Abstract
We address the problem of controlling a small team of robots to estimate the location of a mobile target using nonlinear rangeonly sensors. Our control law maximizes the mutual information between the team’s estimate and future measurements over a finite time horizon. Because the computations associated with such policies scale poorly with the number of robots, the time horizon associated with the policy, and typical nonparametric representations of the belief, we design approximate representations that enable realtime operation. The main contributions of this paper include the control policy, an algorithm for approximating the belief state, and an extensive study of the performance of these algorithms using simulations and real world experiments in complex, indoor environments.
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Acknowledgments
We gratefully acknowledge the support of ONR Grant N000140710829, ARL Grant W911NF0820004, and AFOSR Grant FA95501010567. Benjamin Charrow was supported by a NDSEG fellowship from the Department of Defense.
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Appendix: Proofs
Appendix: Proofs
1.1 Integrating Gaussians over a halfspace
Proving Lemmas 1 and 2 requires integrating Gaussian functions over a halfspace. The following identities will be used several times. Owen (1980) gives these identities for 1dimensional Gaussians, but we have been unable to find a reference for the multivariate case. For completeness, we prove them here.
Lemma 6
If \(f(x)=\mathcal {N}(x; \mu , \Sigma )\) is a \(k\)dimensional Gaussian and \(A=\{x : a^Tx + b > 0\}\) is a halfspace then
where \(p=(a^T\mu +b)/\sqrt{a^T\Sigma a}\) is a scalar, \(\phi \left( x\right) =\mathcal {N}(x;0,1)\) is the PDF of the standard 1dimensional Gaussian and \(\Phi \left( x\right) \) is its CDF.
Proof of (9)
All of these integrals are evaluated by making the substitution \(x=Ry\), where \(R\) is a rotation matrix that makes the halfspace \(A\) axis aligned. Specifically, define \(R\) such that \(a^TR=\Vert a\Vert e_1^T\) where \(e_1\) is a \(k\)dimensional vector whose first entry is 1 and all others are 0. This substitution is advantageous, because it makes the limits of integration for all components of \(y\) except \(y_1 [\infty ,\infty ]\).
Because \(R^Tx=y\), \(y\) is a \(k\)dimensional Gaussian with density \(q(y)=\mathcal {N}(y; R^T\mu , R^T\Sigma R)\). The determinant of the Jacobian of \(y\) is the determinant of a rotation matrix: \(\partial y / \partial x=R^T=1\). Substituting:
where \(q_1(y_1)=\mathcal {N}(y_1; e_1^T (R^T\mu ), e_1^T(R^T\Sigma R)e_1)\) is the density for the first component of \(y\). The final step follows as the limits of integration marginalize \(q(y)\). To simplify the remaining integral, apply the definition of \(R\), \(q_1(y_1) =\mathcal {N}(y_1;\mu ^Ta/\Vert a\Vert , a^T\Sigma a/ \Vert a\Vert ^2)\), and use \(1\Phi \left( x\right) =\Phi \left( x\right) \):
\(\square \)
Proof of (10)
First, we perform a change of variables so that the integral is evaluated over the standard multivariate Gaussian \(g(x)=\mathcal {N}(x;0, I)\). This substitution is \(x=\Sigma ^{1/2}y+\mu \) which can be seen by noting that 1) \(\partial y / \partial x=\Sigma ^{1/2}\) and 2) \({f(x)=\Sigma ^{1/2}g(\Sigma ^{1/2}(x\mu ))}\):
where \(B=\{y : (a^T\Sigma ^{1/2})y +(a^T\mu +b) > 0\}\) is the transformed halfspace.
To evaluate the first term in (12), we calculate \(\int _C yg(y)\,dy\), where \(C=\{y: c^Ty + d > 0\}\) is a new generic halfspace. This integral is easier than the original problem as \(g(y)\) is the density of a zeromean Gaussian with identity covariance. To proceed, substitute \(y=Rz\) where \(R\) is a rotation matrix that makes \(C\) axisaligned (i.e., \(c^TR=\Vert c\Vert e_1^T\)) and observe that \(g(Rz)=g(z)\):
\(e_1\) appears because \(g(x)\) is 0mean; the only nonzero component of the integral is from \(z_1\). The 1dimensional integral follows as \(d\phi (x)/dx=x\phi (x)\).
To finish, (12) can be evaluated using the formula in (9) and (13):
\(\square \)
Proof of (11)
Similar to the last proof, make the substitution \({x=\Sigma ^{1/2}y+\mu }\) with \(g(y)\) as the standard multivariate Gaussian and expand terms.
where \(B=\{y : (a^T\Sigma ^{1/2})y +(a^T\mu +b) > 0\}\) is the transformed halfspace.
To evaluate (14), we only need a formula for the first integral; the previous proofs have expressions for the other 3 integrals. To evaluate the first integral, let \(C=\{y: c^Ty + d > 0\}\) be a new halfspace and use the same rotation substitution \(y=Rz\) as in the last proof.
The previous integral can be evaluated by analyzing each scalar component. \(g\) is the standard multivariate Gaussian, so \(g(z)=\prod _{i=1}^k\phi (z_i)\). There are three types of terms:

\(z_1^2\): The limits of integration marginalize \(g\) and the resulting integral can be solved using integration by parts:
$$\begin{aligned} \int _{z_1\Vert c\Vert +d>0} z_1^2 g(z)\,dz&=\int _{d/\Vert c\Vert }^\infty z_1^2\phi (z_1)\,dz_1 \nonumber \\&=\Phi \left( \frac{d}{\Vert c\Vert }\right) \frac{d}{\Vert c\Vert }\phi \left( \frac{d}{\Vert c\Vert }\right) \end{aligned}$$ 
\(z_i^2; (i > 1)\):
$$\begin{aligned}&\int _{z_1\Vert c\Vert +d>0} z_i^2 g(z)\,dz \nonumber \\ {}&=\int _{d/\Vert c\Vert }^\infty \phi (z_1)\,dz_1 \int _{\infty }^\infty z_i^2 \phi (z_i)\,dz_i = \Phi \left( \frac{d}{\Vert c\Vert }\right) \end{aligned}$$The integral over \(z_i\) is 1 because it’s the variance of the standard normal.

\(z_iz_j\; (i\ne j,i\ne 1)\): These indices cover all nondiagonal elements.
$$\begin{aligned}&\int _{z_1\Vert c\Vert +d>0} z_i z_j g(z)\,dz \nonumber \\&\quad ={\int _{\alpha }^\beta z_j\phi (z_j)\,dz_j}{\int _{\infty }^\infty z_i \phi (z_i)\,dz_i}=0 \end{aligned}$$\(z_i\ne 1\), so its limits are the real line. Because \(g\) is 0 mean the integral is 0.
We now have formula for all of the terms in (14). Recall that \(B=\{x : (\Sigma ^{1/2} a)^Tx + (a^T\mu + b) > 0\}\). Using \(p=(a^T\mu +b)/\sqrt{a^T\Sigma a}\), (9), (10), and (15):
which is the formula we sought. \(\square \)
1.2 Proof of Lemma 1
The norm can be evaluated by splitting the integral up into regions where the absolute value disappears. Let \(A\) be the set where \(f(x) > g(x)\) and \(A^c\) be the complement of \(A\) where \(f(x) \le g(x)\). Noting that \(\int _{A} g(x) = 1  \int _{A^c} g(x)\):
\(f\) and \(g\) have the same covariance, so \(f\) is bigger than \(g\) on a halfspace \(A=\{x : a^Tx + b > 0\}\) where \(a=\Sigma ^{1}(\mu _1\mu _2)\) and \(b=(\mu _1+\mu _2)\Sigma ^{1}(\mu _2\mu _1)/2\). This means (16) can be evaluated using Lemma 6. To do this, we need to evaluate \(p\) for \(\int _A f(x)\,dx\) and \(\int _A g(x)\,dx\). There are three relevant terms:
Using \(\delta =\Vert \mu _1\mu _2\Vert _\Sigma /2\) we get \((a^T\mu _1+b)/\sqrt{a^T\Sigma a}=\delta \) and \((a^T\mu _2+b)/\sqrt{a^T\Sigma a}=\delta \). Plugging these values into Lemma 6 completes the proof.
1.3 Proof of Lemma 2
Let \(X\) be a random variable whose density is \(f(x)g(x)/\Vert fg\Vert _1\). Calculating the entropy of \(X\) is difficult as the expression involves the log of the absolute value of the difference of exponentials. Fortunately, the covariance of \(X\) can be calculated. This is useful because the maximum entropy of any distribution with covariance \(\Sigma \) is \((1/2)\log {((2\pi e)^k \Sigma )}\), the entropy of the multivariate Gaussian (Cover and Thomas 2004, Thm. 8.6.5). By explicitly calculating the determinant of the covariance of \(X\), we prove the desired bound.
To calculate \(X\)’s covariance, use the formula \(\hbox {cov}{X}=\mathbb {E}_{}\left[ XX^T\right] \mathbb {E}_{}\left[ X\right] \mathbb {E}_{}\left[ X\right] ^T\). Similar to the proof of Lemma 1, we evaluate the mean by breaking the integral up into a region \(A\) where \(f(x)>g(x)\) and \(A^C\) where \(g(x)\ge f(x)\) viceversa:
As before, \(f\) and \(g\) have the some covariance, so \(A = \{x : a^Tx + b >0\}\) and \(A^c=\{x : (a)^Tx + (b) \ge 0\}\) are halfspaces with \(a=\Sigma ^{1}(\mu _1\mu _2)\) and \(b=(\mu _1+\mu _2)\Sigma ^{1}(\mu _2\mu _1)/2\).
Each of the terms in (20) are Gaussian functions integrated over a halfspace, which can be evaluated using Lemma 6. To simplify the algebra, we evaluate the integrals involving \(f\) and \(g\) separately. This involves calculating a few terms, three of which are repeats: (17), (18), and (19). The other term is \(\Sigma a = \mu _1\mu _2\). As the difference of the means will arise repeatedly, define \(\Delta =\mu _1\mu _2\). Noting \(2\delta =\Vert \Delta \Vert _\Sigma \) and \(\phi (x)=\phi (x)\), the integrals involving \(f\) are:
Next, evaluate the integrals in (20) involving \(g\). This can be done by pattern matching from (21). The main change is that \(\mu _1\) becomes \(\mu _2\) and the sign of \(p\) in Lemma 6 flips, meaning the sign of the arguments to \(\phi \left( \cdot \right) \) and \(\Phi \left( \cdot \right) \) flip (see (17) and (18)).
Subtracting (22) from (21), recognizing \(\phi (x)=\phi (x)\), and dividing by \(\Vert fg\Vert _1=2(\Phi \left( \delta \right) \Phi \left( \delta \right) )\) simplifies (20):
We now evaluate \(X\)’s second moment. Split the integral up over \(A\) and \(A^c\):
Once again, we evaluate this expression by separately evaluating the integrals involving \(f\) and \(g\).
Starting with \(f\) and using Lemma 6 with \(\Delta \) and \(\delta \):
Taking the difference of (25) and (26):
To evaluate the integrals in (24) involving \(g\), we can pattern match using (25) and (26). As in the derivation of (22), this involves negating the \(p\) terms and replacing \(\mu _1\) with \(\mu _2\).
Note the sign difference of \(\Delta \Delta ^T\) compared to (27).
To finish calculating the second moment, subtract (28) from (27) and divide by \(\Vert fg\Vert _1\), simplifying (24)
We can now express the covariance of \(X\).
Which follows as \((\mu _1\mu _1^T+\mu _2\mu _2^T)/2(\mu _1+\mu _2)(\mu _1+\mu _2)^T/4=\Delta \Delta ^T/4\).
To calculate the determinant of the covariance, factor \(\Sigma \) out of (29) and define \(\alpha =\delta ^2 + \frac{2\phi \left( \delta \right) \delta }{2\Phi \left( \delta \right) 1}\):
The last step follows from the eigenvalues. \(\Sigma ^{1}\Delta \Delta ^T\) only has one nonzero eigenvalue; it is a rank one matrix as \(\Sigma ^{1}\) is full rank and \(\Delta \Delta ^T\) is rank one. The trace of a matrix is the sum of its eigenvalues, so \(tr(\Sigma ^{1}\Delta \Delta ^T) =tr(\Delta ^T\Sigma ^{1}\Delta )=\Vert \Delta \Vert _\Sigma ^2=4\delta ^2\) is the nonzero eigenvalue. Consequently, the only nonzero eigenvalue of \(\alpha \Sigma ^{1}\Delta \Delta ^T/4\delta ^2\) is \(\alpha \). Adding \(I\) to a matrix increases all its eigenvalues by 1 so the only eigenvalue of \(I+\alpha \Sigma ^{1}\Delta \Delta ^T/4\delta ^2\) that is not 1 has a value of \(1 + \alpha \). The determinant of a matrix is the product of its eigenvalues, so \(I+\alpha \Sigma ^{1}\Delta \Delta ^T/4\delta ^2=1+\alpha \). As discussed at the beginning of the proof, substituting (30) into the expression for the entropy of a multivariate Gaussian achieves the desired upper bound.
1.4 Proof of Theorem 2
To prove the theorem, we build on Theorem 1. Unfortunately, it cannot be directly applied because it is difficult to evaluate 1) the \(L_1\) norm between GMMs and 2) the entropy of the normalized difference of mixture models. However, Lemmas 1 and 2 provide a way to evaluate these quantities when the densities are individual Gaussians. To exploit this fact, we split the problem up and analyze the difference in entropies between GMMs that only differ by a single component.
To start, define \(d_j(x)=\sum _{i=1}^j w_i g_i(x) + \sum _{i=j+1}^n w_i f_i(x)\). \(d_j\) is a GMM whose first \(j\) components are the first \(j\) components in \(g\) and last \(nj\) components are the last \(nj\) components in \(f\). Note that \(d_0 (x) = g(x)\) and \(d_n (x) = g(x)\). Using the triangle inequality:
where the last step applied the same trick \(n2\) more times. Because \(d_{j1} (x)  d_{j} (x) = w_j (f_j (x)  g_j (x))\) each term in the summand can be bounded using Theorem 1:
Because \(f_j(x)\) and \(g_j(x)\) are Gaussians with the same covariance, we can apply Lemmas 1 and 2 to complete the proof.
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Charrow, B., Kumar, V. & Michael, N. Approximate representations for multirobot control policies that maximize mutual information. Auton Robot 37, 383–400 (2014). https://doi.org/10.1007/s1051401494112
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DOI: https://doi.org/10.1007/s1051401494112