EEG-Based Person Authentication with Variational Universal Background Model

Abstract

Silent speech is a convenient and natural way for person authentication as users can imagine speaking their password instead of typing it. However there are inherent noises and complex variations in EEG signals making it difficult to capture correct information and model uncertainty. We propose an EEG-based person authentication framework based on a variational inference framework to learn a simple latent representation for complex data. A variational universal background model is created by pooling the latent models of all users. A likelihood ratio of user claimed model to the background model is constructed for testing whether the claim is valid. Extensive experiments on three datasets show the advantages of our proposed framework.

A Appendix

1.1 A.1 KL Divergence Derivations

We state the first proposition without proof.

Proposition 3

Let $f(x)=\mathcal {N}(x;\mu _{f},\varSigma _{f})$ and $g(x)=\mathcal {N}(x;\mu _{g},\varSigma _{g})$ be two Gaussian distributions in $\mathbb {R}^{n}.$ The Kullback-Leibler (KL) divergence between f(x) and g(x) is:

$$\begin{aligned} {D_{KL}}(f(x)\Vert g(x))=&-\frac{n}{2}\log \left( \frac{\det \varSigma _{f}}{\det \varSigma _{g}}\right) -\frac{n}{2}\\&+\frac{1}{2}\text {trace}\left( \varSigma _{f}\varSigma _{g}^{-1}\right) +\frac{1}{2}(\mu _{f}-\mu _{g})\varSigma _{g}^{-1}(\mu _{f}-\mu _{g})^{T} \end{aligned}$$

The following proposition provides the basis for the training objective of our proposed VGM model.

Proposition 4

The variational upper bound of the Kullback-Leibler divergence between a unimodal distribution f(x) and a mixture model $g(x)=\sum _{k}\alpha _{k}g_{k}(x)$  is:

$$ {D_{KL}}(f(x)\Vert g(x))\le -\log \sum _{k}\alpha _{k}\exp \left( -D_{k}\right) $$

where $D_{k}=\int f(x)\log \frac{f(x)}{g_{k}(x)}dx$ is the KL divergence between f(x) and $g_{k}(x)$, the unimodal component distribution of the mixture g(x).

Proof

We have

$$\begin{aligned} {D_{KL}}(f(x)\Vert g(x))=\int f(x)\log \frac{f(x)}{g(x)}dx&=\int f(x)\log f(x)dx-\int f(x)\log g(x)dx\end{aligned}$$

(18)

$$\begin{aligned}&=L_{f}-L_{g} \end{aligned}$$

(19)

where $L_{f}=\int f(x)\log f(x)dx$ and $L_{g}=\int f(x)\log \sum _{k}\alpha _{k}g_{k}(x)dx$, and the index k runs from 1 to K.

We will use variational method to lower bound $L_{g}$. Let us introduce variational variables $a_{1},\dots ,a_{K}$, with $\sum _{k}a_{k}=1$, $a_{k}>0$, $k=1,\dots ,K$,

$$\begin{aligned} L_{g}&=\int f(x)\log \sum _{k}a_{k}\frac{\alpha _{k}g(x)}{a_{k}}dx\end{aligned}$$

(20)

$$\begin{aligned}&\ge \int f(x)\sum _{k}a_{k}\log \frac{\alpha _{k}g_{k}(x)}{a_{k}}dx=L'_{g} \end{aligned}$$

(21)

where we use Jenssen inequality in Eq. 21. We want to find the $a_{k}$ that maximizes this lower bound. Since Eq. 21 is concave in each $a_{k}$ it has a global maximum. Let take partial derivative of $L'_{g}$ w.r.t. each $a_{k}$ then set it to zero and let us denote $D_{k}={D_{KL}}\left( f(x)\Vert g_{k}(x)\right)$ for brevity:

$$\begin{aligned} \frac{\partial L'_{g}}{\partial a_{k}}=0&=\int f(x)\left( \log \frac{\alpha _{k}g_{k}(x)}{a_{k}}+a_{k}(-\frac{1}{a_{k}})\right) dx\end{aligned}$$

(22)

$$\begin{aligned} 0&=\int f(x)\left( \log \alpha _{k}+\log g_{k}(x)-\log a_{k}-1\right) dx\end{aligned}$$

(23)

$$\begin{aligned} \log a_{k}&=\int f(x)\left( \log \alpha _{k}+\log g_{k}(x)-\log f(x) +\log f(x)-1\right) dx\end{aligned}$$

(24)

$$\begin{aligned} \log a_{k}&=\log \alpha _{k}-D_{k}+\int f(x)\left( \log f(x)-1\right) dx\end{aligned}$$

(25)

$$\begin{aligned} a_{k}&=\frac{\alpha _{k}\exp \left( -D_{k}\right) }{\exp \left( \int f(x)\left( 1-\log f(x)\right) dx\right) }\end{aligned}$$

(26)

$$\begin{aligned} \sum _{k=1}^{K}a_{k}=1&=\frac{\sum _{k}\alpha _{k}\exp \left( -D_{k}\right) }{\exp \left( \int f(x)\left( 1-\log f(x)\right) dx\right) }\end{aligned}$$

(27)

$$\begin{aligned} \exp (\int f(x)\left( 1-\log f(x)\right) dx)&=\sum _{k}\alpha _{k}\exp \left( -D_{k}\right) \end{aligned}$$

(28)

where we have used $\int f(x)dx=1$ in Eq. 24.

From Eq. 26 and Eq. 28 we have:

$$\begin{aligned} a_{k}&=\frac{\alpha _{k}\exp \left( -D_{k}\right) }{Z}\end{aligned}$$

(29)

$$\begin{aligned} \text {where}\,\,Z&=\sum _{k=1}^{K}\alpha _{k}\exp \left( -D_{k}\right) \end{aligned}$$

(30)

Plug $L'_{g}$ in Eq. 21 into Eq. 18:

$$\begin{aligned} {D_{KL}}(f(x)\Vert g(x))&=L_{f}-L_{g}\nonumber \\&\le L_{f}-L'_{g}\nonumber \\&=\int f(x)\log f(x)dx-\int f(x)\sum _{k}a_{k}\log \frac{\alpha _{k}g_{k}(x)}{a_{k}}dx\nonumber \\&=\int f(x)\left[ \log f(x)-\sum _{k}a_{k}\log \frac{\alpha _{k}g_{k}(x)}{a_{k}}\right] dx\nonumber \\&=\int f(x)\left[ \log f(x)-\sum _{k}a_{k}\log \frac{g_{k}(x)Z}{\exp \left( -{D_{KL}}_{k}\right) }\right] dx\end{aligned}$$

(31)

$$\begin{aligned}&=-\log Z+\int f(x)\left[ \log f(x)-\sum _{k}a_{k}\log \frac{g_{k}(x)}{\exp \left( -D_{k}\right) }\right] dx\end{aligned}$$

(32)

$$\begin{aligned}&=-\log Z+A \end{aligned}$$

(33)

where we substituted $a_{k}=\frac{\alpha _{k}\exp \left( -D_{k}\right) }{Z}$ in Eq. 31 and move $\log Z$ out of the integral in Eq. 32.

We will show that $A=0$:

$$\begin{aligned} A&=\int f(x)\left[ \log f(x)-\sum _{k}a_{k}\log \frac{g_{k}(x)}{\exp \left( -D_{k}\right) }\right] dx\\&=\int f(x)\left[ \sum _{k}a_{k}\log f(x)-\sum _{k}a_{k}\log \frac{g_{k}(x)}{\exp \left( -D_{k}\right) }\right] dx\\&=\sum _{k}a_{k}\left[ \int f(x)\left( \log f(x)-\log g_{k}(x)\right) \right] dx-D_{k}\\&=\sum _{k}a_{k}D_{k}-D_{k}\\&=0 \end{aligned}$$

where we again used the facts that $\int f(x)dx=1$ and $\sum _{k}a_{k}=1$.

We have shown that ${D_{KL}}(f(x)\Vert g(x))\le L_{f}-L'_{g}=-\log Z$.    $\blacksquare$