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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 ?

sam
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  • The indicator function inside the expectation is measurable with respect to the algebra F. Hence, the conditional expectation of the indicator given the algebra is just the indicator function. – A-B-izi Nov 05 '18 at 11:10
  • Sorry I misread the algebra. The correct answer should be: $\frac{1}{2} \boldsymbol 1_{{(T,H),(T,H) } }$ This is because the only thing you know is that the outcomes of the coin tosses were different, and not the same. Given that, the probability that the first toss was T (which is all is left to know), is 1/2. – A-B-izi Nov 05 '18 at 11:29
  • The question only makes sense if the outcomes of the two coins are distinguishable. If that is not the case then $(H,T)$ is not an outcome of your sample space and consequently $\mathbf1_{{(H,T)}}$ is not measurable. The fact that you use the term "simultaneously" makes me suspect that the outcomes are not distinguishable. What space do you use as outcome space? – drhab Nov 05 '18 at 11:42
  • @drhab: I know that $\boldsymbol 1_{{(H,T)}}$ is not measurable wrt $\mathcal F$. If we consider $\mathcal G$, then $\mathbb E[\boldsymbol 1_{{(T,H)}}\mid \mathcal G]=\boldsymbol 1_{{(H,T)}}$ and is not interesting. The question makes really sense in this context since $\boldsymbol 1_{{(H,T)}}$ is not $\mathcal F-$measurable but $\mathbb E[\boldsymbol 1_{{(H,T)}}\mid \mathcal F]$ is $\mathcal F-$measurable. The space I use is the one written in the question : $\mathcal F=\sigma ({(H,H)},{T,T},{\color{red}{ }(H,T),(T,H){\color{red}}})$ – sam Nov 05 '18 at 15:51
  • I am not concerned about measurability wrt $\mathcal F$ but wrt $\mathcal G$ where $(\Omega,\mathcal G,P)$ is the underlying probability space. My question is: is $(H,T)$ an element of $\Omega$? The use of the word "simultaneously" indicates it is not or at least not necessarily, because then you can do it with $\Omega={(H,H),(T,T),{(H,T),(T,H)}}$ that does not contain $(H,T)$. If ${(H,T)}\notin\mathcal G$ then $1_{(H,T)}$ is not a measurable function and $\mathbb E[1_{(H,T)}\mid\mathcal F]$ is not defined. – drhab Nov 05 '18 at 18:02
  • $\Omega ={(H,T),(T,H),(H,H),(T,T)}$ with event space $\sigma ({(H,H)},{(T,T)},{(H,T),(T,H)})$, $\mathbb P{(H,H)}=\mathbb P{(T,T)}=1/4$ and $\mathbb P{(H,T),(T,H)}=1/4$. I.e. ${(H,T)}$ and ${(T,H)}$ is not measurable. @drhab – sam Nov 05 '18 at 18:05
  • If $(\Omega,\mathcal G,P)$ is the probability space and $\mathcal F\subseteq\mathcal G$ is a sub-sigma-algebra then $\mathbb E[X\mid\mathcal F]$ only makes sense for $\mathcal G$-measurable $X$. Your $X$ is not $\mathcal G$-measurable. – drhab Nov 05 '18 at 18:07
  • @drhab : Take $\mathcal G=2^\Omega $, then it make sense (because $(\Omega ,\mathcal F,\mathbb P)$ can be prolonge in $(\Omega ,2^\Omega , \mathbb P)$ (as you said in a previous space, all discret probability spsace can be taken in the power set). No ? – sam Nov 05 '18 at 18:12
  • That is correct, but in accordance take away the word "simultaneously". The coins must be distinguishable. E.g. a first and a second toss. How else can $(H,T)$ be discerned from $(T,H)$? – drhab Nov 05 '18 at 18:14
  • But If $\Omega ={(H,H),(H,T),(T,H),(T,T)}$ with $\mathcal F$ as I wrote in my original post doesn't it mean that the coin are not distinguishable ? (as I said). And thus, indeed, ${(H,T)}$ and ${(T,H)}$ doesn't occur (since it's not in the event space). And after we prolonge to all $2^\Omega $... no ? @drhab – sam Nov 05 '18 at 18:21
  • @drhab: By the way, with this experiment (i.e. coin are not distinguishable), how would you take $2^\Omega $ as event space ? Because what I did is really take $\mathcal F=\sigma ({(H,H)},{(T,T)},{(T,H),(H,T)})$ and then prolonge it on $2^\Omega $. But how can you write this experiment directly on $2^\Omega $ ? – sam Nov 05 '18 at 18:27

1 Answers1

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

drhab
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  • Thank you for your answer. I really don't get something : 1) We toss two not distinguishable coins. The probability space is $\Omega ={(H,H),(T,H),(T,T),(H,T)}$ s.t. $\mathcal F=\sigma ({(H,H)},{(T,T)},{(H,T),(T,H)})$ and $$\mathbb P{(H,H)}=\mathbb P{(T,T)}=\frac{1}{4}\quad \text{and}\quad \mathbb P{(T,H),(H,T)}=\frac{1}{2}.$$ That is correct ? (it's written in the solution of my exercise, so it must be correct I guess :)). – sam Nov 05 '18 at 19:29
  • 2) Now we can prolonge this space on $(\Omega ,2^\Omega ,\mathbb Q)$ where $\mathbb Q(A)=\mathbb P(A)$ if $A\in \mathcal F$ and $$\mathbb Q{(T,H)}=\mathbb Q{(H,T)}=\frac{1}{4}.$$

    This make sense right ? And now, the question is what is $$\mathbb E[\boldsymbol 1_{{(H,T)}}\mid \mathcal F].$$ And it's written that even if $\boldsymbol 1_{{(H,T)}}$ is not $\mathcal F-$measurable, the conditional expectation $\mathbb E[\boldsymbol 1_{{(H,T)}}\mid \mathcal F]$ is $\mathcal F-$measurable.

     – sam Nov 05 '18 at 19:29
  • 3) At the end, of the solution, my teacher wrote : we could have consider the space $\tilde\Omega ={{H,H},{H,T},{T,T}}$ with $\mathcal G=2^\Omega $ and $\mathbf P$ s.t. $\mathbf P{H,H}=\mathbf P{T,T}=1/4$ and $\mathbf P{T,H}=1/2$, but we can't prolonge such a space to give $\boldsymbol 1_{{(H,T)}}$ measurable, whereas is the previous me can. – sam Nov 05 '18 at 19:29
  • Could you please tel me what do you think about my three previous point ? Many thanks for your time and your answers :) (I'm sure I'm near to understand). – sam Nov 05 '18 at 19:30
  • If we toss two non-distinguishable coins then we can use $\mathcal F$ as $\sigma$-algebra in stead of $\wp(\Omega)$. But doing so that means that event ${(T,H)}$ is not a recognizable event. Then (conditional) expectation of its indicator function is not defined. That was my objection against your setup.
    • – drhab Nov 05 '18 at 19:36
  • I am not familiar with the term prolonge, but if we want to go for $\mathbf E1_{{(T,H)}}$ (conditional or not) then we must go for a probability space that contains ${(T,H)}$ as an event, and we can take $\wp(\Omega)$ instead of $\mathcal F$. This happens in my answer and a $\mathcal F$-measurable expression is found there for $\mathbb E[\mathbf1_{{(T,H)}}\mid\mathcal F]$ while $\mathbf1_{{(T,H)}}$ itself is not $\mathcal F$-measurable.
    • – drhab Nov 05 '18 at 19:42
  • I suspect that concerns the fact that ${H,T}$ cannot be split up in two sets of which one represents $(H,T)$ and the other $(T,H)$.
    • – drhab Nov 05 '18 at 19:46