关于Z~N(0,1)时√t Z的分布及Xt=√t Z是否为布朗运动的问询

Great question! Let's break this down into two clear parts to make it easy to follow.

Distribution of √t Z (t ≥ 0) #

Since ( Z \sim N(0,1) ), we can use basic properties of normal random variables to figure out the distribution of ( \sqrt{t}Z ):

• Mean: The expected value ( \mathbb{E}[\sqrt{t}Z] = \sqrt{t} \cdot \mathbb{E}[Z] = \sqrt{t} \cdot 0 = 0 ), since the mean of a standard normal variable is always 0.
• Variance: The variance ( \text{Var}(\sqrt{t}Z) = (\sqrt{t})^2 \cdot \text{Var}(Z) = t \cdot 1 = t ), because variance scales with the square of the coefficient, and ( \text{Var}(Z) = 1 ) for standard normal.

Since linear transformations of normal random variables remain normal, this means ( \sqrt{t}Z \sim N(0, t) ) — a normal distribution with mean 0 and variance ( t ).

Is ( X_t = \sqrt{t}Z ) a Brownian Motion? #

To answer this, we need to check the core defining properties of a standard Brownian motion (Wiener process). A stochastic process ( {W_t}_{t \geq 0} ) qualifies as Brownian motion if:

1. ( W_0 = 0 ) almost surely.
2. For all ( 0 \leq s < t ), the increment ( W_t - W_s ) is independent of the past (i.e., independent of ( {W_u}_{u \leq s} )) and follows a ( N(0, t-s) ) distribution.
3. The sample paths ( t \mapsto W_t ) are continuous almost surely.

Let's test ( X_t = \sqrt{t}Z ) against these rules:

• Property 1: When ( t=0 ), ( X_0 = \sqrt{0} \cdot Z = 0 ), so this holds.
• Property 2: This is where the process fails. Take any ( 0 < s < t ):
  • The increment ( X_t - X_s = (\sqrt{t} - \sqrt{s})Z ). Its variance is ( (\sqrt{t} - \sqrt{s})^2 = t + s - 2\sqrt{ts} ), which is not equal to ( t-s ) (unless ( s=0 ) or ( s=t )).
  • Also, the increment isn't independent of the past. For example, ( X_1 = Z ) and ( X_2 - X_1 = (\sqrt{2}-1)Z ) — these two are clearly dependent (they're both multiples of the same ( Z )), violating the independent increments requirement.
• Property 3: The path ( t \mapsto \sqrt{t}Z ) is continuous in ( t ) (since ( \sqrt{t} ) is continuous, and ( Z ) is a fixed random variable), so this holds.

Since ( X_t ) fails the critical independent increments property (and the increment variance doesn't match the required form), it is not a Brownian motion.

内容的提问来源于stack exchange，提问作者helen pet