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lines 6-157 of file: example/user/data_mismatch.cpp
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{xrst_begin data_mismatch.cpp}
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Random Effects Variance May Cause Data Mismatch
###############################################

Model
*****

.. math::

    \B{p}( z | \theta ) \sim \B{N} ( \theta , \sigma_z^2 )

.. math::

    \B{p}( y | \theta ) \sim \B{N} [ \exp( u ) \theta , \sigma_y^2 ]

.. math::

    \B{p}( u | \theta ) \sim \B{N} ( 0 , \sigma_u^2 )

The fixed likelihood
:ref:`g(theta)<theory@Fixed Likelihood, g(theta)>`
is

.. math::

    g( \theta ) = \frac{1}{2} \left[
        \log ( 2 \pi \sigma_z^2 ) + ( z - \theta )^2
    \right]

The random likelihood
:ref:`f(theta, u)<theory@Random Likelihood, f(theta, u)>`
is

.. math::

    f(\theta , u ) = \frac{1}{2} \left[
        \log ( 2 \pi \sigma_u^2 ) + u^2 / \sigma_u^2
        +
        \log ( 2 \pi \sigma_y^2 ) + [ y - \exp(u) \theta ]^2 / \sigma_y^2
    \right]

Mismatch
********
In the case where :math:`y = z`, one might expect the solution
:math:`\theta = z` and :math:`u = 0` because
all the data residuals are zero, :math:`\B{p}(u | \theta )`
is maximal, and there is no prior for :math:`\theta`.
This example demonstrates that :math:`\theta = z` and :math:`u = 0`
may not be optimal for the this case.
To be specific it shows that the derivative of
:ref:`L(theta)<theory@Objective@Fixed Effects Objective, L(theta)>`
may be non-zero.

Theory
******
See the :ref:`theory-title` section for the
theory behind the calculations below:

Derivatives
***********

.. math::
    :nowrap:

    \begin{eqnarray}
    g_\theta ( \theta )
    & = &
    ( \theta - z ) \sigma_z^{-2}
    \\
    f_\theta ( \theta , u )
    & = &
    [ \exp(u) \theta - y ] \exp(u) \sigma_y^{-2}
    \\
    f_u ( \theta , u )
    & = &
    u / \sigma_u^2 + [ \exp(u) \theta - y ] \exp(u) \theta \sigma_y^{-2}
    \\
    f_{u,u} ( \theta , u )
    & = &
    \sigma_u^{-2}
    +
    [ 2 \exp(u) \theta - y ] \exp(u) \theta \sigma_y^{-2}
    \\
    f_{u,\theta} ( \theta , u )
    & = &
    [ 2 \exp(u) \theta - y ] \exp(u) \sigma_y^{-2}
    \\
    \hat{u}_\theta ( \theta )
    & = &
    - f_{u,\theta} [ \theta , \hat{u} ( \theta ) ]
    /
    f_{u,u} [ \theta , \hat{u} ( \theta ) ]
    \\
    f_{u,u,u} ( \theta , u )
    & = &
    [ 4 \exp(u) \theta - y ] \exp(u) \theta \sigma_y^{-2}
    \\
    f_{u,u,\theta} ( \theta , u )
    & = &
    [ 4 \exp(u) \theta - y ] \exp(u) \sigma_y^{-2}
    \end{eqnarray}

Objective
*********

.. math::
    :nowrap:

    \begin{eqnarray}
    h( \theta , u )
    & = &
    \frac{1}{2} \log f_{u,u} ( \theta , u )
    +
    f( \theta , u )
    -
    \log( 2 \pi )
    \\
    h_\theta ( \theta , u )
    & = &
    \frac{1}{2} f_{u,u,\theta} ( \theta , u ) / f_{u,u} ( \theta , u )
    +
    f_\theta ( \theta , u )
    \\
    h_u ( \theta , u )
    & = &
    \frac{1}{2} f_{u,u,u} ( \theta , u ) / f_{u,u} ( \theta , u )
    +
    f_u ( \theta , u )
    \\
    L( \theta )
    & = &
    h [ \theta , \hat{u} ( \theta ) ] + g ( \theta )
    \\
    L_\theta ( \theta )
    & = &
    h_\theta [ \theta , \hat{u} ( \theta ) ]
    +
    h_u [ \theta , \hat{u} ( \theta ) ] \hat{u}_\theta ( \theta )
    +
    g_\theta ( \theta )
    \end{eqnarray}

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{xrst_end data_mismatch.cpp}