Proof on Brownian Bridge PROBLEM
Let $U_t$ be a Brownian bridge on $[0,1]$ and let $Z$ be a standard normal random variable independent of $U_t$.
$(a)$ Prove that the process $W_t = U_t + tZ$ is a brownian motion.
$(b)$ Prove that the process $W_t = (1+t)U_{\frac{t}{1+t}}$ on $[0,\infty)$ is a Brownian motion.

I have no idea how to prove these statements, I know that a Brownian motion follows a normal distribution with $\mu = 0$ and  $\sigma^2 = t$.
Can anybody please help me to prove this?
 A: Let's recall that a Brownian bridge $(U_t)_{t \in [0,1]}$ is a Gaussian process with continuous sample paths, mean zero and covariance function
$$C(s,t) := \mathbb{E}(U_s \cdot U_t) = s \wedge t - s \cdot t \tag{1}$$
In order to show that $W_t := U_t+t \cdot Z$ is a Brownian motion we have to check the following properties:


*

*$W_0 = 0$ a.s.

*$t \mapsto W_t(\omega)$ is continuous for almost all $\omega$

*$(W_t)_{t \in [0,1]}$ is a Gaussian process such that the vector $\Gamma := (W_{t_1},\ldots,W_{t_n})$ is Gaussian with mean zero and covariance matrix $(\text{cov}(W_{t_j},W_{t_k}))_{j,k}=(t_j \wedge t_k)_{j,k}$ for any $0<t_1<\ldots<t_n<1$.


Proof:


*

*By definition, we have $U_0=0$ a.s. Obviously, this implies $W_0 = U_0+0=0$ a.s.

*Since $t \mapsto U_t(\omega)$ is continuous almost surely, we find that $t \mapsto W_t(\omega)$ is continuous almost surely.

*Let $0<t_1<\ldots<t_n<1$. Since $(U_t)_{t \in [0,1]}$ is a Gaussian process with mean $0$ and covarianc function $C$, we know that $\Lambda := (U_{t_1},\ldots,U_{t_n})$ is Gaussian with mean $0$ and covariance matrix $(C(t_j,t_k))_{j,k}$. Therefore, we find that $$\Gamma = \Lambda + \begin{pmatrix} t_1 \\ \vdots \\t _n \end{pmatrix} \cdot Z$$ is Gaussian as a sum of two independent Gaussian random variables. It remains to calculate the mean and covariance matrix. We have
$$\mathbb{E}\Gamma = \mathbb{E}\Lambda + \begin{pmatrix} t_1 \\ \vdots \\t _n \end{pmatrix}\cdot \mathbb{E}Z = 0$$ since $\mathbb{E}\Lambda = 0$, $\mathbb{E}Z = 0$. Similarly,
$$\begin{align*} \mathbb{E}(\Gamma_j \cdot \Gamma_k) &= \mathbb{E}(W_{t_j} \cdot W_{t_k}) = \mathbb{E}(U_{t_j} \cdot U_{t_k} + t_j \cdot Z \cdot U_{t_k} + t_k \cdot Z \cdot U_{t_j} + t_j \cdot t_k \cdot Z^2) \\
&= C(t_j,t_k) + t_j \underbrace{\mathbb{E}Z}_{0} \cdot \mathbb{E}U_{t_k} + t_k \cdot \underbrace{\mathbb{E}Z}_{0} \cdot \mathbb{E}U_{t_j}+t_j \cdot t_k \cdot \mathbb{E}(Z^2) \end{align*}$$
where we used the independence of $Z$ and $(U_t)_{t \in [0,1]}$. From $\mathbb{E}(Z^2)=1$ and $(1)$, it follows easily that $\text{cov}(W_{t_j},W_{t_k}) = t_j \wedge t_k$.


This finishes the proof of part (a). The proof of the second part is similarly, I leave it to you.
