Let $b(\theta)$ be a parametric function, let $U$ be a sufficient statistic for $\theta$, let $T$ be an unbiased estimator for $b(\theta)$, and denote $g(U)$ as $g(u)=E[T|U=u]$. I am told that the following is true:


Can someone explain to me why this equality holds? Is there an extra step that could be written which would make it clearer?

  • 1
    $\begingroup$ Note that $g(U)-b(\theta)$ is independent of $T$, since $T$ has been integrated out when taking the expectation $E[T|U=u]$. Therefore it can be moved outside an expectation-with-respect-to-$T$ operation. $\endgroup$
    – jbowman
    Jan 27, 2019 at 20:54

1 Answer 1


(One thing that might help a little bit if you start by using bracketing of different sizes to make it clearer to see where terms start and end. I will do this throughout my working.) Now, since $b(\theta)$ is a constant, and $g(U)$ is a constant when conditioning on $U$, we therefore have:

$$\mathbb{E} \Big[ \big( (g(U)-b(\theta) \big) \cdot f(T,U) \Big| U \Big] = \big( g(U)-b(\theta) \big) \cdot \mathbb{E} \Big[ f(T,U) \Big| U \Big].$$

Substituting $f(T,U) = T-g(U)$ then gives:

$$\mathbb{E} \Big[ (T-g(U))(g(U)-b(\theta)) \Big| U \Big] = (g(U)-b(\theta)) \cdot \mathbb{E} \Big[ T-g(U) \Big| U \Big].$$

Both sides of this expression are random variables that are functions of $U$. Taking expectations of both sides now gives you the formula in your question:

$$\mathbb{E} \Bigg[ \mathbb{E} \Big[ (T-g(U))(g(U)-b(\theta)) \Big| U \Big] \Bigg] = \mathbb{E} \Bigg[ (g(U)-b(\theta)) \mathbb{E} \Big[ T-g(U) \Big| U \Big] \Bigg].$$

  • $\begingroup$ How did you make the big brackets? $\endgroup$ Feb 3, 2019 at 19:34
  • $\begingroup$ You use \Bigg[ and \Bigg]. $\endgroup$
    – Ben
    Feb 3, 2019 at 22:13

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