Find the distribution of the supremum of the Brownian motion Let $A(t)$, $t \in [0,1]$ be a Gaussian process with zero mean and co-variance kernel $\mathrm{Cov}(A(t_1),A(t_2))= \min (t_1,t_2),\, \forall t_1,t_2 \in [0,1]$. Find $$P\left[\sup_{t \in [0,1]}|A(t)|>c\right]$$ where $c= 0.1,0.5,1$.  
Under the same set up, find
$$P\left[\int_0^1 A(t)^2\, dt >c\right]$$
where $c= 0.1,0.5,1$.  
I don't know how to solve this kind problem. I know the definition of Gaussian process only. I am stuck at the first step. Please help.  
 A: Part one of the question:
If this is indeed a Brownian process, $B(t)$, then we can use the following:
$$Pr(\sup |B(t)|> x = 1- \frac{4}{\pi} \sum_{k=0}^{\infty} \exp \left( \frac{-\pi^2(2k+1)^2}{8y^2} \right)$$
and we can calculate this with e.g.
probSupBr <- function(x){
        k <- c(0L, seq.int(1e4))
        1 - (4 / pi) * (sum(((( - 1)^k) / (2 * k + 1)) * 
             exp(-(((pi^2) * (2 * k + 1)^2) / (8 * x^2)))))
}

The use of 1e4 in place of $\infty$ allows for a practical numeric approximation; larger values do not appear to improve this beyond $4$ significant digits.
As to your example values:
> sapply(c(0.1, 0.5, 1), probSupBr)
[1] 1.0000000 0.9908430 0.6292226

I believe the equation was first is attributed to Billingsley (much of this is on Google books.
As to part two, I would be interested to hear if you have managed to solve this with a more precise method than those given below.
As I understand it, the approach used depends on your need for accuracy:


*

*We can use the values of the asymptomtic limits of the distribution, given by Csaki (Springer, paywall) for 'large' or 'small' values of $c$ ($=y$ below): $$ P[B(t)^2 > y] = \frac{4 \sqrt{2}}{\pi^2 \sqrt{y}} \exp \left( -\frac{\pi^2 y}{8} \right) \quad \mathrm{as} \quad y \rightarrow \infty $$
which is:
probSqGr <- function(y){
  min(1,
      (4 * sqrt(2)) / ((pi^2) * sqrt(y)) * (exp(-(pi^2) * y / 8))
      )
}
giving:
sapply(c(0.1, 0.5, 1), probSqGr)
[1] 1.0000000 0.4374169 0.1669114
Note that $y \leq 0.2 \rightarrow P=1$ and so for such small values the lower tail should probably be used instead (same source):
$$ P[B(t)^2 < x] = \frac{4 \sqrt{x}}{\sqrt{\pi}} \exp \left( -\frac{1}{8x} \right) \quad \mathrm{as} \quad x \rightarrow 0 $$


*For values closer to the center of the distribution, we can simulate values of the distribution as per @ThePawn's answer here. As this involves random sampling, this will produce slightly different results each time, even when a very large distribution is generated. Using the suggested cdfW function this poster suggests, with $1 \mathrm{e}^5$ Markov chains and $\mathrm{d}t = 1\mathrm{e}^{-4}$ we have:
1 - sapply(c(0.1, 0.5, 1), cdfW)
[1] 0.83834 0.32298 0.13506
which is not far from the values above.

*There is an exact method for this, but I'm having trouble turning the equation into code; it can be found on the link above or here: Erdos On certain limit theorems of the theory of probability.
