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The Complementary Error Function

The Complementary Error FunctionFrank R. KschischangDepartment of Electrical & Computer EngineeringUniversity of TorontoApril 10, 2017 The Complementary Error Function , erfc(x), is defined, forx 0, aserfc(x) = 2 x1 exp( u2) Complementary Error Function represents the area under the two tails of a zero-meanGaussian probability density Function with variance 2= 1/2, as illustrated in Fig. 1. Theso-called Error Function , erf(x), is defined viaerf(x) = 1 erfc(x).From the fact that a probability density Function has unit integral, we see thaterfc(0) = Complementary Error Function erfc(x) is plotted in Fig.

0 0:5 1 1:5 2 2:5 3 3:5 4 10 8 10 7 10 6 10 5 10 4 10 3 10 2 10 1 100 exp( 2x ) x p ˇ 2exp( 2x ) p ˇ(x+ x2+2) x erfc(x Bounds Upper bound Lower bound erfc(x) Figure 2: The function erfc(x) plotted together with an upper bound and a lower bound as

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Transcription of The Complementary Error Function

1 The Complementary Error FunctionFrank R. KschischangDepartment of Electrical & Computer EngineeringUniversity of TorontoApril 10, 2017 The Complementary Error Function , erfc(x), is defined, forx 0, aserfc(x) = 2 x1 exp( u2) Complementary Error Function represents the area under the two tails of a zero-meanGaussian probability density Function with variance 2= 1/2, as illustrated in Fig. 1. Theso-called Error Function , erf(x), is defined viaerf(x) = 1 erfc(x).From the fact that a probability density Function has unit integral, we see thaterfc(0) = Complementary Error Function erfc(x) is plotted in Fig.

2 2, along with an upper and alower bound (established in Appendix A). The bounds are asymptotically tight, , thedifference between the bound and the actual Function converges to zero for largex. xxu1 exp( u2)Figure 1: The Complementary Error Function is defined as the area under the two Gaussianpdf tails 810 710 610 510 410 310 210 1100exp( x2)x 2 exp( x2) (x+ x2+2)xerfc(x) and BoundsUpper boundLower bounderfc(x)Figure 2: The Function erfc(x) plotted together with an upper bound and a lower bound ( x)xerfc( x)xerfc( x)xerfc( x)

3 X9 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 1: Design Table3 Table 1 gives a mapping from a desired value of erfc( x) to the value ofxthat achieves thisvalue.

4 This table can often be used, in digital communications, to determine the signal-to-noise ratio needed to achieve a target Error Complementary Error Function is part of the standard math library provided with theCprogramming language (simply#include < >) and is also provided by standard mathpackages such digital communications textbooks prefer to define Error probabilities in terms of theso-calledQ- Function , defined, forx 0, viaQ(x) = x1 2 exp( u22) is the area under asingletail of a zero-mean Gaussian of a zero-mean Gaussian prob-ability density Function with unit variance.

5 TheQ- Function and the Complementary errorfunction are obviously closely related; indeedQ(x) =12erfc(x 2)anderfc(x) = 2Q(x 2).A Bounds on the Complementary Error FunctionLetXbe a Gaussian random variable with probability density functionf(x) =1 exp( x2),and, forz 0, leterfc(z) = 2P[X > z] =2 zexp( x2) that erfc(0) = 1. Throughout this appendix, we constrainz non-negative integer, letMn(z) =E[Xn|X > z] denote the conditionalnth momentofX, given thatX > z. ThenM0(z) = 1M1(z) = zx exp( x2) dx12erfc(z)=1 erfc(z) z2xexp( x2) dx=exp( z2) erfc(z),4and, forn 2, integrating by parts, takingu=xn 1du= (n 1)xn 2dxdv= 2xexp( x2) dxv= exp( x2),we getMn(z) = zxn exp( x2) dx12erfc(z)=1 erfc(z) z2xexp( x2)xn 1dx=1 erfc(z)(zn 1exp( z2) +n 12 z2xn 2exp( x2) dx)=zn 1exp( z2) erfc(z)+n 12Mn 2(z)=zn 1M1(z) +n 12Mn 2(z).

6 Thus, for example,M2(z) =zM1(z) +12,M3(z) =z2M1(z) +M1(z) = (z2+ 1)M1(z),M4(z) =z3M1(z) +32M2(z) = (z3+32z)M1(z) +34,M5(z) =z4M1(z) + 2M3(z) = (z4+ 2z2+ 2)M1(z), simple upper bound on erfc(z) arises from the observation thatM1(z)> z, from which itfollows thaterfc(z)<exp( z2) lower bound on erfc(z) arises from the observation that the conditional variance is positive, ,E((X M1(z))2|X > z] =M2(z) M21(z)>0. We then havezM1(z) +12 M1(z)2>0 Since this parabola inM1(z) opens downwards, we have thatM1(z) cannot exceed the largestparabolic zero crossing, ,M1(z)<z+ z2+ 22,from which it follows thaterfc(z)>2 exp( z2) (z+ z2+ 2).)

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