> [!theorem] > > The standard [[Brownian Motion|Brownian motion]] exists. > > *Proof*. Let $(\Omega, \cf, \bp)$ be a [[Probability|probability]] space, and $\bracs{X_{n, m}: n \ge 0, m \ge 1}$ be a family of [[Probabilistic Independence|independent]], [[Normal Distribution|standard Gaussian]] random variables. > > Let $B_0 = 0$, and for integers $m \in \nat$, $B_m^{(0)} = \sum_{j = 1}^mX_{0, j}$. For any $t \in [m - 1, m]$, let $B_t^{(0)} = (1-t)B_{m-1}^{(0)} + tB_m^{(0)}$ such that the definitions on the integers are linearly interpolated. From here, $B^{(0)}$ is piecewise linear and [[Continuity|continuous]], and achieves independent and homogenous increments on the integers. On middle points, > $ > \mu_{B_{m-1/2}^{(0)}} = \mu_{\frac{1}{2}(B_{m-1}^{(0)}+B_{m}^{(0)})} = \gamma_{0,(m-3/4)I} > $ > while the desired variance is $m - 1/2$. > > Define $B_{m}^{(1)} = B_{m}^{(0)}$ and $B_{m-1/2}^{(1)} = B_{m-1/2}^{(0)} + \frac{1}{2}X_{1, m}$, then $B_{m-1/2}^{(1)}$ has the desired variance. Interpolate between the points again to obtain the remaining values. > > Suppose that $\{B_t^{(n)}: t \ge 0\}$ has been constructed such that > 1. $B_t^{(n)}$ has continuous sample paths. > 2. $\mu_{B_{m2^{-n}}^{(n)}} = \gamma_{0, m2^{-n}I}$. > 3. For each $m$ and $m'$, $\xi, \eta \in \real^d$, > > $ > \ev\braks{\angles{B_{m2^{-n}}^{(n)}, \xi} \cdot \angles{B_{m'2^{-n}}^{(n)}, \eta}} = 2^{-n}\min{(m, m')}\angles{\xi, \eta} > $ > > To move on to $n + 1$, let > $ > \begin{align*} > B_{m2^{n}}^{(n+1)} &= B_{m2^{n}}^{(n)}\\ > B_{(2m - 1)2^{n+1}}^{(n+1)} &= B_{(2m - 1)2^{n + 1}}^{(n)} + 2^{-n/2 - 1}X_{n+1, m} > \end{align*} > $ > and linearly interpolate between the points. > > > Now we have a family $\{B^{(n)}: n \in \nat\}$. Let $L \ge 0$, and consider the uniform change. Since the modification is done on the midpoints, the uniform norm is bounded by > $ > \begin{align*} > \norm{B^{(n + 1)} - B^{(n)}}_{u, [0, 2^L]} &= \max_{1 \le m \le 2^{n+L}}\abs{B_{m2^{-(n+L)}}^{(n)} - B_{m2^{-(n+L)}}^{(n)}} \\ > &\le \max_{1 \le m \le 2^{n + L}}2^{-(n+L)}\abs{X_{n+1, m}} > \end{align*} > $ > Taking the expectation yields > $ > \ev\braks{\norm{B^{(n + 1)} - B^{(n)}}_{u, [0, 2^L]}} \le 2^{-(n+L)}\ev\braks{\max_{1 \le m \le 2^{n + L}}\abs{X_{n+1, m}}} > $ > since the $L^1$ norm is bounded by the $L^4$ norm, > $ > \begin{align*} > \ev\braks{\max_{1 \le m \le 2^{n + L}}\abs{X_{n+1, m}}} &\le \ev\braks{\max_{1 \le m \le 2^{n + L}}\abs{X_{n+1, m}}^4}^{1/4} \\ > &\le \ev\braks{\sum_{m = 1}^{2^{n+L}}\abs{X_{n+1, m}}^4}^{1/4} \\ > 2^{-(n+L)}\ev\braks{\max_{1 \le m \le 2^{n + L}}\abs{X_{n+1, m}}} &\le C2^{-n/4}2^{L/4} > \end{align*} > $ > using the same $\eps = 2^{-n/8}$ trick, Markov's inequality, and the Borel-Cantelli lemma yields > $ > \bp\bracs{\norm{B^{(n + 1)} - B^{(n)}}_{u, [0, 2^L]} > 2^{-n/8} \quad i.o.} = 0 > $ > which provides a Cauchy sequence with respect to the uniform norm on finite times. > > Lastly, using joint characteristic functions, > $ > \begin{align*} > \ev\braks{e^{i\angles{\xi, B_r}}e^{t\angles{\eta, B_t - B_s}}} &= \limv{n}\ev\braks{e^{i\angles{\xi, B_{r}^{(n)}}}e^{i\angles{\eta, B_{t}^{(n)} - B_s^{(n)}}}} > \end{align*} > $ > Given the dyadic construction, each time value can be approximated using their dyadic floor $\fl{x}_n$: > $ > ev\braks{e^{i\angles{\xi, B_{\fl{r}_n^{(n)}}}}e^{i\angles{\eta B_{\fl{t}_n}^{(n)} - B_{\fl{s}_n}^{(n)}}}} = e^{-\fl{r}_n\abs{\xi}^2/2}e^{-(\fl{t}_n - \fl{s}_n)\abs{\xi}^2/2} > $ > and the characteristic function can be recovered after taking limits.