Brownian motion (2 Viewers)

He-Mann · May 25, 2017

$\textbf{Question:}$ Consider a standard Brownian motion (BM), $\{ B(t), t \geq0\}$. What is the distribution of $B(s)+B(t), 0 \leq s \leq t?$ \\\textbf{Solution:} Write $B(s)+B(t) = 2B(s) + B(t) - B(s)$. Then by independent increments of BM, $B(s)$ and $B(t) - B(s)$ are independent. Now, we look at their distributions: $2B(s) \sim N(0, 4s)$ and $B(t)-B(s) \sim B(t-s) \sim N(0, t-s)$ by stationary increments of BM. Thus, $\\\\\mathbb{E}(B(s)+B(t)) = \mathbb E (2B(s)) + \mathbb E (B(t) - B(s)) = 0, \\\\$and$\\\\ \mathbb{V}\mathrm{ar}(B(s)+B(t)) = \mathbb{V}\mathrm{ar}(2B(s)) + \mathbb{V}\mathrm{ar}(B(t)-B(s)) = 3s + t. \\\\ $Since, $B(s)$ and $B(t)-B(s)$ are independent and normally distributed, then any linear combination of them is also normal. It follows that, \\$$

$B(s)+B(t) \sim N(0, 3s+t)\quad (=B(3s+t))$

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If B(s) and B(t) were independent, then their sum would be B(s+t). However, when dependence exists, why is there an increase of 2s to the variance? Why does the variance increase in this manner?

Can someone give an intuitive explanation (possibly relating it to concrete things like stock prices?) of why the variance is 3s+t?

seanieg89 · May 25, 2017

Handwavy explanation using a particle moving with Brownian motion:

Note B(t) is the particles position at some intermediate time t. If it ends up at B(s), then B(t) has a tendency to be in the same direction as B(s) from the origin (this is the dependence statement). This means B(s)+B(t) will typically have higher magnitude than B(s+t), where we simply randomly wander for s+t time. The variance statement quantifies this.

He-Mann · May 25, 2017

seanieg89 said:
Handwavy explanation using a particle moving with Brownian motion:

Note B(t) is the particles position at some intermediate time t. If it ends up at B(s), then B(t) has a tendency to be in the same direction as B(s) from the origin (this is the dependence statement). This means B(s)+B(t) will typically have higher magnitude than B(s+t), where we simply randomly wander for s+t time. The variance statement quantifies this.

Thanks for giving me a new perspective on Brownian motion.

seanieg89 · May 25, 2017

No worries, hope it helped

. I haven't actually worked with Brownian motion before but some aspects of it seem like they would be pretty intuitive.

Brownian motion (2 Viewers)

He-Mann

Vexed?

seanieg89

Well-Known Member

He-Mann

Vexed?

seanieg89

Well-Known Member

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