Cited By
View all- Waqas MAhmed STahir MWu JQureshi R(2024)Exploring Multiple Instance Learning (MIL): A brief surveyExpert Systems with Applications10.1016/j.eswa.2024.123893250(123893)Online publication date: Oct-2024
Multiple-instance Learning from Triplet Comparison Bags
Article No.: 90, Pages 1 - 18
Abstract
Multiple-instance learning (MIL) solves the problem where training instances are grouped in bags, and a binary (positive or negative) label is provided for each bag. Most of the existing MIL studies need fully labeled bags for training an effective classifier, while it could be quite hard to collect such data in many real-world scenarios, due to the high cost of data labeling process. Fortunately, unlike fully labeled data, triplet comparison data can be collected in a more accurate and human-friendly way. Therefore, in this article, we for the first time investigate MIL from only triplet comparison bags, where a triplet (Xa, Xb, Xc) contains the weak supervision information that bag Xa is more similar to Xb than to Xc. To solve this problem, we propose to train a bag-level classifier by the empirical risk minimization framework and theoretically provide a generalization error bound. We also show that a convex formulation can be obtained only when specific convex binary losses such as the square loss and the double hinge loss are used. Extensive experiments validate that our proposed method significantly outperforms other baselines.
Appendices
A Generation Process of Triplet Comparison Bags
Recall the assumption that three bags in a triplet are sampling independently. Therefore, for a triplet \((X_{a}, X_{b}, X_{c})\) , the bag labels \((Y_a,Y_b,Y_c)\) can only appear to be one of the following cases:Otherwise, the first bag is more similar to the third bag than to the second bag, and in this case, \((Y_a,Y_b,Y_c)\) appears to be one of the following cases:According to the above distributions \(\mathcal {Y}_{1}\) and \(\mathcal {Y}_{2}\) , we can actually collect two distinct types of datasets as follows:The two types of datasets \(\mathcal {D}_{1}\) and \(\mathcal {D}_{2}\) can be considered to be generated from the following underlying distributions:where \(\theta _{T} = 1- \theta _{+}\theta _{-}\) , \(\theta _{+} = p(y=+1)\) and \(\theta _{-} = p(y=-1)\) and \(p_{+}(X)=p(X|y=+1)\) and \(p_{-}(X)=p(X|y=-1)\) . Then, we haveFurthermore, we denote the pointwise data collected from \(\mathcal {D}_1\) and \(\mathcal {D}_2\) by ignoring the triplet comparison relation as \(\mathcal {D}_{1,a} = \lbrace X_{1,a}\rbrace ^{m_{1}}\) , \(\mathcal {D}_{1,b} = \lbrace X_{1,b}\rbrace ^{m_{1}}\) , \(\mathcal {D}_{1,c} = \lbrace X_{1,c}\rbrace ^{m_{1}}\) , \(\mathcal {D}_{2,a} = \lbrace X_{2,a}\rbrace ^{m_{2}}\) , \(\mathcal {D}_{2,b} = \lbrace X_{2,b}\rbrace ^{m_{2}}\) and \(\mathcal {D}_{2,c} = \lbrace X_{2,c}\rbrace ^{m_{2}}\) . From Theorem 1 in Cui et al. [12], samples in \(\mathcal {D}_{1,a}\) , \(\mathcal {D}_{1,c}\) , \(\mathcal {D}_{2,a}\) and \(\mathcal {D}_{2,c}\) are independently drawn fromsamples in \(\mathcal {D}_{1,b}\) are independently drawn fromand samples in \(\mathcal {D}_{2,b}\) are independently drawn fromThose indicate that from triplet comparison data, we can essentially obtain samples that can be drawn independently from three different distributions. Then, we denote the three aggregated datasets (from respective distributions) aswhereLet \(C = \frac{\theta _{+}^3+2\theta _{+}^2\theta _{-}}{\theta _{T}}\) and \(D = \frac{2\theta _{+}\theta _{-}^2+ \theta _{-}^3}{\theta _{T}}\) , we can express the relationship between these densities as
\(\begin{eqnarray*} \nonumber \mathcal {Y}_{1} = \lbrace (+1,+1,+1),(+1,+1,-1),(+1,-1,-1),(-1,+1,+1),(-1,-1,+1),(-1,-1,-1)\rbrace . \end{eqnarray*}\)
\(\begin{eqnarray*} \nonumber \mathcal {Y}_{2} = \lbrace (+1,-1,+1),(-1,+1,-1)\rbrace . \end{eqnarray*}\)
\(\begin{eqnarray*} \nonumber \mathcal {D}_{1} = \lbrace (X_{a},X_{b},X_{c})|(Y_a,Y_b,Y_c)\in \mathcal {Y}_{1}\rbrace , \quad \mathcal {D}_{2} = \lbrace (X_{a},X_{b},X_{c})|(Y_a,Y_b,Y_c)\in \mathcal {Y}_{2}\rbrace . \end{eqnarray*}\)
\(\begin{eqnarray*} \nonumber p_{1}(X_{a},X_{b},X_{c}) &= \frac{p(X_{a},X_{b},X_{c}, (Y_a,Y_b,Y_c)\in \mathcal {Y}_{1})}{\theta _{T}}, \\ \nonumber p_{2}(X_{a},X_{b},X_{c}) &= \theta _{+}p_{+}(X)p_{+}(X)p_{-}(X) + \theta _{-}p_{-}(X)p_{+}(X)p_{-}(X), \end{eqnarray*}\)
\(\begin{eqnarray*} \nonumber \mathcal {D}_1 = \lbrace (X_{1,a},X_{1,b},X_{1,c})\rbrace ^{m_1}\sim p_{1}(X_{a},X_{b},X_{c}), \quad \mathcal {D}_2 = \lbrace (X_{2,a},X_{2,b},X_{2,c})\rbrace ^{m_2}\sim p_{2}(X_{a},X_{b},X_{c}). \end{eqnarray*}\)
\(\begin{eqnarray*} \nonumber \tilde{p}_{1}(X) = \theta _{+}p_{+}(X) + \theta _{-}p_{-}(X), \end{eqnarray*}\)
\(\begin{eqnarray*} \nonumber \tilde{p}_{2}(X) = \frac{(\theta _{+}^3+2\theta _{+}^2\theta _{-})p_{+}(X) + (2\theta _{+}\theta _{-}^2+ \theta _{-}^3)p_{-}(X)}{\theta _{T}}, \end{eqnarray*}\)
\(\begin{eqnarray*} \nonumber \tilde{p}_{3}(X) = \theta _{-}p_{+}(X) + \theta _{+}p_{-}(X). \end{eqnarray*}\)
\(\begin{eqnarray*} \nonumber \tilde{\mathcal {D}}_{1} =\lbrace \tilde{X}_{i}^{1}\rbrace ^{n_1}_{i=1}= \mathcal {D}_{1,a}\cup \mathcal {D}_{1,c}\cup \mathcal {D}_{2,a}\cup \mathcal {D}_{2,c}, \quad \tilde{\mathcal {D}}_{2} =\lbrace \tilde{X}_{i}^{2}\rbrace ^{n_2}_{i=1} = \mathcal {D}_{1,b}, \quad \tilde{\mathcal {D}}_{3} =\lbrace \tilde{X}_{i}^{3}\rbrace ^{n_3}_{i=1} = \mathcal {D}_{2,b}, \end{eqnarray*}\)
\(\begin{eqnarray*} \nonumber \tilde{\mathcal {D}}_{1} \sim \tilde{p}_{1}(X), \quad \tilde{\mathcal {D}}_{2} \sim \tilde{p}_{2}(X), \quad \tilde{\mathcal {D}}_{3} \sim \tilde{p}_{3}(X). \end{eqnarray*}\)
\(\begin{eqnarray*} \nonumber \begin{bmatrix} \tilde{p}_{1}(X)\\ \tilde{p}_{2}(X)\\ \tilde{p}_{3}(X) \end{bmatrix} = \begin{bmatrix} \theta _{+} &\theta _{-}\\ C & D\\ \theta _{-} &\theta _{+} \end{bmatrix}\begin{bmatrix} p_{+}(X)\\ p_{-}(X) \end{bmatrix}. \end{eqnarray*}\)
B Proof of Theorem 1
Recall that by using the loss function that satisfies the linear-odd condition, \(\widehat{R}_{\mathrm{Trip}}(g)\) can be also represented asIn this way, we can represent \({R}_{\mathrm{Trip}}(g)\) aswhere we assumed that the collected data \(\lbrace X^1_{i}\rbrace _{i=1}^{n_1}\) are independently sampled from \(\widetilde{p}_{1}(X)\) , the collected data \(\lbrace X^2_{i}\rbrace _{i=1}^{n_2}\) are independently sampled from \(\widetilde{p}_{2}(X)\) and the collected data \(\lbrace X^3_{i}\rbrace _{i=1}^{n_3}\) are independently sampled from \(\widetilde{p}_{3}(X)\) . Let us further introduceIn this way, we haveThus,Hence, the problem becomes how to find an upper bound of each term in the right hand size of the inequality.
\(\begin{align*} \widehat{R}_{\mathrm{Trip}}(g) =&\,\, \frac{1}{n_1}\sum \limits _{i=1}^{n_1}\Big ((\lambda _1+\lambda _2)\ell _+\left(g\left(X_i^1\right)\right)+\lambda _2g(X_i^1)\Big) +\frac{1}{n_2}\sum \limits _{i=1}^{n_2}\Big ((\lambda _3+\lambda _4)\ell _+\left(g\left(X_i^2\right)\right)+\lambda _4 g\left(X_i^2\right)\!\Big)\\ &+\frac{1}{n_3}\sum \limits _{i=1}^{n_3}\Big ((\lambda _5+\lambda _6)\ell _-\left(g\left(X_i^3\right)\right)-\lambda _5 g\left(X_i^3\right)\!\Big). \end{align*}\)
\(\begin{align*} {R}_{\mathrm{Trip}}(g) =&\,\, \mathbb {E}_{\widetilde{p}_{1}(X)}\Big [ (\lambda _1+\lambda _2)\ell _+(g(X^1))+\lambda _2 g(X^1)\Big ] +\mathbb {E}_{\widetilde{p}_{2}(X)}\Big [ (\lambda _3+\lambda _4)\ell _+(g(X^2))+\lambda _4 g(X^2)\Big ] \\ &+\mathbb {E}_{\widetilde{p}_{3}(X)}\Big [ (\lambda _5+\lambda _6)\ell _-(g(X^3))-\lambda _5 g(X^3)\Big ], \end{align*}\)
\(\begin{eqnarray*} \nonumber \widehat{R}_{1}(g) =\frac{1}{n_1}\sum \limits _{i=1}^{n_1}\left((\lambda _1+\lambda _2)\ell _+\left(g\left(X_i^1\right)\!\right) +\lambda _2g\left(X_i^1\right)\!\right)\!, \quad R_{1}(g) =\mathbb {E}_{\widetilde{p}_{1}(X)}\Big [(\lambda _1+\lambda _2)\ell _+(g(X^1))+\lambda _2 g(X^1)\Big ],\\ \nonumber \widehat{R}_{2}(g) = \frac{1}{n_2}\sum \limits _{i=1}^{n_2}\left((\lambda _3+\lambda _4)\ell _+\left(g\left(X_i^2\right)\!\right)+\lambda _4 g\left(X_i^2\right)\!\right)\!, \quad R_{2}(g) =\mathbb {E}_{\widetilde{p}_{2}(X)}\Big [ (\lambda _3+\lambda _4)\ell _+(g(X^2))+\lambda _4 g(X^2)\Big ],\\ \nonumber \widehat{R}_{3}(g) = \frac{1}{n_3}\sum \limits _{i=1}^{n_3}\left((\lambda _5+\lambda _6)\ell _-\left(g\left(X_i^3\right)\right)-\lambda _5 g\left(X_i^3\right)\!\right)\!, \quad R_{3}(g) =\mathbb {E}_{\widetilde{p}_{3}(X)}\Big [ (\lambda _5+\lambda _6)\ell _-(g(X^3))-\lambda _5 g(X^3)\Big ]. \end{eqnarray*}\)
\(\begin{eqnarray*} \nonumber \widehat{R}_{\mathrm{Trip}}(g) =\widehat{R}_{1}(g) + \widehat{R}_{2}(g) + \widehat{R}_{3}(g),\quad {R}_{\mathrm{Trip}}(g) = R_{1}(g) + R_{2}(g) + R_{3}(g). \end{eqnarray*}\)
\(\begin{eqnarray*} \nonumber \sup _{g\in \mathcal {G}}\left|{R}_{\mathrm{Trip}}(g)-\widehat{R}_{\mathrm{Trip}}(g)\right|\le \sup _{g\in \mathcal {G}}\left|{R}_{1}(g)-\widehat{R}_{1}(g)\right|+\sup _{g\in \mathcal {G}}\left|{R}_{2}(g)-\widehat{R}_{2}(g)\right| +\sup _{g\in \mathcal {G}}\left|{R}_{3}(g)-\widehat{R}_{3}(g)\right|. \end{eqnarray*}\)
With the introduced definitions and conditions in Theorem 1, for any \(\delta \gt 0\) , with probability at least \(1-\delta\) , we have
\(\begin{eqnarray*} \nonumber \sup _{g\in \mathcal {G}}\left| R_{1}(g)-\widehat{R}_{1}(g)\right| \le & (\left|\lambda _1\right|+\left|\lambda _2\right|)\left(\frac{2C_{\mathcal {G}}}{\sqrt {n_1}}+C_{\boldsymbol {w}}C_{\boldsymbol {\phi }}\sqrt {\frac{\log \frac{2}{\delta }}{2n_1}}\right)\!. \end{eqnarray*}\)
First, it is easy to verify that the double hinge loss \(\ell _{\mathrm{DH}}\) is 1-Lipschitz. Suppose an example in \(\widehat{R}_{1}(g)\) is replaced by another arbitrary example, then the change of \(\sup _{g\in \mathcal {G}}\big (R_{1}(g)-\widehat{R}_{1}(g)\big)\) is no greater than \((\left|\lambda _1\right|+\left|\lambda _2\right|)C_{\boldsymbol {w}}C_{\boldsymbol {\phi }}/n_{1}\) . Then, by applying McDiarmid’s inequality [28], for any \(\delta \gt 0\) , with probability at least \(1-\frac{\delta }{2}\) ,Besides, it is routine [30] to showwhere we have used the Talagrand’s lemma (Lemma 4.2 in Mohri et al. [30]), i.e., \(\mathfrak {R}_{n}(\ell \circ \mathcal {G})\le \rho \mathfrak {R}_n(\mathcal {G})\) if \(\ell\) is a \(\rho\) -Lipschitz loss function. By considering \(\mathfrak {R}_n(\mathcal {G})\le C_{\mathcal {G}}/\sqrt {n}\) , we haveBy further taking into account the other side \(\sup _{g\in \mathcal {G}}\big (\widehat{R}_{1}(g)-R_{1}(g)\big)\) , we have for any \(\delta \gt 0\) , with probability at least \(1-\delta\) ,which completes the proof of Lemma 1. □
\(\begin{eqnarray*} \nonumber \sup _{g\in \mathcal {G}}\big (R_{1}(g)-\widehat{R}_{1}(g)\big) &\le \mathbb {E}\Big [\sup _{g\in \mathcal {G}}\big (R_{1}(g)-\widehat{R}_{1}(g)\big)\Big ]+ (\left|\lambda _1\right|+\left|\lambda _2\right|)C_{\boldsymbol {w}}C_{\boldsymbol {\phi }}\sqrt {\frac{\log \frac{2}{\delta }}{2n_1}}. \end{eqnarray*}\)
\(\begin{eqnarray*} \nonumber \mathbb {E}\Big [\sup _{g\in \mathcal {G}}\big (R_{1}(g)-\widehat{R}_{1}(g)\big)\Big ]\le 2(\left|\lambda _1\right|+\left|\lambda _2\right|)\mathfrak {R}_{n_1}(\mathcal {G}), \end{eqnarray*}\)
\(\begin{eqnarray*} \nonumber \sup _{g\in \mathcal {G}}\big (R_{1}(g)-\widehat{R}_{1}(g)\big) \le & (\left|\lambda _1\right|+\left|\lambda _2\right|)\left(\frac{2C_{\mathcal {G}}}{\sqrt {n_1}}+C_{\boldsymbol {w}}C_{\boldsymbol {\phi }}\sqrt {\frac{\log \frac{2}{\delta }}{2n_1}}\right)\!. \end{eqnarray*}\)
\(\begin{eqnarray*} \nonumber \sup _{g\in \mathcal {G}}\left|R_{1}(g)-\widehat{R}_{1}(g)\right| \le & (\left|\lambda _1\right|+\left|\lambda _2\right|)\left(\frac{2C_{\mathcal {G}}}{\sqrt {n_1}}+C_{\boldsymbol {w}}C_{\boldsymbol {\phi }}\sqrt {\frac{\log \frac{2}{\delta }}{2n_1}}\right)\!, \end{eqnarray*}\)
With the introduced definitions and conditions in Theorem 1, for any \(\delta \gt 0\) , with probability at least \(1-\delta\) , we have
\(\begin{eqnarray*} \nonumber \sup _{g\in \mathcal {G}}\left| R_{2}(g)-\widehat{R}_{2}(g)\right| \le & (\left|\lambda _3\right|+\left|\lambda _4\right|)\left(\frac{2C_{\mathcal {G}}}{\sqrt {n_2}}+C_{\boldsymbol {w}}C_{\boldsymbol {\phi }}\sqrt {\frac{\log \frac{2}{\delta }}{2n_2}}\right)\!. \end{eqnarray*}\)
With the introduced definitions and conditions in Theorem 1, for any \(\delta \gt 0\) , with probability at least \(1-\delta\) , we have
\(\begin{eqnarray*} \nonumber \sup _{g\in \mathcal {G}}\left| R_{2}(g)-\widehat{R}_{2}(g)\right| \le & (\left|\lambda _5\right|+\left|\lambda _6\right|)\left(\frac{2C_{\mathcal {G}}}{\sqrt {n_3}}+C_{\boldsymbol {w}}C_{\boldsymbol {\phi }}\sqrt {\frac{\log \frac{2}{\delta }}{2n_3}}\right)\!. \end{eqnarray*}\)
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Publication History
Published: 12 February 2024
Online AM: 02 January 2024
Accepted: 15 December 2023
Revised: 12 May 2023
Received: 08 June 2022
Published in TKDD Volume 18, Issue 4
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View all- Waqas MAhmed STahir MWu JQureshi R(2024)Exploring Multiple Instance Learning (MIL): A brief surveyExpert Systems with Applications10.1016/j.eswa.2024.123893250(123893)Online publication date: Oct-2024
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