)
>
>     The idea is that u-uhat(xhat) is independent of xhat  i.e the only
>     correlation is between
>     xhat and uhat(x)
>
>     Put in transposes where necessary.
>
>     For any function of x,u the covariance is computed by the delta
>     method.
>
>
>
>
>
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