If $X$ and $X'$ are independent Gaussian random variable with values in $\mathbb{R}^d$ and distribution $\mathcal{N}(0,C)$, the distribution of $X+X'$ (and also $X-X'$) is $\mathcal{N}(0,2C)$, so $X+X'$ (and also $X-X'$) has the same distribution as $\sqrt{2}X$. The same holds for centered Gaussian processes since their distribution is determined by finite-dimensional distributions. In particular, this holds for Brownian bridge, Brownian motion, Ornstein Uhlenbeck processes.

See projecteuclid.org/journals/annals-of-probability/volume-30/…

It is an heuristic argument, not a proof. Writing $(1-\epsilon)^{𝑛^2} \to e^{-\epsilon n^2}$ does not make sense.