Brownian Motion: Distribution of B(t)+B(s)?

[Bridgeport]

B(r) is a Brownian Motion. For fixed t and s find distribution of B(t)+B(s)?

Where should I look for solutions of similar problems?

Spasibo!

[Henrylee]

Just use the fact that vector $\left(B(s),B(t)\right)$ is gaussian with well-known matrix of covariations.

[Bridgeport]

Ogromnoe Spasibo!

So I guess the solution would look like $X:=B(t)+B(s)\quad P(X\leq \alpha)= \int_{-\infty}^{\alpha}\int_{-\infty}^{\alpha-x}f(x,y)dydx$ where $f(x,y)$ is a joint pdf.

Would it be possible to compute pdf for $X$ explicitly (given the joint pdf of BM is not very nice)?

[PAV]

Let $t<s$. Then $B(s)=B(t)+\Delta_{ts}$, where $B(t)$ and $\Delta_{ts}$ are independent gaussian variables with zero mean and known covariance. Then $B(s)+B(t)=2B(t)+\Delta_{ts}$. Like any other linear combination of independent gaussian variables this variable also has gaussian distribution. Parameters of this distribution can be easily calculated using the known properties of mean and covariance.

[Bridgeport]

Da... Velikolepnoe reshenie! Spasibo