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How can you find this expected value? $$ \mathbb{E}[|W_{t}^2 - t|] $$

where $W_{t}$ is a brownian motion.

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3 Answers 3

up vote 4 down vote accepted

$W_t$ is a normal random variable with mean $0$ and variance $t$. If $f(x)$ is the density of a standard normal distribution, you're looking at $t \int_{-\infty}^\infty |x^2 - 1| f(x)\ dx$, which according to Maple is $ 2 t e^{-1/2} \sqrt{2/\pi}$

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$W_{t}^{2}-t$ is a martingale. So $\mathbb{E}[W_{t}^{2}-t]=\mathbb{E}[W_{0}^{2}-0]=0$

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It is in absolute value –  Julius Jun 11 '12 at 19:29

The above expression is a martingale, just use Ito calulus to produce a formula that does not include an integral which is integrating w.r.t time, hence it does not change with a change in time. Therefore the expected value of any martingale is 0.

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Duplicate of a wrong answer already on this page. –  Did Sep 24 '12 at 11:44

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