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Let $P_R$ be a probability on $R$, let $\mathcal{Y}$ be the $\sigma$-algebra generated by the intervals that contain the positive reals $(x\geq 0)$, and let $\mathcal{Z}$ be the $\sigma$-algebra generated by the intervals that contain the negative reals $(x<0)$. Suppose $P_{R/\mathcal{Z}}(z)(A)=P_R[A|\mathcal{Z}](z)$ is a probability on $R$ (in fact it is) for each $z$. We can therefore condition that probability by $\mathcal{Y}$ and thus define $P_{R/\mathcal{Y}/\mathcal{Z}}(y,z)(A)=P_R[A|\mathcal{Y}|\mathcal{Z}](y,z)$. What would $P_{R/\mathcal{Y}/\mathcal{Z}}(y,z)(A)$ be?
And $E[X|\mathcal{Y}|\mathcal{Z}](y,z)\quad (X(x)=x)$?

I would bet on: \begin{align*} &\hspace{-6cm}P_{R/\mathcal{Y}/\mathcal{Z}}(y,z)(A)=I_{R^{^-}}(z)I_{A^{^-}}(y)+I_{R^{^+}}(y)I_{A^{^+}}(z)\\[10pt] &\hspace{-6cm}E[X|\mathcal{Y}|\mathcal{Z}](y,z)=\begin{cases} y& \text{ if }\ y,z< 0 \\ z& \text{ if }\ y,z\geq 0\\ indifferent&\ otherwise \end{cases} \end{align*}

Speltzu
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    A $\sigma$-algebra in $\mathbb R$ is closed under taking the complement. If and interval $[a,+\infty)$ is in your $\sigma$-algebra ${\cal Y}$ does that mean $a\ge 0$? Which complement of this interval do you want to be in ${\cal Y},$? $(-\infty,a)$ or $[0,a),?$ In other words, on which space is your ${\cal Y}$ a $\sigma$-algebra? More clarity please. These won't be my last questions. – Kurt G. Feb 17 '23 at 13:29
  • The question is perfectly clear, don't you know whicht is the $\sigma$-algebra generated by the intervals that contain the interval $[0,\infty)$?. It's easy: it is formed by the borel sets that contain all the positive reals or do not contain any. – Speltzu Feb 17 '23 at 14:30

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