The variational approach to concentration

June 10, 2024

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Let \((S_n)\) be some stochastic process in, say, \(\Re^d\). For instance, \(S_n = \sum_{i=1}^n X_i\) for multivariate observations \(X_i\in\Re^d\). We are aiming to generate a high probability bound on

\[\norm{S_n} = \sup_{v:\norm{v}=1} \la v,S_n\ra.\]There are several well-known ways to attempt this. Among them are covering, chaining, and Doob decomposition (done, e.g., here). Here we’ll explore a separate (and relatively new) approach.

The idea is to use the variational inequality which is at the core of the PAC-Bayesian methodology (so we could equivalently call this the PAC-Bayes approach to concentration). This lets us simultaneously bound \(\la v,S_n\ra\) in each direction \(v\). Recall that a PAC-Bayes bound has the form

\[\begin{equation} \label{eq:pb-basic-bound} \Pr( \forall \rho \in \Mspace{\Theta}: \text{Something holds}) \geq 1-\delta, \tag{1} \end{equation}\]where \(\Theta\) is some parameter space and \(\Mspace{\Theta}\) is the set of all probability measures over that space.
A PAC-Bayes bound provides a high probability bound *simultaneously* over all posteriors. The variational approach to concentration translates this into a high probability bound simultaneously over all directions.

It’s worth taking a moment to understand why this simultaneity property is valuable. A natural thought is to treat \(\la v,S_n\ra\) as a scalar-valued process and apply well understood concentration results for real-valued functions to it. Doing this in the naive way would give a separate bound for each \(v\in\sd\). So we’d have a result of the form:

\[\forall v\in\sd, \Pr(\la v,S_n\ra \geq B_n) \leq \delta.\]But now we’re stuck. This is not a bound on \(\sup_{v\in\sd}\la v,S_n\ra\). One’s first thought is to take a union bound over \(v\) in order to move the “\(\forall v\in\sd\)” inside the probability statement, which would give us the result. But there are uncountably many such vectors. The approach explored here solves this problem by translating the “\(\forall \rho\in\Mspace{\Theta}\)” in \eqref{eq:pb-basic-bound} into “\(\forall v\in\sd\)”.

This variational approach was pioneered by Catoni and Giulini (here and here), and has now been used by a few authors to prove bounds in a variety of settings:

- Zhivotovskiy in 2024 for bounding the singular values of random matrices,
- Nakakita et al. in 2024 for bounding the mean of high-dimensional random matrices under heavy-tails,
- Giulini in 2018 for estimating the Gram operator in Hilbert spaces,
- Myself and others in 2024 for estimating the mean of random vectors.

# A general variational inequality

Different authors use seemingly different PAC-Bayesian inequalities to achieve their results. However, we recently showed that all of these inequalities are specific instantiations of the following more general result.

Let \(\Theta\) be some measurable parameter space and let \(N(\theta)\) be nonnegative and have expected value at most 1 (it’s an e-value, if you like) for all \(\theta\in\Theta\). Then

\[\Pr(\forall\rho\in\Mspace{\Theta}: \E_\rho \log N(\theta) \leq \kl(\rho\|\nu) + \log(1/\delta))\geq 1-\delta,\]where, as above, \(\Mspace{\Theta}\) is the set of all probability measures on \(\Theta\). The variational approach to concentration involves (i) finding some appropriate family of random variables \(N(\theta)\), (ii) choosing \(\nu\) and a family of distributions \(\{\rho_\theta:\theta\in\Theta\}\) such that \(\sup_\theta \kl(\rho_\theta \|\nu)\) is small, and (iii) applying this “master theorem.”

# Example 1: Sub-Gaussian random vectors

This comes from our paper on time-uniform confidence spheres. Consider \(n\) iid copies \(X_1,\dots,X_n\) of a \(\Sigma\)-sub-Gaussian random vector \(X\in\Re^d\). That is,

\[\E\exp(\lambda \la \theta, X\ra) \leq \exp\left(\frac{\lambda^2}{2}\la\theta,\Sigma\theta\ra\right),\]for all \(\lambda\in\Re\) and \(\theta\in\Re^d\). This implies that

\[N(\theta) = \exp\left\{\lambda\sum_{i\leq n}\la \theta, X_i\ra - \frac{n\lambda^2}{2}\la\theta,\Sigma\theta\ra\right\},\]has expectation at most 1. Let \(\nu\) be a Gaussian with mean 0 and covariance \(\beta^{-1}I\) for some \(\beta>0\). Consider the family of distributions \(\{\rho_u:\norm{u}=1\}\) where \(\rho_u\) is a Gaussian with mean \(u\) and covariance \(\beta^{-1}I\). Then the KL divergence between \(\rho_u\) and \(\nu\) is \(\kl(\rho_u\|\nu) = \beta/2\). Using the master theorem above, we obtain that, with probability \(1-\delta\), *simultaneously for all distributions \(\rho\)*,

Now, for \(\rho=\rho_u\), \(\E_\rho \la \theta, X_i\ra = \la u,X_i\ra\) and

\[\E_\rho \la \theta, \Sigma\theta\ra = \la u,\Sigma u\ra + \beta^{-1}\Tr(\Sigma) \leq \norm{\Sigma} + \beta^{-1}\Tr(\Sigma),\]using basic properties of the expectation of quadratic forms under Gaussian distributions (see e.g. here), and definition of the operator norm as \(\norm{A} = \sup_{u,v:\norm{u}=\norm{v}=1}\la u,\Sigma v \ra\). Since this holds simultaneously for all \(\rho_u\), we obtain that, with probability \(1-\delta\),

\[\sup_u \lambda \sum_{i\leq n} \la u,X_i\ra \leq \frac{n\lambda^2}{2}(\norm{\Sigma} + \beta^{-1}\Tr(\Sigma)) + \frac{\beta}{2} + \log(1/\delta).\]The left hand side is equal to \(\lambda \norm{\sum_{i\leq n}X_i}\), which gives us our concentration result. One can then optimize \(\lambda\) using some calculus. This gives us state-of-the-art concentration up to an additive factor of \((\Tr(\Sigma^2)/n)^{1/4}\).

# Example 2: Random matrices with finite Orlicz-norm

This example is adapted from Zhivotovskiy (2024). Let \(M_1,\dots,M_n\) be iid copies of a random matrix \(M\) with finite sub-exponential Orlicz norm, in the sense that, for some \(C>0\),

\[\norm{\la \theta, M\phi\ra}_{\psi_1} \leq C \la\theta, \Sigma\phi\ra,\]for all \(\theta, \phi\in\Re^d\) where \(\Sigma = \E M\). Here

\[\norm{Y}_{\psi_1} = \inf\left\{u>0: \E\exp(|Y|/u)\leq 2\right\}.\]We take our parameter space in the master theorem above to be \(\Theta = \Re^d\times \Re^d\). Let \(\nu\) again be Gaussian with mean 0 and covariance \(\beta^{-1}\Sigma\) and let \(\mu_u\) be a *truncated* Gaussian mean \(u\), covariance \(\beta^{-1}\Sigma\) and radius \(r\). Being slightly loose with notation and writing \(\d\mu\) for the density of \(\mu\), the density of the truncated Gaussian can be written as

where \(Z\) is some normalizing constant and \(\rho_u\) is a non-truncated Gaussian. For a vector \(u\in \Sigma^{1/2}\mathbb{S}^{d-1}\), the KL-divergence between a truncated normal \(\mu_u\) and \(\nu\) is therefore

\[\begin{align*} \kl(\mu_{u} \| \nu) &= \int\log\left(\frac{1}{Z}\frac{\d \rho_u}{\d\nu}(\theta)\right) {\mu}_{u}(\d\theta) \\ &= \log\left(\frac{1}{Z}\right) + \frac{1}{2}\int (\la \theta-u,\beta\Sigma^{-1}(\theta-u)\ra + \la \theta, \beta\Sigma^{-1}\theta\ra )\mu_{u}(\d\theta) \\ &= \log\left(\frac{1}{Z}\right) + \frac{\beta}{2}\int (2\la \theta,\Sigma^{-1}u\ra - \la u, \Sigma^{-1}u\ra )\mu_{u}(\d\theta) \\ &= \log\left(\frac{1}{Z}\right) + \frac{\beta\la u,\Sigma^{-1} u\ra}{2} \leq \log\left(\frac{1}{Z}\right) + \frac{\beta}{2}. \end{align*}\]Here \(Z = \Pr(\norm{\theta - u}\leq r)\) where \(\theta\sim \rho_{u}\). Equivalently, \(Z=\Pr(\norm{Y}\leq r)\) where \(Y\) is a normal with mean \(0\) and covariance \(\beta^{-1}\Sigma\). Hence \(1 - Z = \Pr(\norm{Y}>r)\leq \E\norm{Y}^2/r^2 = \beta^{-1}\Tr(\Sigma)/r^2\). Thus, taking \(r = \sqrt{2 \beta^{-1}\Tr(\Sigma)}\) yields \(Z\geq 1/2\), and we obtain

\[\begin{align} \kl(\mu_u\|\nu) \leq \log(2) + \frac{\beta}{2}. \end{align}\]Therefore, since the KL divergence is additive on product measures, we have that

\[\kl(\rho_u\times\rho_v\|\nu\times\nu) \leq 2\log(2) + \beta.\]Now it remains to construct a relevant quantity to use in the PAC-Bayes theorem. Consider

\[N(\theta,\phi) = \exp\left\{\lambda\sum_{i\leq n}\la\theta, \Sigma^{-1/2}M_i\Sigma^{-1/2}\phi\ra - n\log\E\exp(\lambda\la \theta, \Sigma^{-1/2}M\Sigma^{-1/2}\phi\ra)\right\},\]where the expectation is over \(M\). It’s easy to see that this has expectation at most 1 (it can be written as the product of terms each with expectation exactly one). We apply the master theorem with the product distribution \(\mu_u\times \mu_v\) for \(u,v\in\Sigma^{1/2}\sd\) where \(\sd = \{x:\norm{x}=1\}\) is the unit sphere. Therefore, \(u = \Sigma^{1/2}u'\) for \(v = \Sigma^{1/2}v'\) for some \(u',v'\in\sd\). We obtain that with probability \(1-\delta\), for all \(u,v\),

\[\begin{align} &\lambda \sum_{i\leq n}\E_{\mu_u\times \mu_v} \la \theta, \Sigma^{-1/2}M_i\Sigma^{-1/2}\phi\ra \\ &\leq n \E_{\mu_u\times\mu_v} \log \E\exp(\lambda\la\theta, \Sigma^{-1/2}M\Sigma^{1/2}\phi\ra) + \frac{\beta}{2} + \log(2/\delta).\label{eq:matrix_bound1}\tag{2} \end{align}\]The truncated Gaussian is symmetric about its mean, so the left hand side above becomes

\[\E_{\mu_u\times\mu_v}\la \theta, \Sigma^{-1/2}M_i\Sigma^{-1/2}\phi\ra = \la \Sigma^{-1/2}u,M_i\Sigma^{-1/2}v\ra = \la u', M_i v'\ra,\]where, as before, \(u',v'\in\mathbb{S}^{d-1}\). It remains to bound the right hand side of \eqref{eq:matrix_bound1}. For this we appeal to a result which bounds the MGF of a random variable in terms of its \(\psi_1\)-norm. In particular, we appeal to an exponential inequality, which states that for a random variable \(Y\),

\[\begin{equation} \label{eq:exp_ineq} \E[\exp(\lambda (Y - \E Y))]\leq \exp(4\lambda^2\norm{Y-\E Y}_{\psi_1}^2),\quad \forall |\lambda| \leq \frac{1}{2\norm{Y-\E Y}_{\psi_1}}.\tag{3} \end{equation}\]Applying this with \(Y = \la \theta,\Sigma^{-1/2} M\Sigma^{-1/2}\phi\ra\) and noting that \(\E Y = \la \theta, \phi\ra\), we have

\[\begin{align} & \norm{\la \theta, \Sigma^{-1/2}M \Sigma^{-1/2} \phi\ra - \la \theta, \Sigma^{-1/2}\E M \Sigma^{1/2} \phi\ra}_{\psi_1} \\ &\leq \norm{\la \theta, \Sigma^{-1/2}M \Sigma^{-1/2} \phi\ra}_{\psi_1} + \norm{\la \theta, \Sigma^{-1/2}\E M \Sigma^{-1/2} \phi\ra}_{\psi_1} \\ &\leq \norm{\la \theta, \Sigma^{-1/2}M \Sigma^{-1/2} \phi\ra}_{\psi_1} + \E\norm{\la \theta, \Sigma^{-1/2} \Sigma^{-1/2} \phi\ra}_{\psi_1} \\ &= 2 \norm{\la \Sigma^{-1/2}\theta, M \Sigma^{-1/2} \phi\ra}_{\psi_1} \\ &\leq 2C \la \Sigma^{-1/2}\theta, \Sigma\Sigma^{-1/2}\phi\ra = 2C\la \theta, \phi\ra \leq C(\norm{\theta}^2 + \norm{\phi}^2). \end{align}\]Therefore, \eqref{eq:exp_ineq} yields

\[\begin{align} \E\exp(\lambda \la\theta,\Sigma^{-1/2}M\Sigma^{-1/2}\phi\ra) &\leq \exp(\lambda \la \theta, \phi\ra + 4C^2\lambda^2(\norm{\theta}^2 + \norm{\phi}^2)^2). \label{eq:Eexp} \tag{4} \end{align}\]Note that \(\norm{\theta - u}\leq r\) and \(\norm{u} = \norm{\Sigma^{1/2}u'} \leq \norm{\Sigma^{1/2}} \leq \sqrt{\norm{\Sigma}}\), so

\[\norm{\theta}^2 \leq (r + \norm{u})^2 \leq \left(\sqrt{2\beta^{-1}\Tr(\Sigma)} + \sqrt{\norm{\Sigma}}\right)^2.\]The same bound holds for \(\norm{\phi}^2\). Therefore, \eqref{eq:Eexp} gives

\[\begin{align} &\E_{\rho_u\times\rho_v} \log \E \exp(\lambda\la\theta, \Sigma^{-1/2} M\Sigma^{-1/2}\phi\ra ) \\ &\leq \lambda \la u,v\ra + 8C^2\lambda^2\left(\sqrt{2\beta^{-1}\Tr(\Sigma)} + \sqrt{\norm{\Sigma}}\right)^4 \\ &= \lambda \la u',\Sigma v'\ra + 8C^2\lambda^2\left(\sqrt{2\beta^{-1}\Tr(\Sigma)} + \sqrt{\norm{\Sigma}}\right)^4, \end{align}\]assuming that

\[|\lambda| \leq \frac{1}{4C(\sqrt{2\beta^{-1}\Tr(\Sigma)} + \sqrt{\norm{\Sigma}})^2}.\]Choosing \(\beta = 2\Tr(\Sigma)/\norm{\Sigma}\) and putting everything together, \eqref{eq:matrix_bound1} gives that with probability \(1-\delta,\)

\[\sup_{u',v'\in\sd}\lambda\sum_{i\leq n} \la u', (M_i-\Sigma) v'\ra \lesssim C^2\lambda^2 \norm{\Sigma}^2 + \frac{\Tr(\Sigma)}{\norm{\Sigma}} + \log(1/\delta).\]Dividing by \(n\) and optimizing over \(\lambda\) gives us a final bound that matches state-of-the-art concentration bounds for random matrices.

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