Let $\displaystyle X \sim f(x\mid \theta) = (\theta/2)^{\lvert x \rvert}(1-\theta)^{1-\lvert x \rvert}$

Question

Let $\displaystyle X \sim f(x\mid \theta) = (\theta/2)^{\lvert x \rvert}(1-\theta)^{1-\lvert x \rvert}1_{(x\in \{ -1,0,1 \})}1_{(0 \le \theta \le 1)}$ and let $T(X) = 2\cdot1_{(X=1)}$

I will omit the indicator functions for sake of brevity.

I am trying to find $E[T(X) \mid \lvert X \rvert ]$

For the conditional probabilities:

$\displaystyle P(T(X) = k \mid \lvert X \rvert = j ) = \frac{P(T(X) =k,\lvert X \rvert= j)}{P(\lvert X \rvert= j)}, k = 0,2; j = -1,0,1$

I am having difficulty computing an expression for this conditional probability for the conditional expectation. I know $T(X)$ is unbiased for $\theta$, i.e. $E[T(X)]=\theta$ so it should be that $E[E[T(X) \mid \lvert X \rvert ]] = \theta$

The issue is that $T(X)$ is dependent on $X$ so having trouble with $P(T(X) =k,\lvert X \rvert= j)$

Answer 1

For ease of notation, let $Y = |X|$. We have \begin{align*} \mathbb{E}[T(X) | Y] &= \underbrace{\mathbb{E}[T(X) | Y = 0]}_{=0}\mathbb{P}(Y = 0) + \mathbb{E}[T(X) | Y = 1]\underbrace{\mathbb{P}(Y = 1)}_{=\theta} \\ &=\theta\cdot\mathbb{E}[T(X) | Y = 1] \end{align*} But then $P(X = -1|Y=1) = P(X = 1|Y=1) = \frac{1}{2}$. So $\mathbb{E}[T(X) | Y = 1] = 2\mathbb{P}(X=1|Y=1) = 1$. Therefore, $\mathbb{E}[T(X) | Y] = \theta$.