Intensity modulation and direct detection quantum key distribution based on quantum noise

Figure 4 shows the distribution of Eve's measured light intensity. It has two peaks, corresponding to Alice's bits 0 and 1, respectively, that partially overlap because of quantum noise and optical classical noise. For this signal distribution, Eve sets a threshold d_E between the two peaks, and creates bit 0 or 1 when the received signal is lower or higher than d_E, respectively. When Alice launches the light intensity I₀ into a fiber line with transmittance T, Eve's mean light intensity ${I}_{{{\rm{E}}}_{0}}$ and it's variance ${\sigma }_{{{\rm{E}}}_{0}}^{2}$ are given by

$$\begin{eqnarray}&&{I}_{{{\rm{E}}}_{0}}=(1-T){I}_{0},\end{eqnarray} \tag{ 14 }$$

$$\begin{eqnarray}&&{\sigma }_{{{\rm{E}}}_{0}}^{2}=B\{a{(1-T)}^{2}{{I}_{0}}^{2}+b(1-T){I}_{0}\}.\end{eqnarray} \tag{ 15 }$$

When Alice transmits the light intensity $(1+\delta ){I}_{0}$, on the other hand, Eve's mean light intensity ${I}_{{{\rm{E}}}_{1}}$ and it's variance ${\sigma }_{{{\rm{E}}}_{1}}^{2}$ are expressed as

$$\begin{eqnarray}&&{I}_{{{\rm{E}}}_{1}}=(1-T)(1+\delta ){I}_{0},\end{eqnarray} \tag{ 16 }$$

$$\begin{eqnarray}&&{\sigma }_{{{\rm{E}}}_{1}}^{2}=B\{a{(1-T)}^{2}{(1+\delta )}^{2}{I}_{0}^{2}+b(1-T)(1+\delta ){I}_{0}\}.\end{eqnarray} \tag{ 17 }$$

With these parameters and the threshold d_E, the joint probabilities between Alice and Eve are obtained using a procedure similar to that used in section 3.2. These probabilities are summarized in the table 2.

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Table 2.  Joint probabilities between Alice and Eve for BSA.

Alice's bit a	Eve's bit e	Joint probability ${P}_{{\rm{A}},{\rm{E}}}(a,e)$
0	0	$\displaystyle \frac{1}{2}-\displaystyle \frac{1}{4}{\rm{erfc}}[\displaystyle \frac{{d}_{{\rm{E}}}-(1-T){I}_{0}}{\sqrt{2}{\sigma }_{{{\rm{E}}}_{0}}}]$
0	1	$\displaystyle \frac{1}{4}{\rm{erfc}}[\displaystyle \frac{{d}_{{\rm{E}}}-(1-T){I}_{0}}{\sqrt{2}{\sigma }_{{{\rm{E}}}_{0}}}]$
1	0	$\displaystyle \frac{1}{4}{\rm{erfc}}[\displaystyle \frac{(1-T){I}_{1}-{d}_{{\rm{E}}}}{\sqrt{2}{\sigma }_{{{\rm{E}}}_{1}}}]$
1	1	$\displaystyle \frac{1}{2}-\displaystyle \frac{1}{4}{\rm{erfc}}[\displaystyle \frac{(1-T){I}_{1}-{d}_{{\rm{E}}}}{\sqrt{2}{\sigma }_{{{\rm{E}}}_{1}}}]$

To actually calculate the joint probability summarized in table 2, we need to know Eve's threshold d_E, which is determined as follows. In judging Alice's bit, Eve sets a threshold at a value at which the tails of the two peaks corresponding to Alice's bits 0 and 1 intersect, as shown in figure 4. This condition is expressed as

$$\begin{eqnarray}&&\displaystyle \frac{1}{\sqrt{2\pi }{\sigma }_{{{\rm{E}}}_{0}}}\mathrm{exp}[-\displaystyle \frac{{\{{d}_{{\rm{E}}}-(1-T){I}_{0}\}}^{2}}{2{\sigma }_{{{\rm{E}}}_{0}}^{2}}]=\displaystyle \frac{1}{\sqrt{2\pi }{\sigma }_{{{\rm{E}}}_{1}}}\mathrm{exp}[-\displaystyle \frac{{\{{d}_{{\rm{E}}}-(1-T){I}_{1}\}}^{2}}{2{\sigma }_{{{\rm{E}}}_{1}}^{2}}].\end{eqnarray} \tag{ 18 }$$

From this equation, we obtain Eve's threshold as:

$$\begin{eqnarray}&&{d}_{{\rm{E}}}=\displaystyle \frac{1-T}{{\sigma }_{{{\rm{E}}}_{1}}^{2}-{\sigma }_{{{\rm{E}}}_{0}}^{2}}({\sigma }_{{{\rm{E}}}_{1}}^{2}{I}_{0}-{\sigma }_{{{\rm{E}}}_{0}}^{2}{I}_{1}+{\sigma }_{{{\rm{E}}}_{0}}{\sigma }_{{{\rm{E}}}_{1}}\sqrt{{({I}_{1}-{I}_{0})}^{2}-({\sigma }_{{{\rm{E}}}_{1}}^{2}-{\sigma }_{{{\rm{E}}}_{0}}^{2})\mathrm{ln}\displaystyle \frac{{\sigma }_{{{\rm{E}}}_{0}}^{2}}{{\sigma }_{{{\rm{E}}}_{1}}^{2}}}).\end{eqnarray} \tag{ 19 }$$

When ${\sigma }_{{{\rm{E}}}_{0}}^{2}\approx {\sigma }_{{{\rm{E}}}_{1}}^{2}$, (19) can be approximated as

$$\begin{eqnarray}&&{d}_{{\rm{E}}}=(1-T)\displaystyle \frac{{\sigma }_{{{\rm{E}}}_{1}}{I}_{0}+{\sigma }_{{{\rm{E}}}_{0}}{I}_{1}}{{\sigma }_{{{\rm{E}}}_{0}}+{\sigma }_{{{\rm{E}}}_{1}}}.\end{eqnarray} \tag{ 20 }$$

In this subsection, we derive the joint probabilities between Bob and Eve. The deriving proceedure is somewhat different from that in the previous subsections, because the optical classical noise has a correlation between Bob and Eve. To take this correlation into account, we separately consider the probability densities caused by the classical noise and by the quantum noise.

When Alice sends bit 0, the p.d. of her classical light intensity I_BS, ${p}_{{\rm{A}}}({I}_{\mathrm{BS}})$, is given by

$$\begin{eqnarray}&&{p}_{{\rm{A}}}({I}_{\mathrm{BS}})=\displaystyle \frac{1}{\sqrt{2\pi }{\sigma }_{\mathrm{CA}}}\mathrm{exp}\{-\displaystyle \frac{{({I}_{\mathrm{BS}}-{I}_{0})}^{2}}{2{\sigma }_{\mathrm{CA}}^{2}}\},\end{eqnarray} \tag{ 21 }$$

which originates from the optical classical noise with a variance of ${\sigma }_{\mathrm{CA}}^{2}={{BaI}}_{0}^{2}$. In the above equation, I_BS fluctuates because of the quantum noise, which has no correlation between the two beam-splitter outputs. Taking this into account, the conditional probability that Eve creates bit 0 from this signal intensity I_BS is given by

$$\begin{eqnarray}&&{P}_{{\rm{E}}| {\rm{A}}}(0| {I}_{\mathrm{BS}})={\displaystyle \int }_{-\infty }^{{d}_{{\rm{E}}}}\displaystyle \frac{1}{\sqrt{2\pi }{\sigma }_{\mathrm{Eq}}}\mathrm{exp}\{-\displaystyle \frac{{({I}_{{\rm{E}}}-(1-T){I}_{\mathrm{BS}})}^{2}}{2{\sigma }_{\mathrm{Eq}}^{2}}\}{\rm{d}}{I}_{{\rm{E}}},\end{eqnarray} \tag{ 22 }$$

where I_E is Eve's received light intensity, and ${\sigma }_{\mathrm{Eq}}^{2}={Bb}(1-T){I}_{\mathrm{BS}}\approx {Bb}(1-T){I}_{0}$ is her quantum noise. On the other hand, the conditional probability that Bob creates bit 0 when Alice transmits I_BS is given by

$$\begin{eqnarray}&&{P}_{{\rm{B}}| {\rm{A}}}(0| {I}_{\mathrm{BS}})={\displaystyle \int }_{-\infty }^{{d}_{0}}\displaystyle \frac{1}{\sqrt{2\pi }{\sigma }_{{{\rm{B}}}_{0}}}\mathrm{exp}\{-\displaystyle \frac{{({i}_{{\rm{B}}}-\alpha T\eta {I}_{\mathrm{BS}})}^{2}}{2{\sigma }_{{{\rm{B}}}_{0}}^{2}}\}{\rm{d}}{i}_{{\rm{B}}},\end{eqnarray} \tag{ 23 }$$

where i_B is Bob's received current signal, and ${\sigma }_{{{\rm{B}}}_{0}}^{2}={\alpha }^{2}{\sigma }_{Q}^{2}+{\sigma }_{{\rm{T}}}^{2}$ is his current variance.

From (21)–(23), the joint probability ${P}_{{\rm{B}},{\rm{E}}| {\rm{A}}}(0,0| 0)$, that Bob's bit 0 coincides with Eve's bit 0 when Alice sends bit 0, is given by

$$\begin{eqnarray}{P}_{{\rm{B}},{\rm{E}}| {\rm{A}}}(0,0| 0) & = & {\displaystyle \int }_{-\infty }^{\infty }{\rm{d}}{I}_{\mathrm{BS}}{P}_{{\rm{B}}| {\rm{A}}}(0| {I}_{\mathrm{BS}}){P}_{{\rm{E}}| {\rm{A}}}(0| {I}_{\mathrm{BS}}){p}_{{\rm{A}}}({I}_{\mathrm{BS}}),\\ & = & {\displaystyle \int }_{-\infty }^{\infty }{\rm{d}}{I}_{\mathrm{BS}}[{\displaystyle \int }_{-\infty }^{{d}_{0}}\displaystyle \frac{1}{\sqrt{2\pi }{\sigma }_{{{\rm{B}}}_{0}}}\mathrm{exp}\{-\displaystyle \frac{{({i}_{{\rm{B}}}-\alpha T\eta {I}_{\mathrm{BS}})}^{2}}{2{\sigma }_{{{\rm{B}}}_{0}}^{2}}\}{\rm{d}}{i}_{{\rm{B}}}]\\ & & \times [{\displaystyle \int }_{-\infty }^{{d}_{{\rm{E}}}}\displaystyle \frac{1}{\sqrt{2\pi }{\sigma }_{\mathrm{Eq}}}\mathrm{exp}\{-\displaystyle \frac{{({I}_{{\rm{E}}}-(1-T){I}_{\mathrm{BS}})}^{2}}{2{\sigma }_{\mathrm{Eq}}^{2}}\}{\rm{d}}{I}_{{\rm{E}}}]\\ & & \times \displaystyle \frac{1}{\sqrt{2\pi }{\sigma }_{\mathrm{CA}}}\mathrm{exp}\{-\displaystyle \frac{{({I}_{\mathrm{BS}}-{I}_{0})}^{2}}{2{\sigma }_{\mathrm{CA}}^{2}}\}.\end{eqnarray} \tag{ 24 }$$

By integrating over I_BS and then transforming variables as ${I}_{{\rm{E}}}^{\prime }={I}_{{\rm{E}}}/(1-T)$, ${i}_{{\rm{B}}}^{\prime }={i}_{{\rm{B}}}/(\alpha T\eta )$, ${\sigma }_{X}^{2}={\sigma }_{\mathrm{Eq}}^{2}/{(1-T)}^{2}+{\sigma }_{\mathrm{CA}}^{2}$, ${\sigma }_{Y}^{2}={\sigma }_{{{\rm{B}}}_{0}}^{2}/{(\alpha T\eta )}^{2}+{\sigma }_{\mathrm{CA}}^{2}$, and ${\rho }_{{XY}}=\displaystyle \frac{{\sigma }_{\mathrm{CA}}^{2}}{\sqrt{{\sigma }_{{{\rm{B}}}_{0}}^{2}{\sigma }_{\mathrm{CA}}^{2}/{(\alpha T\eta )}^{2}+{\sigma }_{\mathrm{Eq}}^{2}{\sigma }_{\mathrm{CA}}^{2}/{(1-T)}^{2}+{\sigma }_{\mathrm{Eq}}^{2}{\sigma }_{{{\rm{B}}}_{0}}^{2}/{\{(1-T)\alpha T\eta \}}^{2}+{\sigma }_{\mathrm{CA}}^{4}}}$, this joint probability is rewritten as

$$\begin{eqnarray}{P}_{{\rm{B}},{\rm{E}}| {\rm{A}}}(0,0| 0) & = & {\displaystyle \int }_{-\infty }^{\frac{{d}_{{\rm{E}}}}{1-T}}{\rm{d}}{I}_{{\rm{E}}}^{\prime }{\displaystyle \int }_{-\infty }^{\frac{{d}_{0}}{\alpha T\eta }}{\rm{d}}{i}_{{\rm{B}}}^{\prime }\displaystyle \frac{1}{2\pi {\sigma }_{X}{\sigma }_{Y}\sqrt{1-{\rho }_{{XY}}^{2}}}\\ & & \times \mathrm{exp}[-\displaystyle \frac{1}{2(1-{\rho }_{{XY}}^{2})}\{\displaystyle \frac{{({I}_{{\rm{E}}}^{\prime }-{I}_{0})}^{2}}{{\sigma }_{X}^{2}}+\displaystyle \frac{{({i}_{{\rm{B}}}^{\prime }-{I}_{0})}^{2}}{{\sigma }_{Y}^{2}}-\displaystyle \frac{2{\rho }_{{XY}}({I}_{{\rm{E}}}^{\prime }-{I}_{0})({i}_{{\rm{B}}}^{\prime }-{I}_{0})}{{\sigma }_{X}{\sigma }_{Y}}\}].\end{eqnarray} \tag{ 25 }$$

The joint probabilities of other bit combinations between the three parties (Alice, Bob, and Eve) can be similarly derived, the results of which are schematically shown in figure 5. The horizontal axis and vertical axis represent Bob's transformed signal, ${i}_{{\rm{B}}}^{\prime }$, and Eve's transformed signal, ${I}_{{\rm{E}}}^{\prime }$, respectively. The six regions divided by the thresholds of Bob and Eve represent their bits. We obtain the other joint probabilities by integrating the distributions in corresponding regions, as in (25).

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Applying these results to (2), we can evaluate the mutual information.

Figure 6 shows the setup of the partial IRA, where Eve occasionally extracts the transmitted signal via an optical switch, and performs an IRA on the extracted signal. In this eavesdropping strategy, the switching rate r is an important parameter for Eve to efficiently steal the information. When she employs a high r value, she can measure a large fraction of the transmitted signal, but will induce a large error-rate at Bob, and therefore, has a high risk of being revealed. When employing a low r value, on the other hand, the risk of being revealed is low but the obtainable information amount is small. Thus, there is an optimum switching rate, which is discussed in the following.

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First, we suppose that the information amount when Eve conducts full-IRA without an optical switch is given. Using this assumption, we discuss Eve's strategy for optimizing her switching ratio. The full-IRA information amount between Alice, Bob, and Eve is discussed in the next subsection. We assume that Bob creates a sifted key string of n bits, performs error-correction on it, and sets an error-rate threshold e_th for it. When the error rate of the key string e, calculated from the error-corrected bits, is larger than e_th, Bob discards the key, considering that Eve has eavesdropped on a large number of key bits. When the error-rate e is lower than e_th, on the other hand, he assumes that the amount of leaked information is small, and therefore, performs privacy amplification on the corrected key.

Here, we discuss the error-rate distribution when the partial IRA is conducted, assuming that γ out of n bits of Bob's sifted key are intercepted and resent by Eve, and Bob's and Eve's original mean error-rates are e_Bob and e_Eve, respectively. Note that γ is a stochastic variable because whether an intercepted-resent signal exceeds the Bobs thresholds is probabilistic. Unintercepted $(n-\gamma )$ bits have a mean error-rate $\bar{y}$ equal to e_Bob, while the mean error-rate of the intercepted-resent γ bits is $\bar{x}={e}_{\mathrm{Bob}}+{e}_{\mathrm{Eve}}-2{e}_{\mathrm{Bob}}{e}_{\mathrm{Eve}}$. Because these probabilities are independent of each other, the p.d. of the error-rate in intercepted-resent γ bits, p_IR(x), and that of the untouched $n-\gamma$ bits, p_uIR(y), respectively, follow binomial distributions. With Gaussian approximation, they are given by

$$\begin{eqnarray}&&{p}_{\mathrm{IR}}(x)=\displaystyle \frac{1}{\sqrt{2\pi }{\sigma }_{\mathrm{IR}}}\mathrm{exp}\{-\displaystyle \frac{{(x-\bar{x})}^{2}}{2{\sigma }_{\mathrm{IR}}^{2}}\},\end{eqnarray} \tag{ 26 }$$

$$\begin{eqnarray}&&{p}_{u\mathrm{IR}}(y)=\displaystyle \frac{1}{\sqrt{2\pi }{\sigma }_{u\mathrm{IR}}}\mathrm{exp}\{-\displaystyle \frac{{(y-\bar{y})}^{2}}{2{\sigma }_{u\mathrm{IR}}^{2}}\},\end{eqnarray} \tag{ 27 }$$

where ${\sigma }_{\mathrm{IR}}^{2}=\bar{x}(1-\bar{x})/\gamma$ and ${\sigma }_{u\mathrm{IR}}^{2}=\bar{y}(1-\bar{y})/(n-\gamma )$. The total error rate of n bits, $z=\displaystyle \frac{\gamma }{n}x+\displaystyle \frac{n-\gamma }{n}y$, is a linear combination of x and y. Thus, the p.d. of the total error rate, p_error(z), is given by

$$\begin{eqnarray}&&{p}_{\mathrm{error}}(z)=\displaystyle \frac{1}{\sqrt{2\pi }{\sigma }_{z}({r}^{\prime })}\mathrm{exp}\{-\displaystyle \frac{{(z-\bar{z}({r}^{\prime }))}^{2}}{2{\sigma }_{z}^{2}({r}^{\prime })}\},\end{eqnarray} \tag{ 28 }$$

where ${r}^{\prime }=\gamma /n$, $\bar{z}({r}^{\prime })=r\bar{x}+(1-{r}^{\prime })\bar{y}$, and ${\sigma }_{z}^{2}({r}^{\prime })=\{{r}^{\prime }\bar{x}(1-\bar{x})+(1-{r}^{\prime })\bar{y}(1-\bar{y})\}/n$.

Next, we estimate Eve's net amount of eavesdropping information, R_netE, which is determined by two factors: the information amount that Eve obtains from the intercepted signal and the probability that Bob does not discard a sifted key. Denoting ${R}_{{\rm{E}}}^{\prime }$ as the information amount that Eve obtains from an intercepted signal, the net eavesdropping information amount when γ bits out of n bits of the sifted key are intercept-resent bits is given by

$$\begin{eqnarray}&&{R}_{\mathrm{netE}}(\gamma )=\gamma /n\cdot {R}_{{\rm{E}}}^{\prime }\cdot P(z\leqslant {e}_{\mathrm{th}}),\end{eqnarray} \tag{ 29 }$$

where $P(z\leqslant {e}_{\mathrm{th}})={\displaystyle \int }_{-\infty }^{{e}_{\mathrm{th}}}{p}_{\mathrm{error}}(z){\rm{d}}z$ is the probability of Bob employing a sifted key. In the above expression, γ is a stochastic variable. Assuming that γ follows a binomial distribution with Gaussian approximation determined by the switching rate r, we obtain the net eavesdropping information amount, R_netE, as

$$\begin{eqnarray}&&{R}_{\mathrm{netE}}={\displaystyle \int }_{-\infty }^{\infty }\{{r}^{\prime }{R}_{{\rm{E}}}^{\prime }P(z\leqslant {e}_{\mathrm{th}})\}\cdot \displaystyle \frac{1}{\sqrt{2\pi }{\sigma }_{{r}^{\prime }}}\mathrm{exp}\{-\displaystyle \frac{{({r}^{\prime }-r)}^{2}}{2{\sigma }_{{r}^{\prime }}^{2}}\}{\rm{d}}{r}^{\prime },\end{eqnarray} \tag{ 30 }$$

where ${\sigma }_{{r}^{\prime }}^{2}=r(1-r)/n$ is the variance of ${r}^{\prime }=\gamma /n$. It is efficient for Eve to maximize the net eavesdropping information amount, R_netE, and Alice and Bob conduct privacy amplification assuming the maximized R_netE.

The mutual information between Alice and Bob, on the other hand, is determined by Bob's error-rate and the probability that he discards a sifted key because of its large error-rate. Denoting the information amount shared between Alice and Bob without discarding as ${R}_{\mathrm{AB}}^{\prime }$, the net information amount between them, R_netAB, is given by

$$\begin{eqnarray}&&{R}_{\mathrm{netAB}}={\displaystyle \int }_{-\infty }^{\infty }\{{R}_{\mathrm{AB}}^{\prime }P(z\leqslant {e}_{\mathrm{th}})\}\cdot \displaystyle \frac{1}{\sqrt{2\pi }{\sigma }_{{r}^{\prime }}}\mathrm{exp}\{-\displaystyle \frac{{({r}^{\prime }-r)}^{2}}{2{\sigma }_{{r}^{\prime }}^{2}}\}{\rm{d}}{r}^{\prime }.\end{eqnarray} \tag{ 31 }$$

Therefore, the net final key creation rate on IRA, ${R}_{\mathrm{net}f}$, is given by

$$\begin{eqnarray}{R}_{\mathrm{net}f} & = & {\displaystyle \int }_{-\infty }^{\infty }[\{{R}_{\mathrm{AB}}^{\prime }-{r}^{\prime }{R}_{{\rm{E}}}^{\prime }\}P(z\leqslant {e}_{\mathrm{th}})]\cdot \displaystyle \frac{1}{\sqrt{2\pi }{\sigma }_{{r}^{\prime }}}\mathrm{exp}\{-\displaystyle \frac{{({r}^{\prime }-r)}^{2}}{2{\sigma }_{{r}^{\prime }}^{2}}\}{\rm{d}}{r}^{\prime }\\ & = & {R}_{\mathrm{netAB}}-{R}_{\mathrm{netE}}.\end{eqnarray} \tag{ 32 }$$

The previous subsection derives the final key creation rate, assuming that Eve's information amount ${R}_{{\rm{E}}}^{\prime }$ and the mutual information amount between Alice and Bob ${R}_{\mathrm{AB}}^{\prime }$ are given. In this subsection, we discuss these parameters. In IRA, Eve measures all of Alice's signal in the intercepting stage. Thus, the intensity variance of Eve's received signal when Alice sends bit 0, ${\sigma }_{{\rm{E}}0}^{2}$, is given by

$$\begin{eqnarray}&&{\sigma }_{{\rm{E}}0}^{2}=B({{aI}}_{0}^{2}+{{bI}}_{0}).\end{eqnarray} \tag{ 33 }$$

The variance when Alice sends bit 1 is similarly determined. Then, the joint probabilities between Alice and Eve in IRA are obtained using the same procedure as in section 3.3.1, the results of which are summarized in table 3. Using these joint probabilities, we can estimate ${R}_{{\rm{E}}}^{\prime }$.

Table 3.  Joint probabilities between Alice and Eve for IRA.

Alice's bit a	Eve's bit e	Joint probability ${P}_{{\rm{A}},{\rm{E}}}(a,e)$
0	0	$\displaystyle \frac{1}{2}-\displaystyle \frac{1}{4}{\rm{erfc}}(\displaystyle \frac{{d}_{{\rm{E}}}-{I}_{0}}{\sqrt{2}{\sigma }_{{\rm{E}}0}})$
0	1	$\displaystyle \frac{1}{4}{\rm{erfc}}(\displaystyle \frac{{d}_{{\rm{E}}}-{I}_{0}}{\sqrt{2}{\sigma }_{{\rm{E}}0}})$
1	0	$\displaystyle \frac{1}{4}{\rm{erfc}}(\displaystyle \frac{{I}_{1}-{d}_{{\rm{E}}}}{\sqrt{2}{\sigma }_{{\rm{E}}1}})$
1	1	$\displaystyle \frac{1}{2}-\displaystyle \frac{1}{4}{\rm{erfc}}(\displaystyle \frac{{I}_{1}-{d}_{{\rm{E}}}}{\sqrt{2}{\sigma }_{{\rm{E}}1}})$

Regarding Eve as a transmitter, we can estimate the conditional probabilities of Bob's bits. To deceive Bob, Eve resends a fake signal with the same properties as Alice's signal. For such signals, the probabilities of Bob's bits when Eve resends bit 0 or 1 are the same as those when Alice sends bit 0 or 1, which are summarized in table 4.

Table 4.  Conditional probabilities between Eve and Bob for IRA.

Eve's bit a	Bob's bit b	Conditional probability ${P}_{{\rm{B}}| {\rm{E}}}(b| e)$
0	0	$\displaystyle \frac{1}{2}{\rm{erfc}}(\displaystyle \frac{{i}_{0}-{d}_{0}}{\sqrt{2}{\sigma }_{0}})$
0	X	$1\ -\displaystyle \frac{1}{2}{\rm{erfc}}(\displaystyle \frac{{i}_{0}-{d}_{0}}{\sqrt{2}{\sigma }_{0}})-\displaystyle \frac{1}{2}{\rm{erfc}}(\displaystyle \frac{{d}_{1}-{i}_{0}}{\sqrt{2}{\sigma }_{0}})$
0	1	$\displaystyle \frac{1}{2}{\rm{erfc}}(\displaystyle \frac{{d}_{1}-{i}_{0}}{\sqrt{2}{\sigma }_{0}})$
1	0	$\displaystyle \frac{1}{2}{\rm{erfc}}(\displaystyle \frac{{i}_{1}-{d}_{0}}{\sqrt{2}{\sigma }_{1}})$
1	X	$1\ -\displaystyle \frac{1}{2}{\rm{erfc}}(\displaystyle \frac{{d}_{1}-{i}_{1}}{\sqrt{2}{\sigma }_{1}})-\displaystyle \frac{1}{2}{\rm{erfc}}(\displaystyle \frac{{i}_{1}-{d}_{0}}{\sqrt{2}{\sigma }_{1}})$
1	1	$\displaystyle \frac{1}{2}{\rm{erfc}}(\displaystyle \frac{{d}_{1}-{i}_{1}}{\sqrt{2}{\sigma }_{1}})$

Here, however, there is a difference between Alice and Eve in their bit creation probabilities. Eve's bit creation probabilities can be estimated from the joint probabilities between Alice and Eve that are summarized in table 3. Taking these into account, a joint probability between Eve and Bob in IRA, ${P}_{{\rm{B}},{\rm{E}}}$, is obtained as

$$\begin{eqnarray}&&{P}_{{\rm{B}},{\rm{E}}}(0,0)={P}_{{\rm{E}}}(0){P}_{{\rm{B}}| {\rm{E}}}(0| 0)=({P}_{{\rm{A}},{\rm{E}}}(0,0)+{P}_{{\rm{A}},{\rm{E}}}(1,0))\cdot {P}_{{\rm{B}}| {\rm{E}}}(0| 0),\end{eqnarray} \tag{ 34 }$$

The other joint probabilities are similarly obtained. From these joint probabilities between Eve and Bob, we obtain ${R}_{{\rm{E}}}^{\prime }$, which was given a temporary value in the previous subsection.

The mutual information between Alice and Bob in IRA, ${R}_{\mathrm{AB}}^{\prime }$, is separable into those with and without Eve's interception, ${R}_{\mathrm{onAB}}$ and ${R}_{\mathrm{offAB}}$, as

$$\begin{eqnarray}&&{R}_{\mathrm{AB}}^{\prime }={{rR}}_{\mathrm{onAB}}+(1-r){R}_{\mathrm{offAB}}.\end{eqnarray} \tag{ 35 }$$

${R}_{\mathrm{offAB}}$ is equal to the mutual information without Eve, which is discussed in section 3.1. On the other hand, ${R}_{\mathrm{onAB}}$ is evaluated based on the joint probabilities between Alice and Bob via Eve. These values are obtained from the joint probabilities between Alice and Eve summarized in table 3 and Bob's bit probabilities for Eve's bits are summarized in table 4, as

$$\begin{eqnarray}{P}_{{\rm{A}},{\rm{B}}}(0,0) & = & {P}_{{\rm{A}}| {\rm{E}}}(0| 0)\cdot {P}_{{\rm{B}}| {\rm{E}}}(0| 0)\cdot {P}_{{\rm{E}}}(0)+{P}_{{\rm{A}}| {\rm{E}}}(0| 1)\cdot {P}_{{\rm{B}}| {\rm{E}}}(0| 1)\cdot {P}_{{\rm{E}}}(1)\\ & = & \displaystyle \frac{{P}_{{\rm{A}},{\rm{E}}}(0,0)}{{P}_{{\rm{E}}}(0)}\displaystyle \frac{{P}_{{\rm{B}},{\rm{E}}}(0,0)}{{P}_{{\rm{E}}}(0)}{P}_{{\rm{E}}}(0)+\displaystyle \frac{{P}_{{\rm{A}},{\rm{E}}}(0,1)}{{P}_{{\rm{E}}}(1)}\displaystyle \frac{{P}_{{\rm{B}},{\rm{E}}}(0,1)}{{P}_{{\rm{E}}}(1)}{P}_{{\rm{E}}}(1)\\ & = & \displaystyle \frac{{P}_{{\rm{A}},{\rm{E}}}(0,0){P}_{{\rm{B}},{\rm{E}}}(0,0)}{{P}_{{\rm{E}}}(0)}+\displaystyle \frac{{P}_{{\rm{A}},{\rm{E}}}(0,1){P}_{{\rm{B}},{\rm{E}}}(0,1)}{{P}_{{\rm{E}}}(1)}.\end{eqnarray} \tag{ 36 }$$

Futhermore, we can obtain the error-rates, e_Bob and e_Eve, from these joint probabilities. From table 1, Bob's original error-rate, e_Bob, is given by

$$\begin{eqnarray}&&{e}_{\mathrm{Bob}}=\displaystyle \frac{{P}_{{\rm{A}},{\rm{B}}}(0,1)+{P}_{{\rm{A}},{\rm{B}}}(1,0)}{{P}_{{\rm{A}},{\rm{B}}}(0,0)+{P}_{{\rm{A}},{\rm{B}}}(0,1)+{P}_{{\rm{A}},{\rm{B}}}(1,0)+{P}_{{\rm{A}},{\rm{B}}}(1,1)}.\end{eqnarray} \tag{ 37 }$$

Similarly, Eve's error-rate, e_Eve, is estimated from table 3.