What is the probability that the graph of two independent Brownian motions on $\mathbb{R}^d$ starting at origin on $t \in [0,n]$ are within $R$-distance to each other. In other words, what is $P(dist(B^0[0,n], B^1[0,n]) < R))$?
My idea: the difference of the Brownian motions is also a Brownian motion. So first take the difference and find the probability of hitting time to the ball of radius $\frac{R}{2}$ for $B^0-B^1$ being smaller than $n$. Thus we can find a upper bound for the probability by calculating both $P(|B^0-B^1)[0,n]| < \frac{R}{2}))$ and $P(|B^0[0,n]| < \frac{R}{2}))$ to approxmate by triangular inequality. And then I'm stuck. Any hints will be much appreciated
$$\textrm{d}(W_{1,t},W_{2,t})^2=\|W_{1,t}-W_{2,t}\|_2^2=\|X_{t}\|_2^2$$ We have $X_t \sim \mathcal{N}_d(0,2t\mathbf{I}_d)$ and it can be modeled as $$dX_t=\begin{bmatrix}dX_{1,t} \\ dX_{2,t} \\ .. \\ dX_{d,t} \end{bmatrix}=\begin{bmatrix}\sqrt{2} & 0 & ... & 0 \\ 0 & \sqrt{2} & ... & 0 \\ ... & ... & ... & ... \\0 & 0 & ... & \sqrt{2} \end{bmatrix}\begin{bmatrix}dB_{1,t} \\ dB_{2,t} \\ ... \\ dB_{d,t} \end{bmatrix}$$ where $B_t$ is a $d$-dimensional Brownian motion. Now we want the distribution of $$X_{1,t}^2+...+X_{d,t}^2=2t(Z_{1,t}^2+...+Z_{d,t}^2)$$ where $Z_{j,t} \sim \mathcal{N}(0,1)$. We have $Z_{1,t}^2+...+Z_{d,t}^2 \sim\chi^2_d$ therefore $$X_{1,t}^2+...+X_{d,t}^2 \sim \Gamma(k=d/2,\theta=4t)$$ So $$P(\textrm{d}(W_{1,t},W_{2,t})<R)=P(\textrm{d}(W_{1,t},W_{2,t})^2<R^2)=F_\Gamma(R^2)$$ where $F_\Gamma$ is the Gamma cdf.