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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 $|W_t|$?

Fabian
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    the derivation of $M_t$ comes from reflection principle. At moment, i cannot see how it would work for $\tilde{M}$ – Lost1 Apr 10 '13 at 09:32

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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 $E_x(\max \ldots) = \int_0^\infty P_x(\max\ldots)\,dy$.

user3371583
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  • What is $x$ here? – Greg P Jan 06 '15 at 19:47
  • @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: $x= W_0 = 0$. – user3371583 Jan 06 '15 at 23:10
  • 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 $y$? – Greg P Jan 08 '15 at 02:31
  • @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. – user3371583 Jan 08 '15 at 11:08
  • You could put it in your answer, since it seems relevant to the question ;-) – Greg P Jan 09 '15 at 01:14
  • Is there a simple expression for the expectation value of $\max |W_s|$? – Greg P Jan 09 '15 at 02:09
  • @GregP I've added the part that you've asked me to, to the answer and I'll get back to you for the expectation. – user3371583 Jan 16 '15 at 17:32
  • @GregP In the case $x=0$, $E(\ldots)=t\sqrt{\pi/8}$. (May have an error but there's a simple answer). To calculate: cut the sum on integers in two sums on strictly negative and on positive integers. Simple change of variable will give that both sums are equal. Integrate now and by using a simple integration by parts (i.e. a $y^\prime$ artificially to derivate the Erfc). From then it's easy, dont forget the series expansion or arctan to reduce the sum. – user3371583 Jan 16 '15 at 23:36
  • Hi, can you please take a look at a similar problem? https://math.stackexchange.com/questions/3126681/max-of-absolute-value-of-a-brownian-motion/3126708?noredirect=1 – Susan_Math123 Feb 26 '19 at 00:51