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I recently started learning about Brownian motions/Wiener processes. I was reading this post to understand what is the variance of the square of the increments of a Brownian motion. In the post, it is computed as

$$\text{Var}((W_t-W_s)^2) = \text{Var}(W_{t-s}^2) = \mathbb{E}\left[(W_{t-s}^2-(t-s))^2 \right] \stackrel{\ast}{=} (t-s)^2 \cdot \underbrace{\mathbb{E}((W_1^2-1)^2)}_{2} = 2(t-s)^2.$$

However, I don't understand what trick the person answering used to say directly that $\text{Var}((W_t-W_s)^2) = \text{Var}(W_{t-s}^2)$ and also $\mathbb{E}((W_1^2-1)^2) = 2$.

edamondo
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1 Answers1

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  1. $\mathbb V\left[\left(W_t-W_s\right)^2\right] = \mathbb V\left[W_{t-s}^2\right]$ because the Brownian motion has stationary increments.
  2. $\mathbb E\left[\left(W_1^2-1\right)^2\right] = 2$ because $W_1^2\sim\chi(1)$.
foreignvol
  • 926
  • Thank you for your answer. I didn't get how you deduce the expectation of $(W_1^2-1)^2$ just by knowing that $W_1^2$ has a chi-square distribution. Just to be sure, stationary increments means that the distribution of $W_t-W_s$ has the same distribution as $W_{t-s}$? – edamondo Aug 23 '21 at 20:52