# Example: $\tilde{B}(t)$ is not a brownian motion clarification [duplicate]

I started reading the book Brownian motion by Mörters and Peres where I don't quite understand the following example. For completeness I've provided the definitions for brownian motion and finite dimensional distributions used in the book.

Definition $$1.1$$$$. \quad$$ A real-valued stochastic process $$\{B(t): t \geqslant 0\}$$ is called a (linear) Brownian motion with start in $$x \in \mathbb{R}$$ if the following holds:

$$\bullet\hspace{0,2cm} B(0)=x,$$

$$\bullet\hspace{0,2cm}$$the process has independent increments, i.e. for all times $$0 \leqslant t_{1} \leqslant t_{2} \leqslant \ldots \leqslant t_{n}$$ the increments $$B\left(t_{n}\right)-B\left(t_{n-1}\right), B\left(t_{n-1}\right)-B\left(t_{n-2}\right), \ldots, B\left(t_{2}\right)-B\left(t_{1}\right)$$ are independent random variables,

$$\bullet\hspace{0,2cm}$$ for all $$t \geqslant 0$$ and $$h>0,$$ the increments $$B(t+h)-B(t)$$ are normally distributed with expectation zero and variance $$h$$

$$\bullet\hspace{0,2cm}$$ almost surely, the function $$t \mapsto B(t)$$ is continuous.

We say that $$\{B(t): t \geqslant 0\}$$ is a standard Brownian motion if $$x=0$$.

By the finite-dimensional distributions of a stochastic process $$\{B(t): t \geqslant 0\}$$ we mean the laws of all the finite dimensional random vectors $$\left(B\left(t_{1}\right), B\left(t_{2}\right), \ldots, B\left(t_{n}\right)\right), \text { for all } 0 \leqslant t_{1} \leqslant t_{2} \leqslant \ldots \leqslant t_{n}$$ To describe these joint laws it suffices to describe the joint law of $$B(0)$$ and the increments $$\left(B\left(t_{1}\right)-B(0), B\left(t_{2}\right)-B\left(t_{1}\right), \ldots, B\left(t_{n}\right)-B\left(t_{n-1}\right)\right), \text { for all } 0 \leqslant t_{1} \leqslant t_{2} \leqslant \ldots \leqslant t_{n}$$

Example 1.2 Suppose that $$\{B(t): t \geqslant 0\}$$ is a Brownian motion and $$U$$ is an independent random variable, which is uniformly distributed on [0,1] . Then the process $$\{\tilde{B}(t): t \geqslant 0\}$$ defined by $$\tilde{B}(t)=\left\{\begin{array}{ll} B(t) & \text { if } t \neq U \\ 0 & \text { if } t=U \end{array}\right.$$ has the same finite-dimensional distributions as a Brownian motion, but is discontinuous if $$B(U) \neq 0,$$ i.e. with probability one, and hence this process is not a Brownian motion.

Now my questions are

1. Why does $$\tilde{B}(t)$$ have the same finite-dimensional distributions as a Brownian motion ?
2. Why is $$B(U)\neq 1$$ with probability one ?
• Have a look at this question and this question
– saz
Commented Mar 28, 2020 at 15:06
• Totally answers it. Thank you Commented Mar 29, 2020 at 7:18