Unformatted text preview: all of each color. At time
n we draw out a ball chosen at random. We return it to the urn and add one
more of the color chosen. Let Xn be the fraction of red balls at time n. To
check that Xn is a martingale note that at time n there are n + k balls, so if
Rn = (n + k )Xn is the number of red balls then
P (Rn+1 = Rn + 1) = Xn P (Rn+1 = Rn ) = 1 Xn 173 5.5. CONVERGENCE
Letting Av = {Xn = xn , . . . X0 = x0 } we have
Rn + 1
E (Xn+1 Av ) =
n+k+1
1
=
n+k+1 ✓
◆
Rn
Rn
Rn
·
+
1
n+k n+k+1
n+k
Rn
Rn
n+k
·
+
·
= Xn
n+k n+k n+k+1 Since Xn 0, Theorem 5.17 implies that Xn ! X1 .
Suppose that initially there is one ball of each color. To ﬁnd the distribution
of X1 we note that There are 2 balls at time 0, . . . n + 1 at time n 1 so the
probability the probability that red balls are drawn on the ﬁrst j draws and
then green balls are drawn on the next n j is
1 · · · j · 1 · · · (n j )
j !(n j )!
=
2···j + 1 · j + 2···n + 1
(n + 1)!
A little thought shows that each outcome with j red and n j balls drawn has
the same probabi...
View
Full
Document
This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.
 Spring '10
 DURRETT
 The Land

Click to edit the document details