$\mathbb E[\boldsymbol 1_{\{(H,T)\}}\mid \mathcal F]$ here?

Question

We consider the experience : toss two fair coin simultaneously. Let $$\mathcal F=\sigma\Big(\big\{(H,T),(T,H)\big\},\{(H,H)\},\{(T,T)\}\Big),$$ and $$\mathbb P\{(H,H)\}=\mathbb P\{(T,T)\}=1/4\quad \text{and}\quad \mathbb P\{(H,T),(T,H)\}=1/2.$$

What is $$\mathbb E[\boldsymbol 1_{\{(T,H)\}}\mid \mathcal F]\ \ ?$$

I know that it's the best approximation of $\boldsymbol 1_{\{(H,T)\}}$ knowing $\mathcal F$, but I really don't know how to compute it.

I know that $$\mathbb E[X\mid Y=y_i]=\frac{1}{\mathbb P\{Y=y_i\}}\mathbb E[X\boldsymbol 1_{Y=y_1}],$$ or $$\mathbb E[X\mid Y]=\sum_{y}\frac{1}{\mathbb P\{Y=y\}}\mathbb E[X\boldsymbol 1_{\{Y=y\}}]$$ but I can't use this formula here, do I ?

Answer 1

Two fair coins are thrown and $X$ denotes the number of heads thrown in total.

In this answer I use probability space $(\Omega,\wp(\Omega),P)$ where $\Omega=\{(H,H),(H,T),(T,H),(T,T)\}$ and $P(\{\omega\})=\frac14$ for every $\omega\in\Omega$.

Further $\mathcal F$ denotes the $\sigma$-algebra generated by $\{\{(H,H)\},\{(T,T)\},\{(H,T),(T,H)\}\}\subseteq\wp(\Omega)$

Then your question can be interpreted as: $$\text{"what is }\mathbb E[\mathbf1_{\{(T,H)\}}\mid X]\text{?"}$$

This because: $\sigma(X)=\mathcal F$.

We find:

• $\mathbb E[\mathbf1_{\{(T,H)\}}\mid X=0]=0$
• $\mathbb E[\mathbf1_{\{(T,H)\}}\mid X=1]=P(\{(H,T)\}\mid X=1)=\frac12$
• $\mathbb E[\mathbf1_{\{(T,H)\}}\mid X=2]=0$

This allows the conclusion:$$\mathbb E[\mathbf1_{\{(T,H)\}}\mid X]=\frac12\mathbf1_{\{X=1\}}=\frac12\mathbf1_{\{(H,T),(T,H)\}}$$