Unformatted text preview: lope = ∞ 0 (ro ) = α/n
∞ (ro ) ∞ (ro ) − ro α/n Figure 7.4: Graphical minimization of ∞ (r) − (α/n)r. For any r, ∞ (r) − (α/n)r is found by drawing a line of slope (α/n) from the point (r, ∞ (r)) to the vertical axis. The
minimum occurs when the line of slope α/n is tangent to the curve. The ﬁrst of these inequalities shows how Pr {Sn ≥ α} decreases exponentially with n for ﬁxed
α/n = ∞ 0 (ro ) and the second shows how it decreases with α for the same ratio α/n = ∞ 0 (ro ).
We now give a graphical interpretation in Figure 7.4 of what these exponents mean, and
return subsequently to discuss whether α/n = ∞ 0 (r) actually has a solution.
The function ∞ (r) has a strictly positive second derivative, and thus any tangent to the
function must lie below the function everywhere except at the point of tangency. The
particular tangent shown is tangent at the point r = ro where ∞ 0 (r) = α/n. Thus this
tangent line has slope α/n = ∞ 0 (ro ) and meets the vertical axis at the point ∞ (ro ) − ro ∞ 0 (ro ).
As illustrated, this vertical axis intercept is smaller than ∞ (r) − (α/n)r for any other choice
of r. This is the exponent in (7.20). This exponent is negative and shows that for a ﬁxed
ratio α/n, Pr {Sn ≥ α} decays exponentially in n.
Our primary interest is in the probability that Sn exceeds a positive threshold, α > 0,
but it can be seen that both the algebraic and graphical arguments above apply whenever
α > nE [X ]. Since E [X ] < 0, we might also be interested in the probability that Sn exceeds
the mean by some amount, while also being negative.
Figure 7.4 also gives a geometric interpretation of (7.21) for the case α > 0. The exponent
in α is given by (7.21) to be −ro + ∞ (ro )/∞ 0 (ro ) where ro satisﬁes ∞ 0 (ro ) = α/n. The negative 290 CHAPTER 7. RANDOM WALKS, LARGE DEVIATIONS, AND MARTINGALES of this is seen to be the horizontal axis intercept of the tangent to ∞ (r) at ro , and thus this
intercept gives the exponential decay rate of Pr {Sn ≥ α} in α for ﬁxed α/n.
It is interesting to observe what happens to (7.21) as n is changed while holding α > 0 ﬁxed.
This is an important question for threshold crossings, since it provides an upper bound on
crossing a ﬁxed α for diﬀerent values of n. For the α and n illustrated in Figure 7.4, note
that as n increases with ﬁxed α, the slope of the tangent decreases, moving the horizontal
axis intercept to the right, i.e., increasing the exponential decay rate in α.
Conversely, as n is decreased, the intercept moves to the left, decreasing the exponential
decay rate. Note, however, that when the slope increases to the point where the intercept
reaches the point where ∞ (r) = 0, i.e., the point labelled r∗ in Figure 7.4, then further
reductions in n move the tangent point to where ∞ (r) is positive. At this point, the intercept
starts to move to the right again. This means that for all n, an upper bound to Pr {Sn ≥ α}
is given by
Pr {Sn ≥ α} ≤ exp(−r∗ α) for arbitrary α > 0, n ≥ 1. (7.22) We now must return to the question of whether the equation α/n = ∞ 0 (r) has a solution.
From the assumption that E [X ] < 0, we know that ∞ 0 (0) < 0. We have not yet shown why
∞ 0 (r) should become positive as r increases. To see this in the simplest case, assume that X
is discrete and assume that X takes on positive values (if X were a nonpositive rv, there
would be no point to discussing the probability of crossing a positive threshold).
P
Let xmax be the largest such value. Then g (r) = x p(x)erx ≥ p(xmax )erxmax . It follows
that ∞ (r) ≥ rxmax + ln(p(xmax )). Since ∞ has a positive second derivative, it follows that
∞ 0 (r) must be increasing with r and must approach xmax in the limit as r → 1. Thus
α/n = ∞ 0 (r) has a solution whenever α/n < xmax . It is also clear that Pr {Sn ≥ α} = 0 for
α/n > xmax . Thus ∞ 0 (r) = α/n has a solution over the range of interest. One can extend
this argument to the case where X has an arbitrary distribution function with negative
mean.
Although we have only established (7.20, 7.21, 7.22) as upper bounds, Exercise 7.10 shows
that for any ﬁxed ratio a = α/n, and any ε > 0, there is an n0 (ε) such that for all n ≥ n0 (ε),
Pr {Sn ≥ n(α − ε)} > exp{−n[ra − ∞ (r) + ε]} where r satisﬁes ∞ 0 (r) = a. This means that
for ﬁxed a = α/n, (7.20) is exponentially tight, i.e., Pr {Sn ≥ na} decays exponentially with
increasing n at the asymptotic rate −ra + ∞ (r) where r satisﬁes ∞ 0 (r) = a.
The above discussion has treated only the case where r+ = 1. Figure 5 illustrates the
minimization of (7.17) for the case where r+ < 1. We have assumed that ∞ (r) < 0 for
r < r+ , since the previous argument applies if ∞ (r) crosses 0 at some r∗ ≤ r+ .
To include this case, (7.20) is generalized to
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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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