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Unformatted text preview: the rv’s Xi in a process {Xn; n ≥ 1 have
a ﬁnite number of p ossible sample values. Then
any (possibly defective) stopping trial J can b e rep
resented as a rooted tree where the trial at which
each sample path stops is represented by a terminal
node.
Example: X is binary and stopping o ccurs when the
pattern (1, 0) ﬁrst o ccurs.
1
0
1
0
1
0 ✈
✈
✈
✈
✈
✈
✘
✘ 14 Wald’s equality
Theorem (Wald’s equality) Let {Xn; n ≥ 1} b e a se
quence of I ID rv’s, each of mean X . If J is a stop
ping trial for {Xn; n ≥ 1} and if E [J ] < ∞, then the
sum SJ = X1 + X2 + · · · + XJ at the stopping trial J
satisﬁes E [SJ ] = X E [J ]
Prf:
SJ = X1IJ ≥1 + X2IJ ≥2 + · · · + XnIJ ≥n + · · · E [SJ ] = E
n XnIJ ≥n =
n E XnIJ ≥n The essence of the proof is to show that Xn and
IJ ≥n are independent. 15 To show that Xn and IJ ≥n are independent, note
that IJ ≥n = 1 − IJ<n. Also IJ<n is a function of
X1, . . . , Xn−1. Since the Xi are I ID, Xn is independent
of X1, . . . , Xn−1, and thus IJ<n, and thus of IJ ≥n.
This is surprising, since Xn is certainly not indepen
dent of IJ =n, nor of IJ =n+1, etc.
The resolution of this ‘paradox’ is that, given that
J ≥ n (i.e., that stopping has not o ccured b efore
trial n), the trial at which stopping o ccurs depends
on Xn, but whether or not J ≥ n o ccurs depends
only on X1, . . . , Xn−1.
Now we can ﬁnish the proof. 16 E [SJ ] =
=
n
n =X
=X E XnIJ ≥n
E [Xn] E IJ ≥n n
n E IJ ≥n Pr {J ≥ n} = X E [J ] In many applications, this gives us one equation
in two quantities neither of which is known. Fre
quently, E [SJ ] is easy to ﬁnd and this solves for E [J ].
The following example shows, among other things,
why E [J ] < ∞ is required for Wald’s equality. 17 Stop when you’re ahead
Consider tossing a coin with probability of heads
equal to p. $1 is b et on each toss and you win on
heads, lose on tails. You stop when your winnings
reach $1.
If p > 1/2, your winnings (in the absence of stop
ping) would grow without b ound, passing through
1, so J must b e a rv. SJ = 1 WP1, so E [SJ ] = 1.
Thus, Wald says that E [J ] = 1/X = 2p1 1 . Let’s
−
verify this in another way.
Note that J = 1 with probability p. If J > 1, i.e., if
S1 = −1, then the only way to reach Sn = 1 is to
go from S1 = −1 to Sm = 0 for some m (requiring
J steps on average); J more steps on average then
gets to 1. Thus J = 1 + (1 − p)2J = 2p1 1 .
− 18 Next consider p < 1/2. It is still p ossible to win and
stop (for example, J = 1 with probability p and J = 3
with probability p2(1 − p)). It is also p ossible to head
South forever.
Let θ = Pr {J < ∞}. Note that Pr {J = 1} = p. Given
that J > 1, i.e., that S1 = −1, the event {J < ∞}
requires that Sm − S1 = 1 for some m, and then
Sn − Sm = 1 for some n > m. Each of these are
independent events of probability θ, so
θ = p + (1 − p)θ2
There are two solutions, θ = p/(1 − p) and θ = 1,
which is impossible. Thus J is defective and Wald’s
equation is inapplicable. 19 Finally consider p = 1/2. In the limit as p approaches
1/2 from b elow, Pr {J < ∞} = 1. We ﬁnd other more
convincing ways to see this later. However, as p
approaches 1/2 from above, we see that E [J ] = ∞.
Wald’s equality does not hold here, since E [J ] = ∞,
and in fact does not make sense since X = 0.
However, you make your $1 with probability 1 in
a fair game and can continue to repeat the same
feat.
It takes an inﬁnite time, however, and requires ac
cess to an inﬁnite capital. 20 MIT OpenCourseWare
http://ocw.mit.edu 6.262 Discrete Stochastic Processes
Spring 2011 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms ....
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