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Unformatted text preview: 1} be a random walk with Sn = X1 + · · · + Xn . We assume throughout
that E [X ] exists and is ﬁnite. The reader should focus on the case E [X ] = X < 0 on a
ﬁrst reading, and consider X = 0 and X > 0 later. For X < 0 and α > 0, we shall develop
upper bounds on Pr {Sn ≥ α} that are exponentially decreasing in n and α. These bounds,
and many similar results to follow, are examples of large deviation theory, i.e., probabilities
of highly unlikely events.
We assumeR throughout this section that X has a moment generating function g (r) =
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E erX =
erx dFX (x), and that g (r) is ﬁnite in some open interval around r = 0.
As pointed out in Chapter 1, X must then have moments of all orders and the tails of
its distribution function FX (x) must decay at least exponentially in x as x → −1 and as
R1
x → +1. Note that erx is increasing in r for x > 0, so that if 0 erx dFX (x) blows up for
some r+ > 0, it remains inﬁnite for all r > r+ . Similarly, for x < 0, erx is increasing in −r,
R
so that if x≤0 erx dFX (x) blows up at some r− < 0, it is inﬁnite for all r < r− . Thus if r−
and r+ are the smallest and largest values such that g (r) is ﬁnite for r− < r < r+ , then
g (r) is inﬁnite for r > r+ and for r < r− . The end points r− and r+ can each be ﬁnite or
inﬁnite, and the values g (r+ ) and g (r− ) can each be ﬁnite or inﬁnite.
Note that if X is bounded in the sense that Pr {X < −B } = 0 and Pr {X > B } = 0 for
some B < 1, then g (r) exists for all r. Such rv’s are said to have ﬁnite support and include
all discrete rv’s with a ﬁnite set of possible values. Another simple example is that if X is
a nonnegative rv with FX (x) = 1 − exp(−αx) for x ≥ 0, then r+ = α. Similarly, if X is
a negative rv with FX = exp(β x) for x < 0, then r− = −β . Exercise 7.7 provides further
examples of these possibilities. 288 CHAPTER 7. RANDOM WALKS, LARGE DEVIATIONS, AND MARTINGALES The moment generating function of Sn = X1 + · · · + Xn is given by
gSn (r) = E [exp(rSn )] = E [exp(r(X1 + · · · + Xn )]
= {E [exp(rX )]}n = {g (r)}n . (7.15) It follows that gSn (r) is ﬁnite in the same interval (r− , r+ ) as g (r).
First we look at the probability, Pr {Sn ≥ α}, that the nth step of the random walk satisﬁes
Sn ≥ α for some threshold α > 0. We could actually ﬁnd the distribution of Sn either by
convolving the density of X with itself n times or by going through the transform domain.
This would not give us much insight, however, and would be computationally tedious for
large n. Instead, we explore the exponential bound, (1.38). For any r ≥ 0, in the region
where g (r) is ﬁnite, i.e., for 0 ≤ r < r+ , we have
Pr {Sn ≥ α} ≤ gSn (r)e−rα = [g (r)]n e−rα . (7.16) It is convenient to rewrite (7.16) in terms of the semiinvariant moment generating function
∞ (r) = ln[g (r)].
Pr {Sn ≥ α} ≤ exp[n∞ (r) − rα] ; any r, 0 ≤ r < r+ . (7.17) The ﬁrst two derivatives of ∞ with respect to r are given by
g (r)g 00 (r) − [g 0 (r)]2
.
(7.18)
[g (r)]2
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Recall from (1.32) that g 0 (0) = E [X ] and g 00 (0) = E X 2 . Substituting this into (7.18), we
can evaluate ∞ 0 (0) and ∞ 00 (0) as
∞ 0 (r) = g 0 (r)
;
g (r) ∞ 00 (r) = ∞ 0 (0) = X = E [X ] ; 2
∞ 00 (0) = σX . (7.19) The fact that ∞ 00 (0) is the second central moment of X is why ∞ is called a semiinvariant
moment generating function. Unfortunately, the higherorder derivatives of ∞ , evaluated at
r = 0, are not equal to the higherorder central moments.
Over the range of r where g (r) < 1, it is shown in Exercise 7.8 that ∞ 00 (r) ≥ 0, with strict
inequality except in the very special (and uninteresting) case where X is deterministic. If X
is deterministic, then Sn is also and there is no point to considering a probabilistic model.
We thus assume in what follows that X is nondeterministic and thus ∞ 00 (r) > 0 for all r
between r− and r+ Figure 7.3 sketches ∞ (r) assuming that X < 0 and r+ = 1
We can now minimize the exponent in (7.17) over r ≥ 0. For simplicity, ﬁrst assume that
r+ = 1. Since ∞ 00 (r) > 0, the exponent is minimized by setting its derivative equal to 0.
The minimum (if it exists) occurs at the r, say ro for which ∞ 0 (r) = α/n. As seen from
Figure 7.3, this is satisﬁed with r ≥ 0 only if α/n ≥ X . Thus
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Pr {Sn ≥ α} ≤ exp n ∞ (ro ) − ro ∞ 0 (ro )
where ∞ 0 (ro ) = α/n ≥ E [X ] (7.20)
Ω∑
∏æ
∞ (ro )
= exp α 0
− ro .
(7.21)
∞ (ro ) 7.4. THRESHOLD CROSSING PROBABILITIES IN RANDOM WALKS ❅0
❅
❅
❅
❅
❅ 289 r ∞ (r)
slope = E [X ] Figure 7.3: Semiinvariant moment generating function ∞ (r) for a rv X such that
E [X ] < 0 and r+ = 1. Note that ∞ (r) is tangent to the line of slope E [X ] < 0 at 0
and has a positive second derivative everywhere. 0 ∞ (r) − rα/n r ∞ (r) ro r∗ ro − ∞ (ro )(n/α) s...
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This note was uploaded on 09/27/2010 for the course EE 229 taught by Professor R.srikant during the Spring '09 term at University of Illinois, Urbana Champaign.
 Spring '09
 R.Srikant

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