Solution to Review Question
by Qiang Gao, updated at May 15, 2017
Chapter 1 Finite-Sample Properties of OLS
Section 5 Relation to Maximum Likelihood
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Review Question 1.5.1 (Use of regularity conditions)
Assuming that taking expectations (i.e. taking integrals) and differentiation can be interchanged, prove that the expected value of the score vector,
$$
\mathbf{s} ( \tilde{ \boldsymbol{ \theta } } ) \equiv
\frac{ \partial \log L ( \tilde{ \boldsymbol{ \theta } } ) }
{ \partial \tilde{ \boldsymbol{ \theta } } },
\tag{1.5.9}
$$
if evaluated at the true parameter value $$ \boldsymbol{\theta} $$, is zero.
We start from the identity on pdf,
$$
\int f( \mathbf{z}; \tilde{ \boldsymbol{ \theta } } ) , d \mathbf{z} = 1.
\tag{1}
$$
Differentiate both sides of (1) with respect to $$ \tilde{ \boldsymbol{ \theta } } $$ and use the regularity condition, which allows us to interchange integration and differentiation, we obtain
$$
\int \frac{ \partial f ( \mathbf{z}; \tilde{ \boldsymbol{ \theta } } ) }
{ \partial \tilde{ \boldsymbol{ \theta } } }
, d \mathbf{z} = 0.
\tag{2}
$$
Dividing $$ f( \mathbf{z}; \tilde{ \boldsymbol{ \theta } } ) $$ and multiplying $$ f( \mathbf{z}; \tilde{ \boldsymbol{ \theta } } ) $$ on the integrand of (2),
$$
\int \frac{1}{ f( \mathbf{z}; \tilde{ \boldsymbol{ \theta } } ) }
\frac{ \partial f ( \mathbf{z}; \tilde{ \boldsymbol{ \theta } } ) }
{ \partial \tilde{ \boldsymbol{ \theta } } } \cdot
f( \mathbf{z}; \tilde{ \boldsymbol{ \theta } } )
, d \mathbf{z} = 0.
\tag{3}
$$
From basic calculus, the score vector function (1.5.9) can be written as
$$
\mathbf{s} ( \tilde{ \boldsymbol{ \theta } } ) =
\frac{ \partial \log L ( \tilde{ \boldsymbol{ \theta } } ) }
{ \partial \tilde{ \boldsymbol{ \theta } } } =
\frac{ \partial \log f ( \mathbf{z}; \tilde{ \boldsymbol{ \theta } } ) }{ \partial \tilde{ \boldsymbol{ \theta } } } =
\frac{1}{ f( \mathbf{z}; \tilde{ \boldsymbol{ \theta } } ) }
\frac{ \partial f ( \mathbf{z}; \tilde{ \boldsymbol{ \theta } } ) }
{ \partial \tilde{ \boldsymbol{ \theta } } }.
\tag{4}
$$
Combining (3) and (4), evaluating at the true parameter value $$\boldsymbol{ \theta }$$ , using the definition of expectation,
$$
\int \frac{1}{ f( \mathbf{z}; \boldsymbol{ \theta } ) }
\frac{ \partial f ( \mathbf{z}; \boldsymbol{ \theta } ) }
{ \partial \boldsymbol{ \theta } } \cdot
f( \mathbf{z}; \boldsymbol{ \theta } )
, d \mathbf{z} =
\int \mathbf{s} ( \boldsymbol{ \theta } ) \cdot
f( \mathbf{z}; \boldsymbol{ \theta } )
, d \mathbf{z} =
\mathrm{E} [ \mathbf{s} ( \boldsymbol{ \theta } ) ]
= 0.
$$
Copyright ©2017 by Qiang Gao