Marginal law of z̄ — the sample mean of the z_i is N(0, (1 + σ²/n) I)

Let $n \ge 1$ and $\sigma > 0$. Draw $\theta \sim \mathcal N(0, I_d)$ and then, given $\theta$, $z_1,\dots,z_n$ i.i.d. from $\mathcal N(\theta, \sigma^2 I_d)$. The sample mean $\bar z = \frac1n\sum_{i=1}^n z_i$ then has marginal law

Combined with $\mu' = \frac{n}{\sigma^2+n}\bar z$, this converts the bound $\Xi \ge \frac12\,\mathbb P_{\mathcal M}[\|\mu'\|_\infty \le \varepsilon]$ into a probability about a single standard Gaussian vector.

Formalization Note $n \ge 1$ is added because the mean of zero samples is undefined. The marginal $\mathcal M$ of $(z_1,\dots,z_n)$ is the one in the definitions file. The covariance $(1+\sigma^2/n) I$ is encoded by the standard deviation $\sqrt{1+\sigma^2/n}$.