Effects of measurement dependence on 1-parameter family of Bell tests

In the simplest scenario, two-party, Alice and Bob, have three measurement settings with two outputs respectively. The measurement settings of Alice and Bob are marked by $X_{j}$ and $Y_{k}$ respectively, where $j$, $k\in\{0,1,2\}$. The outputs of Alice and Bob are marked by $a$ and $b$ respectively, where $a$, $b\in\{0,1\}$. After running the Bell experiment many times, conditional probability distribution $p(a,b|X_{j},Y_{k})$ will be obtained. There are two linear constraints based on probability distribution for the Bell tests with three measurement settings. The two linear constraints include the Chain inequality with three measurement bases and the $I_{3322}$ Bell inequality.

The Chain inequality where each party has three measurement settings[29, 31] is defined by

$$I_{Chain}=\left\langle X_{0}Y_{0}\right\rangle+\left\langle X_{1}Y_{1}\right\rangle+\left\langle X_{2}Y_{2}\right\rangle+\left\langle X_{1}Y_{0}\right\rangle+\left\langle X_{2}Y_{1}\right\rangle-\left\langle X_{0}Y_{2}\right\rangle\leq I_{C},$$

(1)

where $\left\langle X_{j}Y_{k}\right\rangle=\displaystyle\sum_{a,b}p(a=b|X_{j},Y_{k})-p(a\neq b|X_{j},Y_{k})$. For classical theory, it is easy to check that for equation (1) the maximum value is $I_{C}=4$. In the quantum mechanic, $p(a,b|X_{j},Y_{k})=\mathrm{tr}(\rho M^{a}_{j}\bigotimes M^{b}_{k})$, where state $\rho$ is shared between Alice and Bob, and $M^{a}_{j}$ $(M^{b}_{k})$ represents the measurement operators of Alice (Bob). In this case, equation (1) can reach the maximum value 3$\sqrt{3}$. For no-signaling theory, the maximum value reaches $6$ in equation (1).

As for $I_{3322}$ Bell inequality, its more general form called 1-PFB inequality has been proposed in Ref. [34]

$$I^{\alpha}_{3322}=\left\langle X_{0}Y_{0}\right\rangle+\left\langle X_{0}Y_{1}\right\rangle+\alpha\left\langle X_{0}Y_{2}\right\rangle+\left\langle X_{1}Y_{0}\right\rangle+\left\langle X_{1}Y_{1}\right\rangle-\alpha\left\langle X_{1}Y_{2}\right\rangle+\alpha\left\langle X_{2}Y_{0}\right\rangle-\alpha\left\langle X_{2}Y_{1}\right\rangle\leq I^{\alpha}_{3322,C},$$

(2)

where $\alpha\in[0,2]$ (the relevant definitions are the same as above). Particularly, when $\alpha=1$, equation (2) is $I_{3322}$ Bell inequality [32, 35]. Obviously, $I^{\alpha}_{3322,C}=\max\{4,4\alpha\}$ is the classical bound of $I^{\alpha}_{3322}$. For quantum theory, the upper bound of $I^{\alpha}_{3322}$ is $I^{\alpha}_{3322,Q}=4+\alpha^{2}$. Similarly, for no-signaling theory, denote the corresponding upper bound as $I^{\alpha}_{3322,NS}=4+4\alpha$. $I^{\alpha}_{3322}$ Bell inequality satisfying $I^{\alpha}_{3322,C}=I^{\alpha}_{3322,Q}$ cannot be used to certify quantum properties, so we only consider the case with $I^{\alpha}_{3322,C}<I^{\alpha}_{3322,Q}$ (i.e., $\alpha\in(0,2)$).

In this paper, we will focus on discussing the effects of measurement dependence on 1-PFB inequality under different input distributions, and use Chain inequality as a comparison.

Generally, there are two black boxes, Alice and Bob. The inputs of Alice and Bob are marked by $X_{j}$ and $Y_{k}$ respectively (with $j,k\in\{0,1,2\}$), and the outputs are marked by $a$ and $b$ respectively (with $a,b\in\{0,1\}$). After many runs, the probability distribution of output conditioned on input will be obtained

$$p(a,b|X_{j},Y_{k})=\dfrac{\sum_{\lambda}p(\lambda)p(X_{j},Y_{k}|\lambda)p(a,b|X_{j},Y_{k},\lambda)}{p(X_{j},Y_{k})},$$

(3)

where $\lambda$ is a local hidden variable or strategy. According to local correlations theory[36], $p(a,b|X_{j},Y_{k},\lambda)$ can be decomposed into $p(a|X_{j},\lambda)p(b|Y_{k},\lambda)$, so equation (3) is further written as

$$p(a,b|X_{j},Y_{k})=\dfrac{\sum_{\lambda}p(\lambda)p(X_{j},Y_{k}|\lambda)p(a|X_{j},\lambda)p(b|Y_{k},\lambda)}{p(X_{j},Y_{k})}.$$

(4)

To make our results more powerful, we adopt the method in Ref. [30] to define the parameters $P$ and $S$ as flexible upper and lower bound of probabilities respectively for a set of selected specific measurement settings

$$\begin{split}\max p(X_{j},Y_{k}|\lambda)=P,\\ \min p(X_{j},Y_{k}|\lambda)=S,\end{split}$$

(5)

for any $\lambda$.

Since each party has three measurement settings, we can easily get $P\in[\dfrac{1}{9},1]$ and $S\in[0,\dfrac{1}{9}]$. Then, we analyze the situation when $P$ and $S$ take different values.
(1) $P=\dfrac{1}{9}$ or $S=\dfrac{1}{9}$, it means that Eve will not obtain information with the local hidden variables $\lambda$, that is, the inputs are entirely random.
(2) $P\in(\dfrac{1}{9},1)$ or $S\in(0,\dfrac{1}{9})$, it means that Eve can obtain part of the input information through the local hidden variables $\lambda$.
(3) $P=1$, it means that Eve can obtain all the input information by using the local hidden variables $\lambda$, that is, Eve completely controls the input by using a deterministic strategy.

Here, we describe the predictability of outputs as guessing probability $G$ with [30, 18, 24]

$$G=\sum_{\lambda}p(\lambda)G(\lambda),$$

(6)

where $G(\lambda)=\displaystyle\max_{a,b,j,k}\{p(a|X_{j},\lambda),p(b|Y_{k},\lambda)\}$. For a given hidden variable $\lambda$, $G(\lambda)$ represents the upper bound of the probability that Eve guesses the output. $G=\dfrac{1}{2}$ $(G=1)$ implies that Eve has an entirely indeterministic (deterministic) strategy.

Firstly, we give the relationship among measurement dependence, guessing probability, and the maximum value of 1-PFB correlation function that Eve can fake under the general input distribution.

Next, we estimate $\left\langle X_{j}Y_{k}\right\rangle$ based on the probability distribution. Let $p_{A}(-1|X_{j},\lambda)=m_{j}$, $p_{B}(-1|Y_{k},\lambda)=n_{k}$, $p(-1,-1|X_{j},Y_{k},\lambda)=c_{j,k}$, we get

$$\begin{split}p(-1,1|X_{j},Y_{k},\lambda)=m_{j}-c_{j,k},\\ p(1,-1|X_{j},Y_{k},\lambda)=n_{k}-c_{j,k},\\ p(1,1|X_{j},Y_{k},\lambda)=1+c_{j,k}-m_{j}-n_{k},\end{split}$$

(13)

so

$$p(a,b|X_{j},Y_{k},\lambda)\in\{c_{j,k},m_{j}-c_{j,k},1+c_{j,k}-m_{j}-n_{k}\}.$$

(14)

As we know, $c_{j,k}$ satisfies

$$\max\{0,m_{j}+n_{k}-1\}\leq c_{j,k}\leq\min\{m_{j},n_{k}\},$$

(15)

where

	$\displaystyle\min\{x,y\}=\frac{1}{2}[x+y-|x-y|],$		(16)
	$\displaystyle\max\{x,y\}=\frac{1}{2}[x+y+|x-y|].$

According to the previous definition of $\left\langle X_{j}Y_{k}\right\rangle$, we gain

	$\displaystyle\left\langle X_{j}Y_{k}\right\rangle=$	$\displaystyle\sum_{a,b}p(a=b|X_{j},Y_{k})-p(a\neq b|X_{j},Y_{k})$		(17)
	$\displaystyle=$	$\displaystyle 1+4c_{j,k}-2(m_{j}+n_{k}).$

By applying equations (15), (16) and (17), the bound of $\left\langle X_{j}Y_{k}\right\rangle$ is given by

$$2|m_{j}+n_{k}-1|-1\leq\left\langle X_{j}Y_{k}\right\rangle\leq 1-2|m_{j}+n_{k}|.$$

(18)

Hence, $\overline{I}^{\alpha}_{3322}$ can be written as

	$\displaystyle\overline{I}^{\alpha}_{3322}=$	$\displaystyle\sum_{\lambda}[p(\lambda|X_{0}Y_{0})\left\langle X_{0}Y_{0}\right\rangle+p(\lambda|X_{0}Y_{1})\left\langle X_{0}Y_{1}\right\rangle+\alpha p(\lambda|X_{0}Y_{2})\left\langle X_{0}Y_{2}\right\rangle+p(\lambda|X_{1}Y_{0})\left\langle X_{1}Y_{0}\right\rangle$		(19)
		$\displaystyle+p(\lambda|X_{1}Y_{1})\left\langle X_{1}Y_{1}\right\rangle-\alpha p(\lambda|X_{1}Y_{2})\left\langle X_{1}Y_{2}\right\rangle+\alpha p(\lambda|X_{2}Y_{0})\left\langle X_{2}Y_{0}\right\rangle-\alpha p(\lambda|X_{2}Y_{1})\left\langle X_{2}Y_{1}\right\rangle]$
	$\displaystyle\leq$	$\displaystyle\sum_{\lambda}[p(\lambda|X_{0}Y_{0})(1-2|m_{0}-n_{0}|)+p(\lambda|X_{0}Y_{1})(1-2|m_{0}-n_{1}|)+\alpha p(\lambda|X_{0}Y_{2})(1-2|m_{0}-n_{2}|)$
		$\displaystyle+p(\lambda|X_{1}Y_{0})(1-2|m_{1}-n_{0}|)+p(\lambda|X_{1}Y_{1})(1-2|m_{1}-n_{1}|)-\alpha p(\lambda|X_{1}Y_{2})(2|m_{1}+n_{2}-1|-1)$
		$\displaystyle+\alpha p(\lambda|X_{2}Y_{0})(1-2|m_{2}-n_{0}|)-\alpha p(\lambda|X_{2}Y_{1})(2|m_{2}+n_{1}-1|-1)]$
	$\displaystyle\leq$	$\displaystyle 4+4\alpha-2\sum_{\lambda}[p(\lambda|X_{0}Y_{0})|m_{0}-n_{0}|+p(\lambda|X_{0}Y_{1})|m_{0}-n_{1}|+\alpha p(\lambda|X_{0}Y_{2})|m_{0}-n_{2}|+p(\lambda|X_{1}Y_{0})$
		$\displaystyle|m_{1}-n_{0}|+p(\lambda|X_{1}Y_{1})2|m_{1}-n_{1}|+\alpha p(\lambda|X_{1}Y_{2})|m_{1}+n_{2}-1|+\alpha p(\lambda|X_{2}Y_{0})|m_{2}-n_{0}|+\alpha p(\lambda|X_{2}Y_{1})$
		$\displaystyle|m_{2}+n_{1}-1|]-2(\alpha-1)\sum_{\lambda}[p(\lambda|X_{0}Y_{2})|m_{0}-n_{2}|+p(\lambda|X_{1}Y_{2})|m_{1}+n_{2}-1|+p(\lambda|X_{2}Y_{0})|m_{2}-n_{0}|$
		$\displaystyle+p(\lambda|X_{2}Y_{1})|m_{2}+n_{1}-1|]$
	$\displaystyle\leq$	$\displaystyle 4+4\alpha-36(2G-1)\sum_{\lambda}p(\lambda)\min p(X_{j},Y_{k}|\lambda),$

when $m_{2}+n_{1}=1$, equality holds.

Based on the results of Ref. [18], we analyze the value of $\min p(X_{j},Y_{k}|\lambda)$:
(1) If $P\geq\frac{1}{8}$, we find that $\min p(X_{j},Y_{k}|\lambda)=0$.
(2) If $\frac{1}{9}\leq P\leq\frac{1}{8}$, let $p(X_{j},Y_{k}|\lambda)=P$, where $(j,k)\neq(j_{1},k_{1})$, so $\min p(X_{j},Y_{k}|\lambda)=p(X_{j_{1}},Y_{k_{1}}|\lambda)=1-8P$. Then equation (19) can be simplified to

$$\\ \overline{I}^{\alpha}_{3322}(G,P)=\begin{cases}4\alpha+4,&\text{$P\geq\dfrac{1}{8}$},\\ 4\alpha+4-36(2G-1)(1-8P),&\text{$\dfrac{1}{9}\leq P<\dfrac{1}{8}$}.\end{cases}\$$

(20)

On this basis, $I^{\alpha}_{3322}(G,S,P)$ can be described as

	$\displaystyle I^{\alpha}_{3322}\leq$	$\displaystyle 4+4\alpha-2\sum_{\lambda}[p(\lambda|X_{0}Y_{0})(|m_{0}-n_{0}|)+p(\lambda|X_{0}Y_{1})(|m_{0}-n_{1}|)+\alpha{p(\lambda|X_{0}Y_{2})(|m_{0}-n_{2}|)}$		(21)
		$\displaystyle+p(\lambda|X_{1}Y_{0})(|m_{1}-n_{0}|)+p(\lambda|X_{1}Y_{1})(|m_{1}-n_{1}|)+p(\lambda|X_{1}Y_{2})(|m_{1}+n_{2}-1|)+\alpha$
		$\displaystyle{p(\lambda|X_{2}Y_{0})(|m_{2}-n_{0}|)}+p(\lambda|X_{2}Y_{1})(|m_{2}+n_{1}-1|)]-2(\alpha-1)\sum_{\lambda}[{p(\lambda|X_{0}Y_{2})}(m_{0}-n_{2})$
		$\displaystyle+p(\lambda|X_{1}Y_{2})(|m_{1}+n_{2}-1|)+\alpha p(\lambda|X_{2}Y_{0})(|m_{2}-n_{0}|)+p(\lambda|X_{2}Y_{1})(|m_{2}+n_{1}-1|)]$
	$\displaystyle\leq$	$\displaystyle 4+4\alpha-36(2G-1)\sum_{\lambda}p(\lambda)\min p(X_{j},Y_{k}|\lambda)-18(\alpha-1)\sum_{\lambda}p(\lambda)\min p(X_{j},Y_{k}|\lambda)$
		$\displaystyle|m_{0}+m_{1}-n_{0}-n_{1}|$
	$\displaystyle\leq$	$\displaystyle 4+4\alpha-36(2G-1)\sum_{\lambda}p(\lambda)\min\frac{p^{{}^{\prime}}(X_{j},Y_{k}|\lambda)-S}{1-9S}-18(\alpha-1)\sum_{\lambda}p(\lambda)\min\frac{p^{{}^{\prime}}(X_{j},Y_{k}|\lambda)-S}{1-9S}$
		$\displaystyle|m_{0}+m_{1}-n_{0}-n_{1}|$
	$\displaystyle\leq$	$\displaystyle(1-9S)[4+4\alpha-36(2G-1)\sum_{\lambda}p^{{}^{\prime}}(\lambda)\min(p(X_{j},Y_{k}|\lambda)-S)]-18(\alpha-1)\sum_{\lambda}p(\lambda)$
		$\displaystyle\min(p^{{}^{\prime}}(X_{j},Y_{k}|\lambda)-S)|m_{0}+m_{1}-n_{0}-n_{1}|.$

According to equations (20) and (21), we obtain that the relationship between $I^{\alpha}_{3322}$ and $\overline{I}^{\alpha}_{3322}$ holds the following form

$$I^{\alpha}_{3322}(G,S,P)=(1-9S)\overline{I}^{\alpha}_{3322}(G,P)+36(\alpha-1)S+36S.$$

(22)

Therefore, we give the following conclusions:
(1) When $\dfrac{P-S}{1-9S}\geq\dfrac{1}{8}$, i.e., $8P+S\geq 1$, we have

	$\displaystyle I^{\alpha}_{3322}(G,S,P)$	$\displaystyle=(1-9S)\overline{I}^{\alpha}_{3322}(G,P)+36(\alpha-1)S+36S$		(23)
		$\displaystyle=4+4\alpha-36S.$

(2) When $\dfrac{P-S}{1-9S}<\dfrac{1}{8}$, i.e., $8P+S<1$, we have

	$\displaystyle I^{\alpha}_{3322}(G,S,P)$	$\displaystyle=(1-9S)\overline{I}^{\alpha}_{3322}(G,P)+36(\alpha-1)S+36S$		(24)
		$\displaystyle=\overline{I}^{\alpha}_{3322}(G,P).$

To summarize the above derivation, we get

$$\\ I^{\alpha}_{3322}(G,S,P)=\begin{cases}4\alpha+4-36S,&\text{$8P+S\geq 1$},\\ 4\alpha+4-36(2G-1)(1-8P),&\text{$8P+S<1$}.\end{cases}\$$

(25)

Here we complete the proof of Theorem 1. ∎

Based on Theorem 1, we take DI randomness expansion as an example and use deterministic strategy (i.e., $G=1$) to analyze the results under different cases.

Case 1. $P=\dfrac{1}{9}$ or $S=\dfrac{1}{9}$. It means that Eve can’t get any information about input with the local hidden variables $\lambda$. Eve attempts to forge inequality violations by using predefined strings as the output of the device (i.e., $G=1$). But from Theorem 1, we can find that the maximum value $I^{\alpha}_{3322}(1,S,P)$ will not exceed $4\alpha$ at this time, so Eve cannot successfully fake the violation of 1-PFB inequality.

Case 2. $8P+S>1$ or $8P+S<1$. It means that Eve can get some information about input through the local hidden variables $\lambda$. In this case, Eve attempts to fake inequality violations by using predefined strings as the output of the device (i.e., $G=1$). We can find $4\alpha\leq I^{\alpha}_{3322}(1,S,P)\leq 4\alpha+4$ from Fig. 1, so Eve can use a deterministic strategy to successfully fake the violation of 1-PFB inequality.

Obviously, true randomness cannot be generated in Case 1, so we are interested in whether true randomness is generated in Case 2. For the sake of more obvious conclusion, we show the relationship between $P$ and $I^{\alpha}_{3322}$ when guessing probability takes different values in Fig. 1. We find that as long as the observed value satisfies $I^{\alpha}_{3322,obv}>I^{\alpha}_{3322}(1,S,P)$, we can ensure that true randomness is generated in the process. Then we will verify the compactness of equation (8) given in Theorem 1 by giving an optimal strategy. In deterministic theory, we admit that

$$\begin{split}p(a|x,\lambda)=\delta_{a,a_{\lambda}(x)},\\ p(b|y,\lambda)=\delta_{b,b_{\lambda}(y)},\end{split}$$

(26)

where $a_{\lambda}(x)$ and $b_{\lambda}(y)$ indicate the output under a certain deterministic strategy. Therefore we get an expression for $I^{\alpha}_{3322}$

$$\begin{split}I^{\alpha}_{3322}=&9\sum_{\lambda}[p(\lambda)p(X_{0},Y_{0}|\lambda)a_{\lambda}(0)b_{\lambda}(0)+p(\lambda)p(X_{0},Y_{1}|\lambda)a_{\lambda}(0)b_{\lambda}(1)\\ &+p(\lambda)p(X_{1},Y_{0}|\lambda)a_{\lambda}(1)b_{\lambda}(0)+p(\lambda)p(X_{1},Y_{1}|\lambda)a_{\lambda}(1)b_{\lambda}(1)]\\ &+9a\sum_{\lambda}[p(\lambda)p(X_{0},Y_{2}|\lambda)a_{\lambda}(0)b_{\lambda}(2)-p(\lambda)p(X_{1},Y_{2}|\lambda)a_{\lambda}(1)b_{\lambda}(2)\\ &+p(\lambda)p(X_{2},Y_{0}|\lambda)a_{\lambda}(2)b_{\lambda}(0)-p(\lambda)p(X_{2},Y_{1}|\lambda)a_{\lambda}(2)b_{\lambda}(1)].\end{split}$$

(27)

Then, we give the optimal strategy corresponding to the maximum value of 1-PFB correlation function that Eve can fake (see Table 1). Based on the strategy in Table 1, we can simplify equation (27) to

	$\displaystyle I^{\alpha}_{3322}=$	$\displaystyle 4+4\alpha-\dfrac{9}{2}(p(X_{0},Y_{0}|\lambda_{0})+p(X_{0},Y_{0}|\lambda_{2})+p(X_{0},Y_{1}|\lambda_{1})+p(X_{0},Y_{1}|\lambda_{3})$		(28)
		$\displaystyle+p(X_{1},Y_{0}|\lambda_{1})+p(X_{1},Y_{0}|\lambda_{3})+p(X_{1},Y_{1}|\lambda_{0})+p(X_{1},Y_{1}|\lambda_{2})),$

where equation (28) is based on $p(\lambda_{n})=\dfrac{1}{4}$, $P(X_{j},Y_{k}|\lambda)=\dfrac{1}{9}$ and $\displaystyle\sum_{\lambda}{p(\lambda)p(X_{j},Y_{k}|\lambda)}=p(X_{j},Y_{k})$. Next, we consider the value of $I^{\alpha}_{3322}$ in different cases:
(1) When $8P+S\geq 1$, let $p(X_{0},Y_{0}|\lambda_{0})=p(X_{0},Y_{0}|\lambda_{2})=p(X_{0},Y_{1}|\lambda_{1})=p(X_{0},Y_{1}|\lambda_{3})=p(X_{1},Y_{0}|\lambda_{1})=p(X_{1},Y_{0}|\lambda_{3})=p(X_{1},Y_{1}|\lambda_{0})=p(X_{1},Y_{1}|\lambda_{2})=S$, we have $\max I^{\alpha}_{3322}=4+4\alpha-36S$.
(2) When $8P+S<1$, we have

	$\displaystyle I^{\alpha}_{3322}=$	$\displaystyle 4+4\alpha-\dfrac{9}{2}(p(X_{0},Y_{0}|\lambda_{0})+p(X_{0},Y_{0}|\lambda_{2})+p(X_{0},Y_{1}|\lambda_{1})+p(X_{0},Y_{1}|\lambda_{3})+p(X_{1},Y_{0}|\lambda_{1})$		(29)
		$\displaystyle+p(X_{1},Y_{0}|\lambda_{3})+p(X_{1},Y_{1}|\lambda_{0})+p(X_{1},Y_{1}|\lambda_{2}))$
	$\displaystyle=$	$\displaystyle 4+4\alpha-\dfrac{9}{2}(1-\sum_{(j,k)\in\{0,1,2\}^{2}/\{0,0\}}p(X_{j},Y_{k}|\lambda_{0})+1-\sum_{(j,k)\in\{0,1,2\}^{2}/\{0,0\}}p(X_{j},Y_{k}|\lambda_{2})$
		$\displaystyle+1-\sum_{(j,k)\in\{0,1,2\}^{2}/\{0,1\}}p(X_{j},Y_{k}|\lambda_{1})+1-\sum_{(j,k)\in\{0,1,2\}^{2}/\{0,1\}}p(X_{j},Y_{k}|\lambda_{3})$
		$\displaystyle+1-\sum_{(j,k)\in\{0,1,2\}^{2}/\{1,0\}}p(X_{j},Y_{k}|\lambda_{1})+1-\sum_{(j,k)\in\{0,1,2\}^{2}/\{1,0\}}p(X_{j},Y_{k}|\lambda_{3})$
		$\displaystyle+1-\sum_{(j,k)\in\{0,1,2\}^{2}/\{1,1\}}p(X_{j},Y_{k}|\lambda_{0})+1-\sum_{(j,k)\in\{0,1,2\}^{2}/\{1,1\}}p(X_{j},Y_{k}|\lambda_{2})),$

let $p(X_{j},Y_{k}|\lambda)=P$ except for $p(X_{0},Y_{0}|\lambda_{0})$, $p(X_{0},Y_{0}|\lambda_{2})$, $p(X_{0},Y_{1}|\lambda_{1})$, $p(X_{0},Y_{1}|\lambda_{3})$, $p(X_{1},Y_{0}|\lambda_{1})$, $p(X_{1},Y_{0}|\lambda_{3})$, $p(X_{1},Y_{1}|\lambda_{0})$ and $p(X_{1},Y_{1}|\lambda_{2})$, we get $\max I^{\alpha}_{3322}=4+4\alpha-36(1-8P)$.

Evidently, it satisfies the maximum value of $I^{\alpha}_{3322}$ we obtained previous. So far, we have found a deterministic strategy that enables Eve to fake maximum value of $I^{\alpha}_{3322}$. In the latter section, we need to compare the difficulty of the maximum quantum violation of 1-PFB inequality and Chain inequality Eve forged by using the relationship between the maximum value of Chain correlation function that Eve can fake and measurement dependence. Using the method shown in the derivation process of Theorem 1, it is easy to get the maximum value of the Chain correlation function faked by Eve. So we no longer give detailed analysis, proof and optimal strategy here, but only give the relationship between the maximum value of the Chain correlation function that Eve can fake under general input distribution in Theorem 2.

Next, we give the relationship among measurement dependence, guessing probability, and the maximum value of 1-PFB correlation function that Eve can fake under the factorizable input distribution.