Discrete-time stochastic processes

One of the uses of this martingale is to provide

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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 non-positive 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 ©£ §™ exp n ∞ (ro ) − ro α n Ω ∑µ ∂ Pr {Sn ≥ α} ...
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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.

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