Transfer learning for Bayesian discovery of multiple Bayesian networks

Abstract

Bayesian network structure learning algorithms with limited data are being used in domains such as systems biology and neuroscience to gain insight into the underlying processes that produce observed data. Learning reliable networks from limited data is difficult; therefore, transfer learning can improve the robustness of learned networks by leveraging data from related tasks. Existing transfer learning algorithms for Bayesian network structure learning give a single maximum a posteriori estimate of network models. Yet, many other models may be equally likely, and so a more informative result is provided by Bayesian structure discovery. Bayesian structure discovery algorithms estimate posterior probabilities of structural features, such as edges. We present transfer learning for Bayesian structure discovery which allows us to explore the shared and unique structural features among related tasks. Efficient computation requires that our transfer learning objective factors into local calculations, which we prove is given by a broad class of transfer biases. Theoretically, we show the efficiency of our approach. Empirically, we show that compared to single-task learning, transfer learning is better able to positively identify true edges. We apply the method to whole-brain neuroimaging data.

Appendix

1.1 Extended proof of Theorem 

We derive the equality in Theorem  starting from the joint probability of data and the feature conditioned on an order given in Eq. 9,

$$\begin{aligned} P(f^{(t)}, \mathcal {D} | \! \prec ) = \sum _{G^{(t)} \subseteq \! \prec } f(G^{(t)}) P(D^{(t)} | G^{(t)}) \left( \sum _{G \subseteq \! \prec } P(G^{(t)}, G | \! \prec ) \prod _{k \ne t} P(D^{(k)} | G) \right) . \end{aligned}$$

Modularity properties are used to substitute $P(D|G) = \prod _{i=1}^n P(x_i | \pi _i)$, and, $f(G) = \prod _{i=1}^n f_i(\pi _i)$, and, $P(G^{(t)}, G | \! \prec ) = \prod _{i=1}^n P(\pi _i^{(t)}, \pi _i | U_i)$, as follows,

$$\begin{aligned} P(f^{(t)}, \mathcal {D} | \! \prec )&= \sum _{\pi _1^{(t)} \subseteq U_1} \cdots \sum _{\pi _n^{(t)} \subseteq U_n} \prod _{i=1}^n \left[ f_i(\pi _i^{(t)}) P(x_i^{(t)} | \pi _i^{(t)}) \right] \\&\times \left( \sum _{\pi _1 \subseteq U_1} \cdots \sum _{\pi _n \subseteq U_n} \prod _{i=1}^n P(\pi _i^{(t)}, \pi _i | U_i) \prod _{k \ne t} P(x_i^{(k)} | \pi _i) \right) . \end{aligned}$$

The goal is to factor the terms so that local calculations at each node, $i$, can be calculated independently. First, we factor terms in the structure bias (the part in round parentheses above) as follows:

$$\begin{aligned}&\sum _{\pi _1 \subseteq U_1} \cdots \sum _{\pi _n \subseteq U_n} \prod _{i=1}^n P(\pi _i^{(t)}, \pi _i | U_i) \prod _{k \ne t} P(x_i^{(k)} | \pi _i)\\&\quad = \sum _{\pi _1 \subseteq U_1} \cdots \sum _{\pi _{n-1} \subseteq U_{n-1}} \prod _{i=1}^{n-1} P(\pi _i^{(t)}, \pi _i | U_i) \prod _{k \ne t} P(x_i^{(k)} | \pi _i) \\&\quad \quad \times \left( \sum _{\pi _{n} \subseteq U_{n}} P(\pi _n^{(t)}, \pi _n | U_n) \prod _{k \ne t} P(x_n^{(k)} | \pi _n) \right) \\&\quad = \left( \sum _{\pi _1 \subseteq U_1} P(\pi _1^{(t)}, \pi _1 | U_1) \prod _{k \ne t} P(x_1^{(k)} | \pi _1) \right) \\&\quad \quad \times \left( \sum _{\pi _2 \subseteq U_2} P(\pi _2^{(t)}, \pi _2 | U_2) \prod _{k \ne t} P(x_2^{(k)} | \pi _2) \right) \times \cdots \\&\quad \quad \times \left( \sum _{\pi _n \subseteq U_n}P(\pi _n^{(t)}, \pi _n | U_n) \prod _{k \ne t} P(x_n^{(k)} | \pi _n) \right) \\&\quad = \prod _{i=1}^n \sum _{\pi _i \subseteq U_i} P(\pi _i^{(t)}, \pi _i | U_i) \prod _{k \ne t} P(x_i^{(k)} | \pi _i).\\ \end{aligned}$$

Plugging the factored structure prior back in to the full equation, we now have:

$$\begin{aligned}&P(f^{(t)}, \mathcal {D} | \! \prec ) \\&\quad = \sum _{\pi _1^{(t)} \subseteq U_1} \!\cdots \! \sum _{\pi _n^{(t)} \subseteq U_n} \prod _{i=1}^n f_i(\pi _i^{(t)}) P(x_i^{(t)} | \pi _i^{(t)}) \left( \sum _{\pi _i \subseteq U_i} P(\pi _i^{(t)}, \pi _i | U_i) \prod _{k \ne t} P(x_i^{(k)} | \pi _i) \right) . \end{aligned}$$

If we perform a similar factoring to that done above on the structure prior, we get achieve the desired result:

$$\begin{aligned} P(f^{(t)}, \mathcal {D} | \! \prec ) = \prod _{i=1}^n \sum _{\pi _i^{(t)} \subseteq U_i} f_i(\pi _i^{(t)}) P(x_i^{(t)} | \pi _i^{(t)}) \left( \sum _{\pi _i \subseteq U_i} P(\pi _i^{(t)}, \pi _i | U_i) \prod _{k \ne t} P(x_i^{(k)} | \pi _i) \right) . \end{aligned}$$

1.2 Normalization constant for structure bias

Calculation of the normalization constant requires summing over all possible combinations of parent sets $(\pi _i^{(t)}, \pi _i^{(k)})$, as follows:

$$\begin{aligned} Z = \sum _{\pi _i^{(t)} \subseteq U_i} \sum _{\pi _i^{(k)} \subseteq U_i} (1-\lambda )^{\Delta _{itk}}. \end{aligned}$$

We can simplify this by first fixing parent set $\pi _i^{(t)}$ while counting how many parent sets $\pi _i^{(k)}$ will give $\Delta _{itk} = 0$ ($\pi _i^{(k)}$ can contain any parents from the set $\{U_i {\setminus } \pi _i^{(t)}\}$ but none from $\pi _i^{(t)}$); then how many $\pi _i^{(k)}$ will give $\Delta _{itk} = 1$ ($\pi _i^{(k)}$ can contain any parents from the set $\{U_i {\setminus } \pi _i^{(t)}\}$ and exactly one from $\pi _i^{(t)}$); etc, up to the maximum of $\Delta _{itk} = |\pi _i^{(t)}|$. This sum turns out to be a binomial expansion, and so we can write it in closed form. The expansion is demonstrated here:

$$\begin{aligned} Z&= \sum _{\pi _i^{(t)} \subseteq U_i} \left[ \sum _{\{\pi _i^{(k)} \subseteq U_i \big | \Delta _{itk} = 0\}}\!\!\!\!\!\!\! 1 + \sum _{\{\pi _i^{(k)} \subseteq U_i \big | \Delta _{itk} = 1\}}\!\!\!\!\!\!\!\!\!\!\!\! (1-\lambda ) + \cdots + \!\!\!\!\!\!\!\!\!\! \sum _{\{\pi _i^{(k)} \subseteq U_i \big | \Delta _{itk} = |\pi _i^{(t)}|\}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (1-\lambda )^{| \pi _i^{(t)}|} \right] \\&= \sum _{\pi _i^{(t)} \subseteq U_i} \left[ \left( {\begin{array}{c}| \pi _i^{(t)}|\\ 0\end{array}}\right) 2^{| U_i | - | \pi _i^{(t)}|} + \left( {\begin{array}{c}| \pi _i^{(t)}|\\ 1\end{array}}\right) 2^{| U_i | - | \pi _i^{(t)}|}(1-\lambda ) + \cdots \right. \\&\left. \quad +\, \left( {\begin{array}{c}| \pi _i^{(t)}|\\ | \pi _i^{(t)}|\end{array}}\right) 2^{| U_i | - | \pi _i^{(t)}|}(1-\lambda )^{| \pi _i^{(t)}|} \right] \\&= \sum _{\pi _i^{(t)} \subseteq U_i} \sum _{j=0}^{| \pi _i^{(t)}|} \left( {\begin{array}{c}| \pi _i^{(t)}|\\ j\end{array}}\right) 2^{| U_i | - | \pi _i^{(t)}|} (1-\lambda )^j \\&= \sum _{\pi _i^{(t)} \subseteq U_i} 2^{| U_i | - | \pi _i^{(t)}|} \sum _{j=0}^{| \pi _i^{(t)}|} \left( {\begin{array}{c}| \pi _i^{(t)}|\\ j\end{array}}\right) (1-\lambda )^j \\&= 2^{| U_i |} \sum _{\pi _i^{(t)} \subseteq U_i} 2^{- | \pi _i^{(t)}|} (1 + 1 - \lambda )^{| \pi _i^{(t)}|}\\&= 2^{| U_i |} \sum _{\pi _i^{(t)} \subseteq U_i} \left( \frac{2-\lambda }{2} \right) ^{| \pi _i^{(t)}|} \\&= 2^{| U_i |} \sum _{\pi _i^{(t)} \subseteq U_i} \left( 1-\frac{\lambda }{2}\right) ^{| \pi _i^{(t)}|} . \end{aligned}$$

Next, we perform a similar expansion of the sum over parent sets $\pi _i^{(t)}$ that have size $|\pi _i^{(t)}| = 0$, and $|\pi _i^{(t)}| = 1$, etc up to the maximum $|\pi _i^{(t)}| = |U_i|$. This sum also turns out to be a binomial expansion and therefore can be simplified into a closed form, as demonstrated:

$$\begin{aligned} Z&= 2^{| U_i |} \sum _{\pi _i^{(t)} \subseteq U_i} \left( 1-\frac{\lambda }{2}\right) ^{| \pi _i^{(t)}|}\\&= 2^{| U_i |} \left[ \sum _{\{ \pi _i^{(t)} \subseteq U_i \big | | \pi _i^{(t)} | = 0\}}\!\!\!\!\!\!\! 1 + \sum _{\{ \pi _i^{(t)} \subseteq U_i \big | | \pi _i^{(t)} | = 1\}}\!\!\!\!\!\!\!\!\!\!\!\! \left( 1 - \frac{\lambda }{2} \right) + \cdots + \!\!\!\!\!\!\!\!\!\!\!\!\!\! \sum _{\{ \pi _i^{(t)} \subseteq U_i \big | | \pi _i^{(t)} | = | U_i |\}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \left( 1 - \frac{\lambda }{2} \right) ^{| U_i |} \right] \\&= 2^{| U_i |} \left[ \left( {\begin{array}{c}|U_i|\\ 0\end{array}}\right) + \left( {\begin{array}{c}|U_i|\\ 1\end{array}}\right) \left( 1-\frac{\lambda }{2} \right) + \cdots + \left( {\begin{array}{c}|U_i|\\ |U_i|\end{array}}\right) \left( 1 - \frac{\lambda }{2} \right) ^ {|U_i|} \right] \\&= 2^{| U_i |} \sum _{j=0}^{|U_i|} \left( {\begin{array}{c}|U_i|\\ j\end{array}}\right) \left( 1 - \frac{\lambda }{2}\right) ^j\\&= 2^{| U_i |} \left( 2-\frac{\lambda }{2}\right) ^{| U_i |} \\&= (4 - \lambda )^{| U_i |}. \end{aligned}$$

1.3 Integration of Bayesian model average

We integrate over all values of the parameter $\lambda$ to obtain the Bayesian model average over all possible structure priors, thus eliminating the need to select a point-estimate for $\lambda$. We use an uninformative, uniform prior for $\lambda$, i.e., $p(\lambda |U_i) = 1$ for $0 \le \lambda \le 1$.

$$\begin{aligned} P(\pi _i^{(t)}, \pi _i^{(k)} | U_i)&= \int _0^1 P(\pi _i^{(t)}, \pi _i^{(k)} | U_i, \lambda ) p(\lambda | U_i) d\lambda \\&= \int _0^1 \frac{(1-\lambda ) ^{\Delta _{itk}}}{(4 - \lambda )^{|U_i|}} p(\lambda | U_i) d\lambda \\&= \int _0^1 \frac{(1-\lambda ) ^ {\Delta _{itk}}}{(4 - \lambda )^{|U_i|}} d\lambda \\ \end{aligned}$$

Next, we use an identity given by Euler in 1748 [1]. If $\beta$ is the beta function and $_2\text {F}_{\!1}$ is the ordinary hypergeometric function, then

$$\begin{aligned} \int _0^1 x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a} dx = \beta (b, c-b) _2\text {F}_{\!1} (a, b; c; z) \quad \text {for } \mathfrak {R}(c) > \mathfrak {R}(b) > 0 . \end{aligned}$$

Let $x = \lambda ,\,a = |U_i|,\,b = 1,\,c = \Delta _{itk} + 2$, and $z = 1/4$. Then the condition, $\Delta _{itk} + 2 > 1 > 0$, holds for any $\Delta _{itk} \ge 0$ which is the valid range for $\Delta _{itk}$ and:

$$\begin{aligned} \int _0^1\!\! \lambda ^{0}(1-\lambda )^{\Delta _{itk}}(1-\lambda /4)^{-|U_i|} d\lambda&= \beta (1, \Delta _{itk} + 1) _2\text {F}_{\!1} (|U_i|, 1; \Delta _{itk} + 2; 1/4) \\ \int _0^1 \frac{4^{|U_i|} (1-\lambda )^{\Delta _{itk}}}{(4-\lambda )^{|U_i|}} d \lambda&= \left( \frac{0! \Delta _{itk}!}{(\Delta _{itk} + 1)!}\right) {_2}\text {F}_{\!1} (|U_i|, 1; \Delta _{itk} + 2; 1/4) \\ \int _0^1 \frac{(1-\lambda )^{\Delta _{itk}}}{(4-\lambda )^{|U_i|}} d \lambda&= \left( \!\frac{1}{4^{|U_i|} (\Delta _{itk} + 1)} \!\right) {_2}\text {F}_{\!1} (|U_i|, 1; \Delta _{itk} + 2; 1/4) ,\\ \end{aligned}$$

giving the result:

$$\begin{aligned} P(\pi _i^{(t)}, \pi _i^{(k)} | U_i) = \frac{_2\text {F}_{\!1}(|U_i|, 1; \Delta _{itk}+2; 1/4)}{4^{|U_i|} (\Delta _{itk} + 1)} \end{aligned}$$

We only need the solution to this formula for a limited number of combinations of integer values of $\Delta _{itk}$ and $|U_i|,\,0 \le \Delta _{itk} \le |U_i| < n$, where $n$ is the number of variables in the network. Therefore, we pre-compute a lookup table of the necessary values using the GNU Scientific Library² hypergeometric function solver.