SERVING THE QUANTITATIVE FINANCE COMMUNITY

 
User avatar
reaverprog
Topic Author
Posts: 194
Joined: October 28th, 2009, 8:53 am

Random walk

August 12th, 2015, 5:17 am

Hi all,I am struggling with that one :Find for a random walk, which starts at 0 and ends at 0 after 2n steps, the probability P2t,2n that in the interval from 0 to 2n along the abscissa, the path has 2t vertices on the positive side (including 0) and 2n-2t vertices on the negative side.Thanks
 
User avatar
wileysw
Posts: 593
Joined: December 9th, 2006, 6:13 pm

Random walk

August 15th, 2015, 5:24 pm

reaverprog,i dont think you mean vertices, e.g., when n=1, there are only 2 steps, there would be no path for t=1, because the last step always ends at 0, one has P(2,2)=P(1,2)=1/2 and P(0,2)=0.the question would make sense if one considers instead the number of steps on the positive side (2t) and negative side (2n-2t). the answer is an equidistribution, i.e., P(2t, 2n)=1/(n+1) for t=0,1,...,n.this is startling if you happen to know the arcsine law in the discrete setup. when one removes the constraint that the final step return to zero, the probability instead is P'(2t,2n)=(2t choose t)*(2n-2t choose n-t)/2^(2n). fixing x=t/n then taking the limit of n->infty would result the arcsine distribution (by using Stirling formula). analogous results hold respectively for the position of maxima as well. a good reference for these results is Feller's textbook vol 1 chapter III.
ABOUT WILMOTT

PW by JB

Wilmott.com has been "Serving the Quantitative Finance Community" since 2001. Continued...


Twitter LinkedIn Instagram

JOBS BOARD

JOBS BOARD

Looking for a quant job, risk, algo trading,...? Browse jobs here...


GZIP: On