Proving that the expectation of supremum of absolute value of Brownian motion over the unit interval is finite.

Question

Let $(B_{t},t\geq0) $ be the standard Brownian Motion. Define $M(t)=\underset{s\in [0,t]}{\mathrm{sup}}|B_{s}|$. I want to show that $$E[M(1)] =\int_{0}^{\infty}P[M(1)>x]dx < \infty $$ My approach so far...$$P[\underset{s\in [0,1]}{\mathrm{sup}}|B_{s}|>x] \leq P[\underset{s\in [0,1]}{\mathrm{sup}}B_{s}>x]+P[\underset{s\in [0,1]}{\mathrm{inf}}B_{s}<-x] = 2P[\underset{s\in [0,1]}{\mathrm{sup}}B_{s}>x]$$ using the fact that $B_{t}$ and $-B_{t}$ have same distribution. Now by the reflection principle we have that $$P[\underset{s\in [0,1]}{\mathrm{sup}}B_{s}>x] = 2P[B_{1}>x]=2[1-N(x)]$$ where $N(x)$ is the cdf of $\mathcal{N}(0,1)$. Using all this I am not able to show that $E[M(1)] < \infty$. I was also thinking to use Doob's maximal inequality as $(|B_{t}|,t\geq0) $ is a submartingale with respect to canonical filtration but I guess it won't help.

Answer 1

From what you proved, we can deduce that $$P[\underset{s\in [0,1]}{\mathrm{sup}}|B_{s}|>x] \le 2P[\underset{s\in [0,1]}{\mathrm{sup}}B_{s}>x]$$ As $\underset{s\in [0,1]}{\mathrm{sup}}B_{s} =|B_1|$ in law and by applying the Chebyshev's inequality we have $$P[\underset{s\in [0,1]}{\mathrm{sup}}|B_{s}|>x] \le 2P[|B_1|>x] < \frac{2}{x^2} \qquad \text{for } x>0$$ Then $$ \begin{align} E[M(1)]&= \int_0^aP[\underset{s\in [0,1]}{\mathrm{sup}}|B_{s}|>x]dx+\int_a^{+\infty}P[\underset{s\in [0,1]}{\mathrm{sup}}|B_{s}|>x]dx \\ &\le \int_0^a1dx+\int_a^{+\infty}\frac{2}{x^2}dx=a+\frac{2}{a}<+\infty \end{align} $$

Q.E.D