Is the Brownian motion increasing?

Question

Let $(B_t)$ the brownian motion. Why $$\mathbb E[\sup_{0\leq s\leq t}(as+cB_s)^2]\leq 2(a^2t^2+4b^2\mathbb E[B_t^2]).$$ Notcie that we have used $(a+b)^2\leq 2a^2+2b^2$. The thing is why $$\mathbb E\sup_{0\leq s\leq t}B_s^2=\mathbb E[B_t^2] \ \ ?$$

Answer 1

First of all: No, Brownian motion is not increasing.

Doob's maximal inequality states that for any square-integrable martingale $(M_t)_{t \geq 0}$ with continuous sample paths, we have

$$\mathbb{E} \left( \sup_{s \leq t} M_s^2 \right) \leq 4 \mathbb{E}(M_t^2), \qquad t \geq 0.$$

Since the Brownian motion $(B_t)_{t \geq 0}$ is a square-integrable martingale with continuous sample paths, this gives the estimate which was used to prove the inequality in your question.

Alternative reasoning: Using the reflection principle, it is possible to show that $$M_t :=\sup_{s \leq t} B_s \stackrel{d}{=} |B_t| \quad \text{and} \quad m_t :=- \inf_{s \leq t} B_s \stackrel{d}{=} |B_t|.$$ Since $$\sup_{s \leq t} B_s^2 \leq M_t^2 + m_t^2$$ we get $$\mathbb{E} \left( \sup_{s \leq t} B_s^2 \right) \leq 2 \mathbb{E}(B_t^2).$$