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Let ($W_t$) denote a standard one-dimensional Brownian motion with $W_0 = 0$, and let $M_t = \max W_s$ denote the running maximum process of $W$. Let $m_t = \min W_s$ denote the minimum of $W$ at time $t$.

Compute the expected range of $W: E(M_t − m_t)$

(Hint: use symmetry).

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$$-m_t = -\inf\limits_{0\leq s \leq t}W_s = \sup\limits_{0\leq s \leq t} (-W_s) \sim M_t \sim |W_t|$$ So, $$E(M_t - m_t) = 2E(|W_t|) =\sqrt{\frac{8t}{\pi}}$$

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  • $\begingroup$ thanks! I know that M_t=max (W_s), but how do you know sup (−Ws)∼M_t? $\endgroup$
    – nikoo18
    May 7, 2018 at 22:46
  • $\begingroup$ One can easily show that $-W_s \sim W_s$. But more, $-W_s$ is again a Brownian motion. $\endgroup$
    – James Yang
    May 8, 2018 at 15:43
  • $\begingroup$ what is the best way to show that −Ws∼Ws? $\endgroup$
    – nikoo18
    May 10, 2018 at 18:05
  • $\begingroup$ $P(-W_s \leq x) = P(W_s \geq -x) = P(W_s \leq x)$, where the last equality follows from symmetry of normal distribution. $\endgroup$
    – James Yang
    May 11, 2018 at 4:12
  • $\begingroup$ It is very clear! Thanks! $\endgroup$
    – nikoo18
    May 11, 2018 at 15:10

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