1.Binomial Distribution 1.1 Definitions and Properties The random variable X is Bernoulli(p),or Bp,if X E{0,1}and Pr[X =1]=p=1-Pr 0].The probability mass function,or discrete probability density function(pdf),is f万(x)=p*(1-p)1-x. (1) The sum Xofn independent and identically distributed(iid)Bp is Bernoulli(n,p),or Bp.Note that The binomial coeficient is ( Standard functions are defined for k∈{0，l,2,.,n}.The pdf is fp(x)=C(n,k)p*(1-p)1-x. (2) The cumulative distribution function(cdf)is Enp(k)=Pr[X≤k= (3) with Fp(x)=0 ifx<0,and Fp(x)=1 if x>n,and right-continuous extension to R,as is customary for discrete distributions.The quantile function(qf)is the usual unique left- continuous pseudo-inverse of the cdf, Qnp(u)=inf {k:Fip(k)u} (4) It is also useful to have an upper-probability version of the cdf, Gnp (k)=Pr[X z k]=Ef=k fp ( (5) So Gnp(k)=1-Fp(k)and Fp(k)=1-Gnp(k +1).Note that Fp(k)increases with k and decreases with norp,and Gnp(k)decreases with k and increases with n or p. Moments are Ex的]=∑*后p0 (6) i=0 So E[X]np,and E[X2]np(1-p)+n2p2,and Var[x]np(1-p). 3
3 1. Binomial Distribution 1.1 Definitions and Properties The random variable 𝑋𝑋 is Bernoulli(𝑝𝑝), or 𝐵𝐵𝑝𝑝, if 𝑋𝑋 ∈ {0,1} and Pr[𝑋𝑋 = 1] = 𝑝𝑝 = 1 − Pr[𝑋𝑋 = 0]. The probability mass function, or discrete probability density function (pdf), is 𝑓𝑓𝑝𝑝 (𝑥𝑥) = 𝑝𝑝𝑥𝑥 (1 − 𝑝𝑝)1−𝑥𝑥. (1) The sum X of n independent and identically distributed (iid) Bp is Bernoulli(𝑛𝑛, 𝑝𝑝), or 𝐵𝐵𝑛𝑛,𝑝𝑝. Note that 𝐵𝐵𝑝𝑝 = 𝐵𝐵1,𝑝𝑝. The binomial coefficient is (𝑛𝑛, 𝑘𝑘) = � 𝑛𝑛 𝑘𝑘� = 𝑛𝑛! 𝑘𝑘!(𝑛𝑛−𝑘𝑘)! . Standard functions are defined for 𝑘𝑘 ∈ {0,1,2, … , 𝑛𝑛}. The pdf is 𝑓𝑓𝑛𝑛,𝑝𝑝 (𝑥𝑥) = 𝐶𝐶(𝑛𝑛, 𝑘𝑘)𝑝𝑝𝑥𝑥 (1 − 𝑝𝑝)1−𝑥𝑥 . (2) The cumulative distribution function (cdf) is 𝐹𝐹𝑛𝑛,𝑝𝑝 (𝑘𝑘) = Pr[𝑋𝑋 ≤ 𝑘𝑘] = �𝑓𝑓𝑛𝑛,𝑝𝑝 (𝑗𝑗) 𝑘𝑘 𝑗𝑗=0 (3) with 𝐹𝐹𝑛𝑛,𝑝𝑝 (𝑥𝑥) = 0 if 𝑥𝑥 ≤ 0, and 𝐹𝐹𝑛𝑛,𝑝𝑝 (𝑥𝑥) = 1 if 𝑥𝑥 ≥ 𝑛𝑛, and right-continuous extension to ℝ, as is customary for discrete distributions. The quantile function (qf) is the usual unique leftcontinuous pseudo-inverse of the cdf, 𝑄𝑄𝑛𝑛,𝑝𝑝 (𝑢𝑢) = inf � 𝑘𝑘: 𝐹𝐹𝑛𝑛,𝑝𝑝 (𝑘𝑘) ≥ 𝑢𝑢 � (4) It is also useful to have an upper-probability version of the cdf, 𝐺𝐺𝑛𝑛,𝑝𝑝 (𝑘𝑘) = Pr[𝑋𝑋 ≥ 𝑘𝑘] = ∑ 𝑓𝑓𝑛𝑛,𝑝𝑝 (𝑗𝑗) 𝑛𝑛 𝑗𝑗=𝑘𝑘 . (5) So 𝐺𝐺𝑛𝑛,𝑝𝑝 (𝑘𝑘) = 1 − 𝐹𝐹𝑛𝑛,𝑝𝑝 (𝑘𝑘) and 𝐹𝐹𝑛𝑛,𝑝𝑝 (𝑘𝑘) = 1 − 𝐺𝐺𝑛𝑛,𝑝𝑝 (𝑘𝑘 + 1). Note that 𝐹𝐹𝑛𝑛,𝑝𝑝 (𝑘𝑘) increases with k and decreases with n or p, and 𝐺𝐺𝑛𝑛,𝑝𝑝 (𝑘𝑘) decreases with k and increases with n or p. Moments are E[𝑋𝑋𝑘𝑘 ] = �𝑗𝑗𝑘𝑘 𝑓𝑓𝑛𝑛,𝑝𝑝 (𝑗𝑗) 𝑛𝑛 𝑗𝑗=0 (6) So E[𝑋𝑋] = 𝑛𝑛𝑛𝑛, and E[𝑋𝑋2] = 𝑛𝑛𝑛𝑛(1 − 𝑝𝑝) + 𝑛𝑛2𝑝𝑝2, and Var[X] = 𝑛𝑛𝑛𝑛(1 − 𝑝𝑝)
The log-likelihood function is C= >(logc(n,)+logp+(n-x)log(1-p)) 71 (7) logC(n,j) k logp +(n-k)log(1-p). The maximum likelihood estimator(MLE)of p is k/n. 1.2 Implementation Issue Reasonable algorithms for F(k)return the appropriate values for large n and small k but succumb to numeric underflow for large k. For example,since Fp(0)=(1-p)"and Fp(n-1)=1-p",set n 100 and p =0.6 to get F10,0.6(0)=(1-0.6)100=0.4100兰1.6×10-40 (8) and F00,0.6(99)=1-0.6100=1-6.5×10-23 (9) But with standard double precision resolution of about 16 decimal places,the result is F100.0.(99)=1 exactly.The cure for lower tail probabilities too close to 1 is to use upper tail probabilities,which will be close to 0,and work with G instead of F,since Gn.p(n)=p"=1- Frtp (n -1),and so G1000.6(100)=0.610=1-F1000.6(99)=1-(1-0.6)100兰6.5×10-23 (10) But naive use of Gnp(k)=1-Fp(k-1)directly will of course give G100.0.6(100)=0.To obtain a useful representation of G(k)with k large in terms of an accurate algorithm for F(k) with k small,let X ~Bn,p and W=n-X.Note that Fw(k)=Pr[Wsk]Pr[n-Xk] Pr[X z n-k]=1-Pr[X n-k]1-Pr[X <n-k-1]=1-Fx(n-k-1).Then Gx(k)=1-Fx(k-1)=1-Fx(n-(n-k)-1)=Fw(n-k),so Gnp(k)=Pr[Bnp <k]=Pr[Bn1-p <n-k]=Fn1-p(n-k). (11) This givesG1o0,0.6(100)=Fo0,0.4(0)=(1-0.4)100=0.6100兰6.5×10-23 as required. 4
4 The log-likelihood function is ℒ = ��log 𝐶𝐶(𝑛𝑛,𝑗𝑗) + 𝑋𝑋𝑗𝑗 log 𝑝𝑝 + �𝑛𝑛 − 𝑋𝑋𝑗𝑗 � log(1 − 𝑝𝑝)� 𝑛𝑛 𝑗𝑗=0 = ��log 𝐶𝐶(𝑛𝑛,𝑗𝑗) 𝑛𝑛 𝑗𝑗=0 � + 𝑘𝑘 log 𝑝𝑝 + (𝑛𝑛 − 𝑘𝑘) log(1 − 𝑝𝑝) . (7) The maximum likelihood estimator (MLE) of p is 𝑘𝑘 𝑛𝑛⁄ . 1.2 Implementation Issue Reasonable algorithms for 𝐹𝐹𝑛𝑛,𝑝𝑝 (𝑘𝑘) return the appropriate values for large n and small k but succumb to numeric underflow for large k. For example, since 𝐹𝐹𝑛𝑛,𝑝𝑝 (0) = (1 − 𝑝𝑝)𝑛𝑛 and 𝐹𝐹𝑛𝑛,𝑝𝑝 (𝑛𝑛 − 1) = 1 − 𝑝𝑝𝑛𝑛 , set 𝑛𝑛 = 100 and 𝑝𝑝 = 0.6 to get 𝐹𝐹100,0.6(0) = (1 − 0.6)100 = 0.4100 ≅ 1.6 × 10−40 (8) and 𝐹𝐹100,0.6(99) = 1 − 0.6100 ≅ 1 − 6.5 × 10−23. (9) But with standard double precision resolution of about 16 decimal places, the result is 𝐹𝐹100,0.6(99) = 1 exactly. The cure for lower tail probabilities too close to 1 is to use upper tail probabilities, which will be close to 0, and work with G instead of F, since 𝐺𝐺𝑛𝑛,𝑝𝑝 (𝑛𝑛) = 𝑝𝑝𝑛𝑛 = 1 − 𝐹𝐹𝑛𝑛,𝑝𝑝 (𝑛𝑛 − 1), and so 𝐺𝐺100,0.6(100) = 0.6100 = 1 − 𝐹𝐹100,0.6(99) = 1 − (1 − 0.6)100 ≅ 6.5 × 10−23. (10) But naïve use of 𝐺𝐺𝑛𝑛,𝑝𝑝 (𝑘𝑘) = 1 − 𝐹𝐹𝑛𝑛,𝑝𝑝 (𝑘𝑘 − 1) directly will of course give 𝐺𝐺100,0.6(100) = 0. To obtain a useful representation of 𝐺𝐺(𝑘𝑘) with k large in terms of an accurate algorithm for 𝐹𝐹(𝑘𝑘) with k small, let 𝑋𝑋 ~𝐵𝐵𝑛𝑛,𝑝𝑝 and 𝑊𝑊 = 𝑛𝑛 − 𝑋𝑋. Note that 𝐹𝐹𝑊𝑊(𝑘𝑘) = Pr[ 𝑊𝑊 ≤ 𝑘𝑘] = Pr[ 𝑛𝑛 − 𝑋𝑋 ≤ 𝑘𝑘] = Pr[𝑋𝑋 ≥ 𝑛𝑛 − 𝑘𝑘] = 1 − Pr[𝑋𝑋 < 𝑛𝑛 − 𝑘𝑘] = 1 − Pr[𝑋𝑋 ≤ 𝑛𝑛 − 𝑘𝑘 − 1] = 1 − 𝐹𝐹𝑋𝑋(𝑛𝑛 − 𝑘𝑘 − 1) . Then 𝐺𝐺𝑋𝑋(𝑘𝑘) = 1 − 𝐹𝐹𝑋𝑋(𝑘𝑘 − 1) = 1 − 𝐹𝐹𝑋𝑋(𝑛𝑛 − (𝑛𝑛 − 𝑘𝑘) − 1) = 𝐹𝐹𝑊𝑊(𝑛𝑛 − 𝑘𝑘), so 𝐺𝐺𝑛𝑛,𝑝𝑝 (𝑘𝑘) = 𝑃𝑃𝑃𝑃�𝐵𝐵𝑛𝑛,𝑝𝑝 ≤ 𝑘𝑘� = Pr�𝐵𝐵𝑛𝑛,1−𝑝𝑝 ≤ 𝑛𝑛 − 𝑘𝑘� = 𝐹𝐹𝑛𝑛,1−𝑝𝑝(𝑛𝑛 − 𝑘𝑘). (11) This gives 𝐺𝐺100,0.6(100) = 𝐹𝐹100,0.4(0) = (1 − 0.4)100 = 0.6100 ≅ 6.5 × 10−23 as required
1.3 Beta Distribution The Gamma function,which has r(n)=(n-1)!for n=1,2,3,...,is given by r(z）=u2-1e-“du. (12) The Beta function,which has B(n,k)=(n-1)!(k-1)!/(n+k-1)!on positive integers,is B(a.b)= a)dx T(a+b) (13) Jo The Beta(a,b)distribution on [0,1],with a >0 and b>0,has pdf and cdf 1 fieta ()(x)=B(a.b)-x)1 Fgeta (ab)(x)=fBeta (a,b)(u)du. (14) The Binomial and Beta cdfs are related by Fp(k-1)=FBeta (n-k+1k)(1-p),so F.p(k)=FBeta (n-kk+1)(1-p),or Fn-k+11-p(k-1)=FBeta (nk)(p) (15) and also Gn.p(k)=FBeta (kn-k+1)(p),or Gn-k+1p(k)=FBeta (n.k)(p). (16) To show Fup(k-1)=FBeta (n-k+1k)(1-p),first note that B(n-k+1,k)=(n-k)!(k- 1)!/n!,and also FBeta (n1)(1-p)=nfx-1dx=(1-p)"=fp(0).Now,integrating by parts
5 1.3 Beta Distribution The Gamma function, which has Γ(𝑛𝑛) = (𝑛𝑛 − 1)! for n = 1,2,3,…, is given by Γ(𝑧𝑧) = � 𝑢𝑢𝑧𝑧−1𝑒𝑒−𝑢𝑢𝑑𝑑𝑑𝑑 ∞ 0 . (12) The Beta function, which has 𝐵𝐵(𝑛𝑛, 𝑘𝑘) = (𝑛𝑛 − 1)! (𝑘𝑘 − 1)!/(𝑛𝑛 + 𝑘𝑘 − 1)! on positive integers, is 𝐵𝐵(𝑎𝑎, 𝑏𝑏) = Γ(𝑎𝑎)Γ(𝑏𝑏) Γ(𝑎𝑎 + 𝑏𝑏) = � 𝑥𝑥𝑎𝑎−1(1 − 𝑥𝑥)𝑏𝑏−1𝑑𝑑𝑑𝑑 1 0 . (13) The Beta (𝑎𝑎, 𝑏𝑏) distribution on [0,1], with 𝑎𝑎 > 0 and 𝑏𝑏 > 0, has pdf and cdf 𝑓𝑓Beta (𝑎𝑎,𝑏𝑏)(𝑥𝑥) = 1 𝐵𝐵(𝑎𝑎, 𝑏𝑏) 𝑥𝑥𝑎𝑎−1(1 − 𝑥𝑥)𝑏𝑏−1 𝐹𝐹Beta (𝑎𝑎,𝑏𝑏)(𝑥𝑥) = � 𝑓𝑓Beta (𝑎𝑎,𝑏𝑏)(𝑢𝑢) 𝑑𝑑𝑑𝑑 𝑥𝑥 0 . (14) The Binomial and Beta cdfs are related by 𝐹𝐹𝑛𝑛,𝑝𝑝 (𝑘𝑘 − 1) = 𝐹𝐹Beta (𝑛𝑛−𝑘𝑘+1,𝑘𝑘)(1 − 𝑝𝑝), so 𝐹𝐹𝑛𝑛,𝑝𝑝 (𝑘𝑘) = 𝐹𝐹Beta (𝑛𝑛−𝑘𝑘,𝑘𝑘+1)(1 − 𝑝𝑝), or 𝐹𝐹𝑛𝑛−𝑘𝑘+1,1−𝑝𝑝(𝑘𝑘 − 1) = 𝐹𝐹Beta (𝑛𝑛,𝑘𝑘)(𝑝𝑝) (15) and also 𝐺𝐺𝑛𝑛,𝑝𝑝 (𝑘𝑘) = 𝐹𝐹Beta (𝑘𝑘,𝑛𝑛−𝑘𝑘+1)(𝑝𝑝), or 𝐺𝐺𝑛𝑛−𝑘𝑘+1,𝑝𝑝(𝑘𝑘) = 𝐹𝐹Beta (𝑛𝑛,𝑘𝑘)(𝑝𝑝). (16) To show 𝐹𝐹𝑛𝑛,𝑝𝑝 (𝑘𝑘 − 1) = 𝐹𝐹Beta (𝑛𝑛−𝑘𝑘+1,𝑘𝑘)(1 − 𝑝𝑝), first note that 𝐵𝐵(𝑛𝑛 − 𝑘𝑘 + 1, 𝑘𝑘) = (𝑛𝑛 − 𝑘𝑘)! (𝑘𝑘 − 1)!/𝑛𝑛!, and also 𝐹𝐹Beta (𝑛𝑛,1) (1 − 𝑝𝑝) = 𝑛𝑛 ∫ 𝑥𝑥𝑛𝑛−1𝑑𝑑𝑑𝑑 = (1 − 𝑝𝑝)𝑛𝑛 = 𝑓𝑓𝑛𝑛,𝑝𝑝 (0) 1 0 . Now, integrating by parts
n! c1-p Beaa-t+11-p）=a-k1k-m,-*1-)-1d a=m=+1+”《-*之 n! xn-k(1-x)k-11-p n-k+1 _n!pk-1(1-p)n-k+1 0k+k-k-2%-k+x-*+41-x水西 n! =k-+an-2Dxa-对s 1 =fn.p(k-1)+FBeta (n-k+2.k-1)(1-p) =fp(k -1)++fup(1)+FBeta (n.1)(1-p) =fn,p(k-1)+…+fn,p(1)+fn,p(0) =Fp(k-1). (17) See,for example,Stuart (1) In fact,the function x≤-1 F(x)= FBeta (n-x+1x)(1-p),-1<x<n (18) 1, n≤x is a continuous cdf on [-1,n],and F(k)=Fp(k)for k =0,1,2,....n.So F serves as a continuous version,or interpolation,of which is sometimes useful.The corresponding quantile function Q(u)=F-1(u)is not related to QBeta,but can be obtained by solving F(x)= u numerically to get x=Q(u).See figure 1. Note that QBeta does provide the solutionp ofu=Fp(k).Since u=Fp(k)=FBeta (n-k,k+1)(1-p),and so QBeta (n-kk+1)(u)=1-p,observe that p =1-QBeta (n-k.k+1)(u). (19) Likewise for u=G(np)(k).Since u=Gnp(k)=FBeta (kn-k+1)(p)and so QBeta (k.n-k+1)(u)= p,it follows that 6
6 𝐹𝐹Beta (𝑛𝑛−𝑘𝑘+1,𝑘𝑘)(1 − 𝑝𝑝) = 𝑛𝑛! (𝑛𝑛 − 𝑘𝑘)! (𝑘𝑘 − 𝑛𝑛)! � 𝑥𝑥𝑛𝑛−𝑘𝑘 (1 − 𝑥𝑥)𝑘𝑘−1𝑑𝑑𝑑𝑑 1−𝑝𝑝 0 = 𝑛𝑛! (𝑛𝑛 − 𝑘𝑘)! (𝑘𝑘 − 𝑛𝑛)! � 𝑥𝑥𝑛𝑛−𝑘𝑘 (1 − 𝑥𝑥)𝑘𝑘−1 𝑛𝑛 − 𝑘𝑘 + 1 � 0 1−𝑝𝑝 + � (𝑘𝑘 − 1)𝑥𝑥𝑛𝑛−𝑘𝑘+1(1 − 𝑥𝑥)𝑘𝑘−2 𝑛𝑛 − 𝑘𝑘 + 1 1−𝑝𝑝 0 � = 𝑛𝑛! 𝑝𝑝𝑘𝑘−1(1 − 𝑝𝑝)𝑛𝑛−𝑘𝑘+1 (𝑛𝑛 − 𝑘𝑘 + 1)! (𝑘𝑘 − 1)! + 𝑛𝑛! (𝑘𝑘 − 2)! (𝑛𝑛 − 𝑘𝑘 + 1)! � 𝑥𝑥𝑛𝑛−𝑘𝑘+1(1 − 𝑥𝑥)𝑘𝑘−2𝑑𝑑𝑑𝑑 1−𝑝𝑝 0 = 𝑓𝑓𝑛𝑛,𝑝𝑝 (𝑘𝑘 − 1) + 1 𝐵𝐵(𝑛𝑛 − 𝑘𝑘 + 2, 𝑘𝑘 − 1) � 𝑥𝑥𝑛𝑛−𝑘𝑘+1(1 − 𝑥𝑥)𝑘𝑘−2𝑑𝑑𝑑𝑑 1−𝑝𝑝 0 = 𝑓𝑓𝑛𝑛,𝑝𝑝 (𝑘𝑘 − 1) + 𝐹𝐹Beta (𝑛𝑛−𝑘𝑘+2,𝑘𝑘−1)(1 − 𝑝𝑝) = 𝑓𝑓𝑛𝑛,𝑝𝑝 (𝑘𝑘 − 1) + ⋯ + 𝑓𝑓𝑛𝑛,𝑝𝑝 (1) + 𝐹𝐹Beta (𝑛𝑛,1) (1 − 𝑝𝑝) = 𝑓𝑓𝑛𝑛,𝑝𝑝 (𝑘𝑘 − 1) + ⋯ + 𝑓𝑓𝑛𝑛,𝑝𝑝 (1) + 𝑓𝑓𝑛𝑛,𝑝𝑝 (0) = 𝐹𝐹𝑛𝑛,𝑝𝑝 (𝑘𝑘 − 1). (17) See, for example, Stuart (1). In fact, the function 𝐹𝐹(𝑥𝑥) = � 0, 𝑥𝑥 ≤ −1 𝐹𝐹Beta (𝑛𝑛−𝑥𝑥+1,𝑥𝑥)(1 − 𝑝𝑝), −1 < 𝑥𝑥 < 𝑛𝑛 1, 𝑛𝑛 ≤ 𝑥𝑥 (18) is a continuous cdf on [−1, 𝑛𝑛], and 𝐹𝐹(𝑘𝑘) = 𝐹𝐹𝑛𝑛,𝑝𝑝(𝑘𝑘) for 𝑘𝑘 = 0,1,2, … , 𝑛𝑛. So F serves as a continuous version, or interpolation, of 𝐹𝐹𝑛𝑛,𝑝𝑝, which is sometimes useful. The corresponding quantile function 𝑄𝑄(𝑢𝑢) = 𝐹𝐹−1(𝑢𝑢) is not related to 𝑄𝑄Beta , but can be obtained by solving 𝐹𝐹(𝑥𝑥) = 𝑢𝑢 numerically to get 𝑥𝑥 = 𝑄𝑄(𝑢𝑢). See figure 1. Note that 𝑄𝑄Beta does provide the solution p of 𝑢𝑢 = 𝐹𝐹𝑛𝑛,𝑝𝑝 (𝑘𝑘). Since 𝑢𝑢 = 𝐹𝐹𝑛𝑛,𝑝𝑝 (𝑘𝑘) = 𝐹𝐹Beta (𝑛𝑛−𝑘𝑘,𝑘𝑘+1)(1 − 𝑝𝑝), and so 𝑄𝑄Beta (𝑛𝑛−𝑘𝑘,𝑘𝑘+1)(𝑢𝑢) = 1 − 𝑝𝑝, observe that 𝑝𝑝 = 1 − 𝑄𝑄Beta (𝑛𝑛−𝑘𝑘,𝑘𝑘+1)(𝑢𝑢). (19) Likewise for 𝑢𝑢 = 𝐺𝐺(𝑛𝑛,𝑝𝑝)(𝑘𝑘). Since 𝑢𝑢 = 𝐺𝐺𝑛𝑛,𝑝𝑝 (𝑘𝑘) = 𝐹𝐹Beta (𝑘𝑘,𝑛𝑛−𝑘𝑘+1)(𝑝𝑝) and so 𝑄𝑄Beta (𝑘𝑘,𝑛𝑛−𝑘𝑘+1)(𝑢𝑢) = 𝑝𝑝, it follows that
p=QBeta(k,-k+1)(u）. (20) Library implementations of QBeta provide for efficient and accurate evaluation of p in these situations. This will be useful later. F(x) Q(u) 1.0 8 0.8 6 0.6 4 0.4 0.2 0.0 0 2 4 8 0.0 0.4 0.8 F(X) Q(u) 1.0 15 0.8 0.6 10 0.4 5 0.2 0.0 5 10 15 0.0 0.4 0.8 Figure 1.Binomial cdf and quantile function(qf)examples with continuous interpolation. 7
7 𝑝𝑝 = 𝑄𝑄Beta (𝑘𝑘,𝑛𝑛−𝑘𝑘+1)(𝑢𝑢). (20) Library implementations of QBeta provide for efficient and accurate evaluation of p in these situations. This will be useful later. Figure 1. Binomial cdf and quantile function (qf) examples with continuous interpolation. 0 2 4 6 8 0.0 0.2 0.4 0.6 0.8 1.0 x F(x) 0.0 0.4 0.8 0 2 4 6 8 u Q(u) 0 5 10 15 0.0 0.2 0.4 0.6 0.8 1.0 x F(x) 0.0 0.4 0.8 0 5 10 15 u Q(u)