The ridge estimate lies in a time-uniform linear confidence ellipsoid

Since $V_0=\lambda I_d$ is diagonal, [F7] gives $\det V_0={\lambda }^d>0$. On ${\mathcal{H}}_s={\mathcal{F}}_{s+1}$, define $\sigma$ as the first $s\in \{0,\dots ,n\}$ with $Z_s\ge 1\mathrm{/}\delta$, and set $\sigma =n$ if there is no such $s$. For $k<n$, $\{\sigma \le k\}={\bigcup }_{s=0}^k\{Z_s\ge 1\mathrm{/}\delta \}\in {\mathcal{H}}_k$, while $\{\sigma \le n\}=\mathrm{\Omega }$. Hence $\sigma$ is a stopping time bounded by $n$.

For each $0\le s<n$, the model reward equation gives $b_s=G_s{\theta }^{\ast }+S_s=(V_s-\lambda I_d){\theta }^{\ast }+S_s$. Therefore ${\widehat{\theta }}_s-{\theta }^{\ast }=V_s^{-1}{S}_s-\lambda V_s^{-1}{\theta }^{\ast }\mathrm{.}$

For fixed $s$, define $⟨x,y{⟩}_{V_s}=x^{\mathsf{T}}{V}_{s}y$. By [F2] this form is symmetric and positive definite, so [F6] gives its induced norm triangle inequality. Also $\mathrm{\parallel }{V}_s^{-1}{S}_s{\mathrm{\parallel }}_{V_s}^2=S_s^{\mathsf{T}}{V}_s^{-1}{S}_s$ and $\mathrm{\parallel }\lambda V_s^{-1}{\theta }^{\ast }{\mathrm{\parallel }}_{V_s}^2={\lambda }^2({\theta }^{\ast }{)}^{\mathsf{T}}{V}_s^{-1}{\theta }^{\ast }\le \lambda \mathrm{\parallel }{\theta }^{\ast }{\mathrm{\parallel }}_2^2\le \lambda m_2^2\mathrm{.}$

By [F4], $\mathbb{E}[Z_{\sigma }]\le \mathbb{E}[Z_0]=1$. On the event that some $Z_s$ crosses $1\mathrm{/}\delta$ by time $n$, the definition of $\sigma$ gives $Z_{\sigma }\ge 1\mathrm{/}\delta$. Thus [F5] yields $\mathbb{P}\{\mathrm{\exists }s\le n:Z_s\ge 1\mathrm{/}\delta \}\le \delta$.

Outside the crossing event bounded in step 2.1, every $0\le s\le n$ has $Z_s<1\mathrm{/}\delta$. Taking logarithms in [F3]'s formula and using $\det V_0={\lambda }^d$ from step 1.1 gives $S_s^{\mathsf{T}}{V}_s^{-1}{S}_s<2\log (1\mathrm{/}\delta )+\log \text{\hspace{0.17em}⁣}\left(\frac{\det V_s}{{\lambda }^d}\right)\mathrm{.}$ All logarithm arguments are positive by [F2].

Apply the triangle inequality from step 1.3 to step 1.2, then use step 3.1. For $t=s+1$ this yields $\mathrm{\parallel }{\widehat{\theta }}_{t-1}-{\theta }^{\ast }{\mathrm{\parallel }}_{V_{t-1}}\le \sqrt{S_{t-1}^{\mathsf{T}}{V}_{t-1}^{-1}{S}_{t-1}}+\sqrt{\lambda }{m}_2\le {\rho }_t\mathrm{.}$ The radius ${\rho }_t$ is measurable: its determinant input is measurable by [F8], the logarithm is continuous by [F9], and $x\mapsto \sqrt{x}$ is continuous on $[0,\mathrm{\infty })$ since $\mathrm{\mid }\sqrt{x}-\sqrt{y}{\mathrm{\mid }}^2\le \mathrm{\mid }x-y\mathrm{\mid }$ there; the radicand is nonnegative by [F2] and $\delta \in (0,1)$. By [F8] and [F10], the quadratic form defining $C_t$ is measurable. The finite intersection of these events is measurable by [F11], so $\{{\theta }^{\ast }\in C_t\text{ for every }1\le t\le n\}$ is measurable. Its probability is at least $1-\delta$ by step 2.1.