I've been given the following question and solution:

Let $W_t$ be a standard Brownian Motion w.r.t. ($\mathbf{P},\mathcal{F}_t)$. Prove that \begin{align} E[|W_t|] < \infty, \forall \text{ } t \end{align}

Solution: \begin{align} E[|W_t|] < E[1+W_t^{2}] < 1 + E[W_t^2] < 1+t <\infty \end{align}

My question is, what allows us to state the following? \begin{align} E[|W_t|] < E[1+W_t^2] \end{align}

Many thanks,



Note that $$|W_t| =|W_t| \cdot 1_{\{|W_t| \leq 1\}} + |W_t| \cdot 1_{\{|W_t|>1\}} \leq 1 + |W_t|^2.$$ This implies $$\mathbb{E}(|W_t|) \leq 1+ \mathbb{E}(W_t^2).$$


  • Please note that any random variable $X \in L^2$ is automatically integrable, i.e. $X \in L^1$. This follows from Jensen's inequality or the Cauchy Schwarz inequality. So if you know that $\mathbb{E}(W_t^2)<\infty$, this proves $W_t \in L^1$.
  • As $W_t$ is a Gaussian random variable, the (absolute) moments can be calculated explicitely, see here.
  • $\begingroup$ Thanks Saz. Could you explain the your subscript please? $\endgroup$ – John Smith Mar 4 '14 at 16:10
  • $\begingroup$ @JohnSmith $1_A$ denotes the indicator function of the set $A$, i.e. $$1_{\{|W_t| \leq 1\}} = \begin{cases} 1 & |W_t| \leq 1 \\ 0 & \text{otherwise} \end{cases}.$$ $\endgroup$ – saz Mar 4 '14 at 16:12
  • $\begingroup$ Much appreciated. Thanks. $\endgroup$ – John Smith Mar 4 '14 at 16:15

Let $x\in \mathbb R$.

If $|x| <1$, $|x| < 1 + x^2$.

If $|x| \ge1$, $|x| \le x^2 \le 1 + x^2$.

Actually, you can make a better majoration with not much more effort:

$$ |x| = \frac 12 \left( x^2 + 1 - (1 - {|x|})^2 \right) \le \frac 12 \left( x^2 + 1 \right) $$

Now apply to $W_t$ and integrate, and you are done.

  • $\begingroup$ Much appreciated. $\endgroup$ – John Smith Mar 4 '14 at 16:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.