Question about definition of brownian motion

The definition of standard Brownian motion I saw in a course I am following is the following:

A stochastic proces $X = \{X_t, t \geq 0\}$ is a standard Brownian motion on some probability space $(\Omega, \mathcal{F}, P)$ if $X_0$ is zero almost surely, $X$ has independent increments, $X$ has stationary increments and $X_{t+s} - X_{s}$ is normally distributed with mean $0$ and variance $t$.

Looking up what was meant by 'stationary increments', I found that this meant that $X_{t+s} - X_s$ has the same distribution as $X_t - X_0 = X_t$. However, if I combine this with the last condition, stating that increments follow a normal distribution, I obtain that $$X_t = X_t - X_0 \sim X_{t+s} - X_s \sim N(0,t)$$ and this seems very odd to me...

Can anyone clarify what I misunderstood in this definition?

• You should add $s\ge t$ to your definition of stationary increments. Where are you feeling odd exactly? I personally don't see anything unnatural in the definition, so it might be hard to help you unless you make clear what is odd to you and for what reason. – Cave Johnson Jan 11 '18 at 8:34
• You have understood the definition correctly at least according to the conclusion you wrote. What is odd in your conclusion ? – Harto Saarinen Jan 11 '18 at 8:44
• @CaveJohnson You probably meant to say that $s\geq0$ must be added to the definition (where $X_{t+s}-X_t$ is used, not $X_s-X_t$). – drhab Jan 11 '18 at 8:45
• As an illustration of how it makes sense, If you look at $X_{t+s}$, you can either say it is $(X_{t+s}-X_s)+X_s$ so its distribution is that of the sum of a random variable with $N(0,t)$ distribution and a random variable with $N(0,s)$ distribution, or you can say directly it has a $N(0,t+s)$ distribution. Happily, these two ways of looking at it are consistent since the increments are independent – Henry Jan 11 '18 at 8:48
• @drhab Yes, thanks for correction :) – Cave Johnson Jan 11 '18 at 8:49