Let $B=\{B_t:t\geq 0\}$ be a standard Brownian motion. Define the Brownian brige $X=\{X_t:t\geq0\}$ as $$ X_t=B_t-tB_1\quad t\in[0,1] $$ Show that $X$ is (i) Gaussian and find its (ii) mean and (iii) covariance.
TWO questions:
- Due to my lack of basics on the subsject, can I see a full proof of (ii) and (iii) ?
- Can anybody check my attempt on (i)?
Attempt on (i).
Given that $X$ is a stochastic process by definition only need to show that $\sum_i^n \lambda_i X_{t_i}$ is Gaussian for some real $\lambda_1,\lambda_2,...,\lambda_n$.
\begin{equation*}
\begin{split}
\sum_i^n \lambda_i( B_{t_i}-t_iB_1) & = \sum_i^n \lambda_i B_{t_i}-\gamma_iB_1 \ \ \text{ where }\gamma_i=\lambda_it_i \\
& = \sum_i^n \phi_iB_{t_i} +\gamma_i(B_{t_i}-B_1)\text{ where }\phi_i=\lambda_i-\gamma_i\\
& = \sum_i^n \phi_iB_{t_i} +\sum_i^n \gamma_i(B_{t_i}-B_1)\\
& = \sum_i^{n }\theta_i(B_{t_i}-B_{t_{i-s}} )+\sum_i^n \gamma_i(B_{t_i}-B_1),
\end{split}
\end{equation*}
scalar multiplication for normal distribution was used in every step. Then using the independence property of BM and the sum of independent normal variable property we satisfy the definition of a Gaussian process (doubts on the last equality/rearranging, namely on $\theta$ and $n$ in the first sum). $\blacksquare$
Sorry that it is a bit messy. Hope it helps. You really need to look at some basic definitions.
You may also find this useful. Notice the tricks for these questions are identical.
https://stats.stackexchange.com/questions/78087/i-want-to-show-e-alpha-tbe2-alpha-t-is-a-gaussian-process-and-i-find/81010#81010
EDIT there should be a minus sign in front of $\lambda_{n+1}$