Brownian covariance is equivalent to independent stationary normal increments

Assume condition 1 and fix $0=t_0<t_1<\cdots <t_n$. By [F1], the increment vector $\mathrm{\Delta }=(X_{t_j}-X_{t_{j-1}}{)}_{j=1}^n$ is multivariate normal. It is centered, and $\mathrm{Var}({\mathrm{\Delta }}_j)=t_j-t_{j-1}\mathrm{.}$ If $i<j$, expanding the four covariance terms and using $t_{i-1}<t_i\le t_{j-1}<t_j$ gives $\mathrm{Cov}({\mathrm{\Delta }}_i,{\mathrm{\Delta }}_j)=t_i-t_i-t_{i-1}+t_{i-1}=0.$ Thus $\mathrm{\Delta }\sim N_n(0,D)$ for $D=\mathrm{diag}(t_j-t_{j-1})$, by [F3].

Conversely assume condition 2. Given any finite time list, discard repetitions only for the construction and write its distinct positive values as $0=r_0<r_1<\cdots <r_k$. Put ${\mathrm{\Delta }}_{\mathrm{\ell }}=X_{r_{\mathrm{\ell }}}-X_{r_{\mathrm{\ell }-1}}$. Then condition 2 makes these independent with ${\mathrm{\Delta }}_{\mathrm{\ell }}\sim N(0,r_{\mathrm{\ell }}-r_{\mathrm{\ell }-1})$, and telescoping with $X_0=0$ gives $X_t={\sum }_{\mathrm{\ell }:r_{\mathrm{\ell }}\le t}{\mathrm{\Delta }}_{\mathrm{\ell }}$ almost surely at every time in the original list.

By [F2], $N_n(0,D)$ is also the joint law of the vector $(\sqrt{t_j-t_{j-1}}{Z}_j{)}_{j=1}^n$ with independent standard normals $Z_j$. The joint-law identity in step 1.1 transfers every measurable-rectangle probability, so [F7] makes the ${\mathrm{\Delta }}_j$ mutually independent; their one-coordinate laws are $N(0,t_j-t_{j-1})$. For $n=0$ this is the vacuous empty family. Hence condition 2 holds.

For coefficients $a_1,\dots ,a_m$ attached to the original time list, step 1.2 rewrites ${\sum }_{j=1}^m{a}_j{X}_{t_j}={\sum }_{\mathrm{\ell }=1}^k{b}_{\mathrm{\ell }}{\mathrm{\Delta }}_{\mathrm{\ell }},b_{\mathrm{\ell }}={\sum }_{j:r_{\mathrm{\ell }}\le t_j}{a}_j\mathrm{.}$ By [F4]–[F5], its characteristic function is ${\prod }_{\mathrm{\ell }=1}^k\exp \text{\hspace{0.17em}⁣}(-\frac{1}{2}{u}^2{b}_{\mathrm{\ell }}^2(r_{\mathrm{\ell }}-r_{\mathrm{\ell }-1}))=\exp \text{\hspace{0.17em}⁣}(-\frac{1}{2}{u}^2{\sum }_{\mathrm{\ell }=1}^k{b}_{\mathrm{\ell }}^2(r_{\mathrm{\ell }}-r_{\mathrm{\ell }-1}))\mathrm{.}$ By [F4] and [F6], this is a centered normal law, including the empty sum and variance zero. Since the list and coefficients were arbitrary, [F1] makes $X$ a centered Gaussian process.

For $0\le s\le t$, condition 2 decomposes $X_t=X_s+(X_t-X_s)$ into independent centered normal variables. Consequently [F8] gives $E[X_s{X}_t]=E[X_s^2]+E[X_s]E[X_t-X_s]=s,$ while both means are zero by step 2.2; symmetry gives $\mathrm{Cov}(X_s,X_t)=\min (s,t)$ for every order of $s,t$. The cases $s=0$, $s=t$, and $t=0$ follow from the same formula and $X_0=0$ almost surely. Thus condition 1 holds. AC is used exactly in [F1]–[F4] and [F6], for the library's normal-law constructions and characteristic-function uniqueness; no path regularity is asserted.