In Wikipedia a formula is given for the distribution of $$M_t = \max_{0\leq s \leq t} W_s$$ even conditioned on $W_t$.

I wonder if there is also a simple expression for (note the absolute value) $$\tilde M_t = \max_{0\leq s \leq t} |W_s|$$ maybe conditioned on $|W_t|$?

  • 1
    $\begingroup$ the derivation of $M_t$ comes from reflection principle. At moment, i cannot see how it would work for $\tilde{M}$ $\endgroup$ – Lost1 Apr 10 '13 at 9:32

I'd recomment Borodin and Salminen's Handbook of Brownian Motion - Facts and Formulae. There's no simple expression for the distribution, but formula 1.1.8 at page 250 gives for $y > \max(x,z)$:

$$P_x(\max_{0<s<t}|W_s|<y, |W_t| \in dz) = $$

$$\frac{1}{\sqrt{2\pi t}} \sum_{k=-\infty}^\infty(-1)^k \left( e^{-(z-x+2ky)^2/2t} + e^{-(z+x+2ky)^2/2t} \right)\,dz $$

Isn't that a beautiful work of art... You can use standard techniques to obtain all the formulas you want (but they'll still look like this).

EDIT (as a reply to the a comment): We have the following for the cdf:

$$P_x(\max_{0<s<t}|W_s| \geq y) = $$

$$\sum_{k=-\infty}^\infty (-1)^k \textrm{sign}(x+(2k+1)y) \textrm{Erfc}\left( \frac{|x+(2k+1)y|}{\sqrt{2}t}\right) $$

and to obtain the expectation on does $E_x(\max \ldots) = \int_0^\infty P_x(\max\ldots)\,dy$.

  • $\begingroup$ What is $x$ here? $\endgroup$ – Greg P Jan 6 '15 at 19:47
  • $\begingroup$ @Greg P it's standard notation to have $P_x(X_t \in A) := P(X_t \in A \,|\,X_0=x)$. Long story stort it's the starting position. In his particular case: $x= W_0 = 0$. $\endgroup$ – user3371583 Jan 6 '15 at 23:10
  • $\begingroup$ Thanks. I have been trying to calculate this quantity (say, with $x=0$) without the conditioning over $|W_t|$. Due to the constraint $y > \max(x,z)$ I'm not certain how to get it. Should I integrate $P$ over $z$ from $0$ to $y$? $\endgroup$ – Greg P Jan 8 '15 at 2:31
  • $\begingroup$ @Greg Yes that is what you should do, but you'll just obtain a series of error functions. The same book I cited also gives it in a simpler form, as the inverse of a Laplace Transform. $P(\max |W_s|\leq y) = \mathcal{L}^{-1}_\gamma (1/\gamma ch(y\sqrt{2\gamma}))$. The book also has it as a sum of erfc but the comment section is a bit small. $\endgroup$ – user3371583 Jan 8 '15 at 11:08
  • $\begingroup$ You could put it in your answer, since it seems relevant to the question ;-) $\endgroup$ – Greg P Jan 9 '15 at 1:14

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