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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,

John

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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 $$\mathbb{E}(|W_t|) \leq 1+ \mathbb{E}(W_t^2).$$

Remarks

  • 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.
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  • $\begingroup$ 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
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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.

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

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