\[M_{\pm} = \frac{1}{2}\left(\mathbb{I} \pm \hat{x} \cdot \vec{\sigma}\right)\]
\[M_{\pm} = \frac{1}{2}\left(\mathbb{I} \pm \hat{x} \cdot \vec{\sigma}\right) \xrightarrow{\text{noise}} M_{\pm}^{r=\frac{1}{2}} = \frac{1}{2}\left(\mathbb{I} \pm \frac{1}{2} \hat{x} \cdot \vec{\sigma}\right)\]
Parent POVM:
\[\Pi_{\hat{\lambda}} = \frac{1}{4\pi}\left(\mathbb{I} + \hat{\lambda} \cdot \vec{\sigma}\right)\]
Response function:
\[p(\pm|\{\hat{x}\},\hat{\lambda}) = \Theta(\pm\hat{x}\cdot\hat{\lambda}) = \begin{cases}1 & \text{if } \pm\hat{x}\cdot\hat{\lambda} >0\\ 0 & \text{otherwise}\end{cases}\]
Checking that \(\Pi_{\hat{\lambda}}\) simulates \(M_{\pm}^{r=\frac{1}{2}}\):
\[\begin{align}\int_S d\hat{\lambda} p(\pm|\{\hat{x}\},\hat{\lambda}) \Pi_{\hat{\lambda}} &= \frac{1}{4\pi} \int_{\pm\hat{x}\cdot\hat{\lambda} >0} d\hat{\lambda} (\mathbb{I} + \hat{\lambda} \cdot \vec{\sigma}) = \\ &= \frac{1}{2}\left(\mathbb{I} \pm \frac{1}{2} \hat{x} \cdot \vec{\sigma}\right) = M_{\pm}^{r=\frac{1}{2}}\end{align}\]
\[M_a = \mu_a\left(\mathbb{I} + \hat{x}_a \cdot \vec{\sigma}\right)\text{, } \sum_a \mu_a = 1\text{, } \sum_a \mu_a \hat{x}_a = 0\text{, } \mu_a \ge 0\]
Coarse-grained POVM:
\[\underbrace{\Pi_{\hat{\lambda}}}_{\substack{\text{parent POVM}\\\text{infinite}}} \xrightarrow{\text{simulates}} \underbrace{\Pi_A}_{\substack{\text{coarse-grained POVM}\\\text{finite}}} \xrightarrow{\text{simulates}} \underbrace{M_a^{r=\frac{1}{2}}}_{\text{target}}\]
6 regions: \(A \in \{\{1\}, \{2\}, \{3\}, \{1, 2\}, \{2, 3\}, \{1, 3\}\}\)
Integrate \(\hat{\lambda}\) over each region \(A\):
\[\Pi_A = \int_S d\hat{\lambda} \prod_{a \in A}\Theta(\hat{x}_a\cdot\hat{\lambda})\prod_{a' \notin A}(1 - \Theta(\hat{x}_{a'}\cdot\hat{\lambda})) \Pi_{\hat{\lambda}}\]
Integrate \(\hat{\lambda}\) over each region \(A\):
\[\Pi_A = \int_S d\hat{\lambda} \quad\text{"1 on the region A"}\quad \Pi_{\hat{\lambda}}\]
\[\begin{align}\Pi_{\{1\}} &= \frac{\theta_1}{2\pi}(\mathbb{I} - \frac{\pi \sin{\frac{\theta_1}{2}}}{2 \theta_1}\hat{\lambda}_1 \cdot \vec{\sigma})\\\Pi_{\{1\}^c}=\Pi_{\{2, 3\}} &= \frac{\theta_1}{2\pi}(\mathbb{I} + \frac{\pi \sin{\frac{\theta_1}{2}}}{2 \theta_1}\hat{\lambda}_1 \cdot \vec{\sigma})\end{align}\]
\[\hat{\lambda}_1 = \frac{\hat{x}_2 + \hat{x}_3}{|\hat{x}_2 + \hat{x}_3|}\text{, } \theta_1 = \arccos{(-\hat{x}_2 \cdot \hat{x}_3)}\text{, }\theta_1 + \theta_2 + \theta_3 = \pi\]
Response function table:
| \(\Pi_{\{2,3\}}\) | \(\Pi_{\{1,3\}}\) | \(\Pi_{\{1,2\}}\) | \(\Pi_{\{3\}}\) | \(\Pi_{\{2\}}\) | \(\Pi_{\{1\}}\) | |
|---|---|---|---|---|---|---|
| \(M_1^{r=\frac{1}{2}}\) | ||||||
| \(M_2^{r=\frac{1}{2}}\) | ||||||
| \(M_3^{r=\frac{1}{2}}\) |
Constant response function:
| \(\Pi_{\{2,3\}}\) | \(\Pi_{\{1,3\}}\) | \(\Pi_{\{1,2\}}\) | \(\Pi_{\{3\}}\) | \(\Pi_{\{2\}}\) | \(\Pi_{\{1\}}\) | |
|---|---|---|---|---|---|---|
| \(M_1^{r=\frac{1}{2}}\) | \(2\mu_1\) | \(2\mu_1\) | \(2\mu_1\) | |||
| \(M_2^{r=\frac{1}{2}}\) | \(2\mu_2\) | \(2\mu_2\) | \(2\mu_2\) | |||
| \(M_3^{r=\frac{1}{2}}\) | \(2\mu_3\) | \(2\mu_3\) | \(2\mu_3\) |
This response function is not normalized:
\[\sum_a p(a|\{\hat{x}_a\},\hat{\lambda}) \neq 1\]
But we can simulate \(M_a^{r=\frac{1}{2}}\): \[\begin{align}\sum_{a \in A} 2\mu_a \Pi_A &= \int_{\hat{x}_a\cdot\hat{\lambda} >0} d\hat{\lambda} 2 \mu_a \Pi_{\hat{\lambda}} =\\&= 2\mu_a\frac{1}{4\pi} \int_{\hat{x}_a\cdot\hat{\lambda} >0} d\hat{\lambda} (\mathbb{I} + \hat{\lambda} \cdot \vec{\sigma}) =\\&= 2\mu_a\frac{1}{2}\left(\mathbb{I} + \frac{1}{2} \hat{x}_a \cdot \vec{\sigma}\right) = M_a^{r=\frac{1}{2}}\end{align}\]
\[\begin{align}\Pi_{\{a\}} &= \frac{\theta_a}{2\pi}(\mathbb{I} - \frac{\pi \sin{\frac{\theta_a}{2}}}{2 \theta_a}\hat{\lambda}_a \cdot \vec{\sigma})\\\Pi_{\{a\}^c} &= \frac{\theta_a}{2\pi}(\mathbb{I} + \frac{\pi \sin{\frac{\theta_a}{2}}}{2 \theta_a}\hat{\lambda}_a \cdot \vec{\sigma})\end{align}\]
Linear dependencies between operators: \[\sum_{a \in \{1,2,3\}} (1 - 2\mu_a)(\Pi_{\{a\}} - \Pi_{\{a\}^c}) = 0\]
\[\frac{1}{\theta_a}(\Pi_{\{a\}} + \Pi_{\{a\}^c}) = \frac{1}{\theta_{a'}}(\Pi_{\{a'\}} + \Pi_{\{a'\}^c})\]
Modified response function:
| \(\Pi_{\{2,3\}}\) | \(\Pi_{\{1,3\}}\) | \(\Pi_{\{1,2\}}\) | \(\Pi_{\{3\}}\) | \(\Pi_{\{2\}}\) | \(\Pi_{\{1\}}\) | |
|---|---|---|---|---|---|---|
| \(M_1^{r=\frac{1}{2}}\) | \(2\mu_1\) | \(2\mu_1\) | \(2\mu_1\) | |||
| \(M_2^{r=\frac{1}{2}}\) | \(2\mu_2 - X\) | \(1 - 2\mu_1\) | \(1\) | \(Y\) | ||
| \(M_3^{r=\frac{1}{2}}\) | \(2\mu_3 - Y\) | \(1 - 2\mu_1\) | \(1\) | \(X\) |
\[X = \frac{\theta_1(1-2\mu_1) - \theta_2(1-2\mu_2) + \theta_3(1-2\mu_3)}{2\theta_1}\] \[Y = \frac{\theta_1(1-2\mu_1) + \theta_2(1-2\mu_2) - \theta_3(1-2\mu_3)}{2\theta_1}\]
\[M_a = \mu_a\left(\mathbb{I} + \hat{x}_a \cdot \vec{\sigma}\right)\text{, } \sum_a \mu_a = 1\text{, } \sum_a \mu_a \hat{x}_a = 0\text{, } \mu_a \ge 0\]
Let's introduce two extra vectors \(\hat{x}_{5^+}\) and \(\hat{x}_{5^-}\):
\[\mu_5\hat{x}_{5^+} = -\mu_5\hat{x}_{5^-} = -\mu_1\hat{x}_1 - \mu_2\hat{x}_2 = \mu_3\hat{x}_3 + \mu_4\hat{x}_4\]
Two 3-outcome POVMs:
\[\begin{align}\operatorname{POVM}_{+} &= \frac{1}{k_+}\{ M_{5^+}^{r=\frac{1}{2}}, M_{1}^{r=\frac{1}{2}}, M_{2}^{r=\frac{1}{2}} \}\\ \operatorname{POVM}_{-} &= \frac{1}{k_-}\{ M_{5^-}^{r=\frac{1}{2}}, M_{3}^{r=\frac{1}{2}}, M_{4}^{r=\frac{1}{2}} \}\end{align}\]
where
\[\begin{aligned}M_{5^\pm}^{r=\frac{1}{2}} &= \mu_5\left(\mathbb{I} + \frac{1}{2}\hat{x}_{5^\pm}\cdot \vec{\sigma}\right)\\k_+ &= \mu_1 + \mu_2 + \mu_5\\\quad k_- &= \mu_3 + \mu_4 + \mu_5\\\end{aligned}\]
We know how to simulate 3-outcome POVMs!
Define coarse-grained POVMs \(\Pi^+_{A_+}\) and \(\Pi^-_{A_-}\): \[\begin{align}\Pi^+_{A_+} \xrightarrow{\text{simulates}} \frac{1}{k_+}\{ M_{5^+}^{r=\frac{1}{2}}, M_{1}^{r=\frac{1}{2}}, M_{2}^{r=\frac{1}{2}} \}\\ \Pi^-_{A_-} \xrightarrow{\text{simulates}} \frac{1}{k_-}\{ M_{5^-}^{r=\frac{1}{2}}, M_{3}^{r=\frac{1}{2}}, M_{4}^{r=\frac{1}{2}} \}\end{align}\]
Merge \(\Pi^+_{A_+}\) and \(\Pi^-_{A_-}\) into \(\Pi_{A = A_+ \cap A_-}\) to simulate \(M_{a}^{r=\frac{1}{2}}\)
Coarse-grained POVM: \[\begin{align} \Pi_{\{i\}} &= \frac{\theta_i}{2\pi}(\mathbb{I} - \frac{\pi \sin{\frac{\theta_i}{2}}}{2 \theta_i}\hat{\lambda}_i \cdot \vec{\sigma}) &= \frac{\theta_i}{2\pi}\mathbb{I} - \frac{1}{4}\hat{\lambda}'_1 \cdot \vec{\sigma}\\ \Pi_{\{i\}^c} &= \frac{\theta_i}{2\pi}(\mathbb{I} + \frac{\pi \sin{\frac{\theta_i}{2}}}{2 \theta_i}\hat{\lambda}_i \cdot \vec{\sigma}) &= \frac{\theta_i}{2\pi}\mathbb{I} + \frac{1}{4}\hat{\lambda}'_i \cdot \vec{\sigma}\end{align}\]
We can simulate \(M_a^{r=\frac{1}{2}}\) with unnormalized response function: \[\begin{align} 2 \mu_1 \Pi_{\{1\}} + 2 \mu_1 \Pi_{\{1, 2\}} + 2 \mu_1 \Pi_{\{1,3\}} &= M_1^{r=\frac{1}{2}} &= \mu_1(\mathbb{I} + \frac{1}{2}\hat{x}_1\cdot\vec{\sigma})\\ 2 \mu_2 \Pi_{\{1,2\}} + 2 \mu_2 \Pi_{\{2\}} + 2 \mu_2 \Pi_{\{2,3\}} &= M_2^{r=\frac{1}{2}} &= \mu_2(\mathbb{I} + \frac{1}{2}\hat{x}_2\cdot\vec{\sigma})\\ 2 \mu_3 \Pi_{\{1,3\}} + 2 \mu_3 \Pi_{\{2,3\}} + 2 \mu_3 \Pi_{\{3\}} &= M_3^{r=\frac{1}{2}} &= \mu_3(\mathbb{I} + \frac{1}{2}\hat{x}_3\cdot\vec{\sigma}) \end{align}\]
We get: \[\begin{align} \mu_1\frac{\theta_1 + \theta_2 + \theta_3}{\pi}\mathbb{I} + \frac{\mu_1}{2}(\hat{\lambda}'_2 + \hat{\lambda}'_3 - \hat{\lambda}'_1)\cdot\vec{\sigma} &= \mu_1\mathbb{I} + \frac{\mu_1}{2} \hat{x}_1 \cdot \vec{\sigma}\\ \mu_2\frac{\theta_1 + \theta_2 + \theta_3}{\pi}\mathbb{I} + \frac{\mu_2}{2}(\hat{\lambda}'_1 + \hat{\lambda}'_3 - \hat{\lambda}'_2)\cdot\vec{\sigma} &= \mu_2\mathbb{I} + \frac{\mu_2}{2} \hat{x}_2 \cdot \vec{\sigma}\\ \mu_3\frac{\theta_1 + \theta_2 + \theta_3}{\pi}\mathbb{I} + \frac{\mu_3}{2}(\hat{\lambda}'_1 + \hat{\lambda}'_2 - \hat{\lambda}'_3)\cdot\vec{\sigma} &= \mu_3\mathbb{I} + \frac{\mu_3}{2} \hat{x}_3 \cdot \vec{\sigma} \end{align}\]
We get: \[\begin{align} \mu_1(\hat{\lambda}'_2 + \hat{\lambda}'_3 - \hat{\lambda}'_1) &= \mu_1 \hat{x}_1\\ \mu_2(\hat{\lambda}'_1 + \hat{\lambda}'_3 - \hat{\lambda}'_2) &= \mu_2 \hat{x}_2\\ \mu_3(\hat{\lambda}'_1 + \hat{\lambda}'_2 - \hat{\lambda}'_3) &= \mu_3 \hat{x}_3 \end{align}\]
\[ (1-2\mu_1)\hat{\lambda}'_1 + (1-2\mu_2)\hat{\lambda}'_2 + (1-2\mu_3)\hat{\lambda}'_3 = \mu_1 \hat{x}_1 + \mu_2 \hat{x}_2 + \mu_3 \hat{x}_3 = 0 \]
Coarse-grained POVM: \[ \Pi_{\{i\}} = \frac{\theta_i}{2\pi}\mathbb{I} - \frac{1}{4}\hat{\lambda}'_1 \cdot \vec{\sigma}\text{, }\quad\Pi_{\{i\}^c} = \frac{\theta_i}{2\pi}\mathbb{I} + \frac{1}{4}\hat{\lambda}'_i \cdot \vec{\sigma} \]
We need to simulate \(M_a^{r=\frac{1}{2}}\) with normalized response function: \[\begin{align} 2 \mu_1 \Pi_{\{1\}} + 2 \mu_1 \Pi_{\{1, 2\}} + 2 \mu_1 \Pi_{\{1,3\}} &= M_1^{r=\frac{1}{2}} &= \mu_1(\mathbb{I} + \frac{1}{2}\hat{x}_1\cdot\vec{\sigma})\\ (1 - 2 \mu_1) \Pi_{\{1,2\}} + \Pi_{\{2\}} + (2 \mu_2 - X) \Pi_{\{2,3\}} + Y \Pi_{\{1\}} &= M_2^{r=\frac{1}{2}} &= \mu_2(\mathbb{I} + \frac{1}{2}\hat{x}_2\cdot\vec{\sigma})\\ (1 - 2\mu_1) \Pi_{\{1,3\}} + (2 \mu_3 - Y) \Pi_{\{2,3\}} + \Pi_{\{3\}} + X \Pi_{\{1\}} &= M_3^{r=\frac{1}{2}} &= \mu_3(\mathbb{I} + \frac{1}{2}\hat{x}_3\cdot\vec{\sigma}) \end{align}\]
Coarse-grained POVM: \[ \Pi_{\{i\}} = \frac{\theta_i}{2\pi}\mathbb{I} - \frac{1}{4}\hat{\lambda}'_1 \cdot \vec{\sigma}\text{, }\quad\Pi_{\{i\}^c} = \frac{\theta_i}{2\pi}\mathbb{I} + \frac{1}{4}\hat{\lambda}'_i \cdot \vec{\sigma} \]
We need to simulate \(M_2^{r=\frac{1}{2}}\) with normalized response function: \[ (1 - 2 \mu_1) \Pi_{\{1,2\}} + \Pi_{\{2\}} + (2 \mu_2 - X) \Pi_{\{2,3\}} + Y \Pi_{\{1\}} = \mu_2(\mathbb{I} + \frac{1}{2}\hat{x}_2\cdot\vec{\sigma}) \]
Coefficients in front of \(\vec{\sigma}\): \[ (1-2\mu_1)\hat{\lambda}'_3 - \hat{\lambda}'_2 + (2\mu_2 - X) \hat{\lambda}'_1 - Y\hat{\lambda}'_1 = 2\mu_2\hat{x}'_2 \]
We already know that: \[ 2\mu_2\hat{\lambda}'_3 - 2\mu_2\hat{\lambda}'_2 + 2\mu_2 \hat{\lambda}'_1 = 2\mu_2\hat{x}'_2 \]
Subtracting, we get: \[ (1-2\mu_1-2\mu_2)\hat{\lambda}'_3 + (2\mu_2-1) \hat{\lambda}'_2 + (-Y - X) \hat{\lambda}'_1 = 0 \]
Which we can rewrite as: \[ (2\mu_3-1)\hat{\lambda}'_3 + (2\mu_2-1) \hat{\lambda}'_2 + (2\mu_1-1) \hat{\lambda}'_1 = 0 \]
Coefficients in front of \(\mathbb{I}\): \[ (1 - 2\mu_1)\frac{\theta_3}{2\pi} + \frac{\theta_2}{2\pi} + (2\mu_2 - X)\frac{\theta_1}{2\pi} + Y\frac{\theta_1}{2\pi} = \mu_2 \]
\[ (1 - 2\mu_1)\theta_3 + \theta_2 + (2\mu_2 - X)\theta_1 + (1 - 2\mu_1 - X)\theta_1 = 2 \mu_2 (\theta_1 + \theta_2 + \theta_3) \]
\[\begin{aligned} X &= \frac{\theta_1(1-2\mu_1) - \theta_2(1-2\mu_2) + \theta_3(1-2\mu_3)}{2\theta_1}\\ Y &= \frac{\theta_1(1-2\mu_1) + \theta_2(1-2\mu_2) - \theta_3(1-2\mu_3)}{2\theta_1} \end{aligned}\]
Coarse-grained POVMs \(\{\Pi^{\pm}_{A_\pm}\}\): \[\begin{array}{l} \Pi^{\pm}_{A_\pm} = \int_S d\hat{\lambda} \prod_{a \in A_{\pm}}\Theta(\hat{x}_a\cdot\hat{\lambda})\prod_{a \notin A_{\pm}}(1 - \Theta(\hat{x}_a\cdot\hat{\lambda})) \Pi_{\hat{\lambda}}\\ A_{\pm} \subset O_\pm\text{, where }O_+ =\{5^+, 1, 2\}\text{, }O_+ =\{5^-, 3, 4\} \end{array}\]
Response function \(p_+(a|A_+)\) for \(\operatorname{POVM}_+\):
| \(\Pi^+_{\{1,2\}}\) | \(\Pi^+_{\{2,5^+\}}\) | \(\Pi^+_{\{1,5^+\}}\) | \(\Pi^+_{\{2\}}\) | \(\Pi^+_{\{1\}}\) | \(\Pi^+_{\{5^+\}}\) | |
|---|---|---|---|---|---|---|
| \(\frac{1}{k_+}M_{5^+}^{r=\frac{1}{2}}\) | \(\frac{2}{k_+}\mu_5\) | \(\frac{2}{k_+}\mu_5\) | \(\frac{2}{k_+}\mu_5\) | |||
| \(\frac{1}{k_+}M_1^{r=\frac{1}{2}}\) | \(\frac{2}{k_+}\mu_1 - X_+\) | \(1 - \frac{2}{k_+}\mu_5\) | \(1\) | \(Y_+\) | ||
| \(\frac{1}{k_+}M_2^{r=\frac{1}{2}}\) | \(\frac{2}{k_+}\mu_2 - Y_+\) | \(1 - \frac{2}{k_+}\mu_5\) | \(1\) | \(X_+\) |
Response function \(p_-(a|A_-)\) for \(\operatorname{POVM}_-\):
| \(\Pi^-_{\{3,4\}}\) | \(\Pi^-_{\{4,5^-\}}\) | \(\Pi^-_{\{3,5^-\}}\) | \(\Pi^-_{\{4\}}\) | \(\Pi^-_{\{3\}}\) | \(\Pi^-_{\{5^-\}}\) | |
|---|---|---|---|---|---|---|
| \(\frac{1}{k_-}M_{5^-}^{r=\frac{1}{2}}\) | \(\frac{2}{k_-}\mu_5\) | \(\frac{2}{k_-}\mu_5\) | \(\frac{2}{k_-}\mu_5\) | |||
| \(\frac{1}{k_-}M_3^{r=\frac{1}{2}}\) | \(\frac{2}{k_-}\mu_3 - X_-\) | \(1 - \frac{2}{k_-}\mu_5\) | \(1\) | \(Y_-\) | ||
| \(\frac{1}{k_-}M_4^{r=\frac{1}{2}}\) | \(\frac{2}{k_-}\mu_4 - Y_-\) | \(1 - \frac{2}{k_-}\mu_5\) | \(1\) | \(X_-\) |
\[ \begin{array}{rcl} &\Pi_{\hat{\lambda}} &\\ &\downarrow &\\ &\Pi_{A = A_+ \cap A_-} &\\ &\swarrow\quad\quad\quad \searrow & \\ \Pi^+_{A_+} & &\Pi^-_{A_-}\\ \swarrow\quad\quad & &\quad\quad\searrow \\ \frac{1}{k_+}\{ M_{5^+}^{r=\frac{1}{2}}, M_{1}^{r=\frac{1}{2}}, M_{2}^{r=\frac{1}{2}} \} & &\frac{1}{k_-}\{ M_{5^-}^{r=\frac{1}{2}}, M_{3}^{r=\frac{1}{2}}, M_{4}^{r=\frac{1}{2}} \} \end{array} \]
\(M_{a}^{r=\frac{1}{2}}\) are simulated with either \(\Pi^+_{A_+}\) or \(\Pi^+_{A_-}\): \[\begin{array}{rlcrl} M_1^{r=\frac{1}{2}} &= k_+ \sum_{A_+ \subset \{5^+, 1, 2\} } p_+(1|A_+) \Pi^+_{A_+} &\quad &M_3^{r=\frac{1}{2}} &= k_- \sum_{A_- \subset \{5^-, 3, 4\} } p_-(3|A_-) \Pi^-_{A_-}\\ M_2^{r=\frac{1}{2}} &= k_+ \sum_{A_+ \subset \{5^+, 1, 2\} } p_+(2|A_+) \Pi^+_{A_+} &\quad &M_4^{r=\frac{1}{2}} &= k_- \sum_{A_- \subset \{5^-, 3, 4\} } p_-(4|A_-) \Pi^-_{A_-} \end{array} \]
\(\Pi^+_{A_+}\) and \(\Pi^+_{A_-}\) can be expressed using \(\Pi_{A = A_+ \cap A_-}\)
\(\Pi_{A = A_+ \cap A_-}\) can be expressed with \(\Pi_{\hat{\lambda}}\)
Why normalization works?
\[\Pi_{A} = \int_S d\hat{\lambda} \prod_{a \in A}\Theta(\hat{x}_a\cdot\hat{\lambda})\prod_{a \notin A}(1 - \Theta(\hat{x}_a\cdot\hat{\lambda})) \Pi_{\hat{\lambda}}\]
where \(A = A_+ \cap A_-\) is one of 18 regions
Let's simulate \(M_1^{r=\frac{1}{2}}\)!
\(M_1^{r=\frac{1}{2}}\) was in \(\operatorname{POVM}_+\), so we can simulate it using \(\Pi^+\): \[\begin{multline}M_1^{r=\frac{1}{2}} = k_+ \sum_{A_+ \subset \{5^+, 1, 2\} } p_+(1|A_+) \Pi^+_{A_+} = \\ = k_+ (\frac{2}{k_+}\mu_1 - X_+) \Pi^+_{\{1,2\}} + (1 - \frac{2}{k_+}\mu_5)) \Pi^+_{\{1,5^+\}} + \\ + \Pi^+_{\{1\}} + Y_+ \Pi^+_{\{5^+\}})\end{multline}\]
And then express \(\Pi^+\) using \(\Pi_{A}\): \[\begin{aligned} \Pi^+_{\{1,2\}} &= \Pi_{\{1,2,5−\}} + \Pi_{\{1,2,3,5−\}} + \Pi_{\{1,2,4,5−\}} \\ \Pi^+_{\{1,5^+\}} &= \Pi_{\{1,3,5+\}} + \Pi_{\{1,4,5+\}} + \Pi_{\{1,3,4,5+\}}\\ \Pi^+_{\{1\}} &= \Pi_{\{1,5−\}} + \Pi_{\{1,4,5−\}} + \Pi_{\{1,3,5−\}}\\ \Pi^+_{\{5^+\}} &= \Pi_{\{4,5+\}} + \Pi_{\{3,5+\}} + \Pi_{\{3,4,5+\}} \end{aligned}\]
\[\begin{aligned}M_1^{r=\frac{1}{2}} &= \sum_{A} p(1|A) \Pi_{A}\\p(1|A) &= k_+ p_+(1|A_+)\text{,}A = A_+ \cap A_-\end{aligned}\]
Response function for \(A = A_+ \cap A_-\): \[\begin{aligned}p(1|A) &= k_+ p_+(1|A_+) & p(2|A) &= k_+ p_+(2|A_+)\\p(3|A) &= k_- p_-(3|A_-) & p(4|A) &= k_- p_-(4|A_-)\end{aligned}\]
Let's check normalization: \[\begin{multline}\sum_{a=1}^4 p(a|A) = k_+ p_+(1|A_+) + k_+ p_+(2|A_+) + \\ + k_- p_-(3|A_-) + k_- p_-(4|A_-)\end{multline}\]
Let's check normalization: \[\begin{multline}\sum_{a=1}^4 p(a|A) = k_+ p_+(1|A_+) + k_+ p_+(2|A_+) + \\ + k_- p_-(3|A_-) + k_- p_-(4|A_-) = \\ = k_+(1 - p_+(5^+|A_+)) + k_-(1 - p_-(5^-|A_-))\end{multline}\]
Let's check normalization: \[\begin{multline}\sum_{a=1}^4 p(a|A) = k_+ p_+(1|A_+) + k_+ p_+(2|A_+) + \\ + k_- p_-(3|A_-) + k_- p_-(4|A_-) = \\ = k_+(1 - p_+(5^+|A_+)) + k_-(1 - p_-(5^-|A_-)) = \\ = k_+ + k_- - 2\mu_5 = \mu_1 + \mu_2 + \mu_5 + \\ + \mu_3 + \mu_4 + \mu_5 - 2\mu_5 = 1\end{multline}\]