Application of Machine Learning Methods in Baikal-GVD: Background Noise Rejection and Selection of Neutrino-Induced Events

Abstract

Baikal-GVD is a large ($\sim$1 km${}^{3}$) underwater neutrino telescope located in Lake Baikal, Russia. In this report, we present two machine learning techniques developed for its data analysis. First, we introduce a neural network for an efficient rejection of noise hits, emerging due to natural water luminescence. Second, we develop a neural network for distinguishing muon- and neutrino-induced events. By choosing an appropriate classification threshold, we preserve $90\%$ of neutrino-induced events, while muon-induced events are suppressed by a factor of $10^{-6}$. Both of the developed neural networks employ the causal structure of events and surpass the precision of standard algorithmic approaches.

APPENDIX

1.1 DERIVATION OF Eq. (3)

In this section we derive distributions, expected values and dispersions of random values in Eq. (2). Then, using them, we evaluate the error of estimating the number of neutrino events using Eq. (2). In this section, we use $P$ to denote probability of some outcome for a random variable, and $M$ and $D$ for its expected value and dispersion, accordingly.

We start by discussing the properties of random variables in the experimental dataset. Let the latter contains $n^{0}$ events in total, $n_{\nu}^{0}$ of which are neutrino-induced and $n_{\mu}^{0}\equiv n^{0}-n_{\nu}^{0}$ are EAS-induced. $n_{\nu}^{0}$ and $n_{\mu}^{0}$ are random variables distributed according to the Poisson law with parameters $\nu$ and $\mu$, respectively. Hence,

$$P(n_{\nu}^{0}=k)=\frac{\nu^{k}e^{-\nu}}{k!},$$

(A1)

$$P(n_{\mu}^{0}=k)=\frac{\mu^{k}e^{-\mu}}{k!}.$$

(A2)

Since $n^{0}$ is a sum of $n^{0}_{\nu}$ and $n^{0}_{\mu}$, it also follows the Poisson distribution,

$$P(n^{0}=k)=\frac{(\nu+\mu)^{k}e^{-(\nu+\mu)}}{k!}.$$

(A3)

Its expected value and dispersion are:

$$M(n^{0})=D(n^{0})=\nu+\mu.$$

(A4)

Let us now address the classification of events by the neural network. A trained neural network can be considered as a black box. As it was discussed in Subsection 4.3, for a fixed classification threshold $\xi$, the network classifies a neutrino-induced event correctly with some probability $E$, and EAS-induced event is identified incorrectly with the probability $S$. Hence, the number of identified true and false neutrino-induced events are independent random variables with binomial distributions:

$$P(n_{\nu}=k|n_{\nu}^{0}=m)=Bin(m,E)(k),$$

(A5)

$$P(n_{\mu}=k|n_{\mu}^{0}=m)=Bin(m,S)(k).$$

(A6)

Here $n_{\nu}(\xi)\equiv n_{\nu}$, $n_{\mu}(\xi)\equiv n_{\mu}$, and $Bin(m,p)(k)$ stands for the binomial distribution with number of experiments $m$ and success probability $p$:

$$Bin(m,p)(k)=C_{m}^{k}p^{k}(1-p)^{m-k}.$$

(A7)

The number of neutrino-induced events identified by the neural network on a test dataset is subject to both of the above-described random processes. Hence, the full probability distributions, $P_{\text{f}}$, of $n_{\nu}$ and $n_{\mu}$ can be obtained by multiplying the corresponding Poisson and binomial distributions,

$$P_{\text{f}}(n_{\nu}=k)$$

$${}=\sum_{m=0}^{\infty}P(n_{\nu}=k|n_{\nu}^{0}=m)P(n_{\nu}^{0}=m),$$

(A8)

$$P_{\text{f}}(n_{\mu}=k)$$

$${}=\sum_{m=0}^{\infty}P(n_{\mu}=k|n_{\mu}^{0}=m)P(n_{\mu}^{0}=m).$$

(A9)

Using Eqs. (A8), (A9), (A1), and Eq. (A2), one can evaluate the expected values and dispersions of $n_{\nu}$ and $n_{\mu}$:

$$M(n_{\nu})=D(n_{\nu})=E\nu,$$

(A10)

$$M(n_{\mu})=D(n_{\mu})=S\mu.$$

(A11)

For the random variable $n$, which is a sum of $n_{\nu}$ and $n_{\mu}$, one has:

$$M(n)=D(n)=E\nu+S\mu.$$

(A12)

Next, let us elaborate on the distribution of $\tilde{E}$ and $\tilde{S}$ considered as random variables. Let the test Monte Carlo dataset contains $\tilde{n}^{0}$ events, $\tilde{n}_{\nu}^{0}$ of which are neutrino-induced and $\tilde{n}_{\mu}^{0}$ are EAS-induced. Since the measured exposure and suppression, $\tilde{E}$ and $\tilde{S}$, are given by Eq. (1), they follow the probability distributions

$$P(\tilde{E}=\alpha)=P(\tilde{n}_{\nu}=\alpha\tilde{n}_{\nu}^{0})$$

$${}=Bin(\tilde{n}_{\nu}^{0},E)(\alpha\tilde{n}_{\nu}^{0}),$$

(A13)

$$P(\tilde{S}=\beta)=P(\tilde{n}_{\mu}=\beta\tilde{n}_{\mu}^{0})$$

$${}=Bin(\tilde{n}_{\mu}^{0},S)(\beta\tilde{n}_{\mu}^{0}),$$

(A14)

where $\alpha$ and $\beta$ are discrete variables so that $\alpha\tilde{n}_{\nu}$ and $\beta\tilde{n}_{\mu}$ are integers. Hence,

$$M(\tilde{E})=E;\quad D(\tilde{E})=\frac{E(1-E)}{\tilde{n}_{\nu}^{0}},$$

(A15)

$$M(\tilde{S})=S;\quad D(\tilde{S})=\frac{S(1-S)}{\tilde{n}_{\mu}^{0}},$$

(A16)

proving that $\tilde{E}$ and $\tilde{S}$ are unbiased estimates of true values of $E$ and $S$, respectively.

Now, we are ready to evaluate the dispersion of $N_{\xi}$ estimated using Eq. (2). For this purpose, we used the standard formula for the dispersion of a function of random variables,

$$\sigma_{N_{\xi}}^{2}=\sum_{v}\left(\frac{\partial N_{\xi}}{\partial v}\right)^{2}\sigma_{v}^{2}$$

$${}+2\sum_{v\neq u}\left(\frac{\partial N_{\xi}}{\partial v}\right)\left(\frac{\partial N_{\xi}}{\partial u}\right)Cov_{v,u}.$$

(A17)

Here, $v$ and $u$ denote arguments of the function $N_{\xi}$, which are $n(\xi)\equiv n$, $n(0)\equiv n^{0}$, $\tilde{E}(\xi)$ and $\tilde{S}(\xi)$; $\sigma_{v}^{2}$ stands for squared variance of $v$, and $Cov_{v,u}$ denotes covariance between $v$ and $u$.

Let us explicitly write out the estimation of the variances and covariances. According to Eq. (A4), $\sigma_{n^{0}}^{2}$ can be estimated as

$$\sigma_{n^{0}}^{2}=n^{0}.$$

(A18)

Further, from Eq. (A12), one gets

$$\sigma_{n}^{2}=n.$$

(A19)

Next, $\sigma_{\tilde{E}}^{2}$ and $\sigma_{\tilde{S}}^{2}$ can be obtained from Eq. (A15) and Eq. (A16),

$$\sigma_{\tilde{E}}^{2}=\frac{\tilde{E}(1-\tilde{E})}{\tilde{n}_{\nu}^{0}},\quad\sigma_{\tilde{S}}^{2}=\frac{\tilde{S}(1-\tilde{S})}{\tilde{n}_{\mu}^{0}}.$$

(A20)

Finally, note that there are only two dependent random variables in Eq. (2)—$n$ and $n^{0}$. Their covariance can be calculated using Eqs. (A3), (A8), and (A9),

$$Cov_{n^{0},n}=E\nu+S\mu=M(n).$$

(A21)

Therefore this covariance can be estimated as $n$.

By calculating the partial derivatives of $N_{\xi}$ in Eq. (A17) and substituting the obtained expressions for variances and covariances, we arrive at the final result:

$$\sigma_{N_{\xi}}^{2}=\frac{(n-n^{0}\tilde{S})^{2}}{(\tilde{E}-\tilde{S})^{4}}\frac{\tilde{E}(1-\tilde{E})}{\tilde{n}_{\nu}^{0}}$$

$${}+\frac{(n-n^{0}\tilde{E})^{2}}{(\tilde{E}-\tilde{S})^{4}}\frac{\tilde{S}(1-\tilde{S})}{\tilde{n}_{\mu}^{0}}$$

$${}+\frac{n-2n{\tilde{S}}+n^{0}(\tilde{S})^{2}}{(\tilde{E}-\tilde{S})^{2}}.$$

(A22)