Question

(b) At time n = 0, an urn contains 2m balls, of which m are red and m are blue. At each time n = 1, ..., 2m, a single ball is randomly selected and taken away with no replacement. Hence, at time n, the urn has 2m – n balls. Let Rn denotes the number of red balls remaining in the urn at time n. For n= 0,..., 2m – 1, let B Rn Pn = 2m - in be the fraction of red balls remaining after time n. Assume we have infor- mation about the color of balls being selected up to and including the time n - 1. Is Pn a martingale?

Answer 1

No. Date: Need to prove ff pri pn-1 =$]=pi so Pn = Rn - (2m-n) - now Rn = Rnt wp (1-P) a Rn-knit wp p I this happen when a blue ball is selected traction of flue ball at time nt=1 IP (blue ball is selected at stage n) = 1- P - L - 2mon similarly RA-ray when a red ball is select this happens wp k as fraction of red ball is p trn | Pn-takt Rnt (1-b) + (R1-1-1) 11 (2m-n (2m-hi / wp 11-6) Rn=kna wp b Rn = Rn. BA-L-Rn 1p + Rn-ip-p - amunt now p = pnte Rnd 2m-nti SMO = b (2m-ntl)

IVO. Date : 1 1 (2m-h71cb 20ch = b. Hence E (Pn lint= b] =p r =) Pn is a m