# Conditional-expectation operator inside of expectation operator

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:

$$E[E[(T-g(U))(g(U)-b(\theta))|U]]=E[(g(U)-b(\theta))E[T-g(U)|U]].$$

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

• 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. Jan 27, 2019 at 20:54

(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:

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

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