这是另一种解决方案:我们定义
I(γ)=∫∞−∞Φ(ξx+γ)N(x|0,σ2)dx,
γ=−ξμI(γ)I(0)=0γ
dIdγ=∫∞−∞N((ξx+γ)|0,1)N(x|0,σ2)dx=∫∞−∞12π−−√exp(−12(ξx+γ)2)12πσ2−−−−√exp(−x22σ2)dx.
(ξx+γ)2+x2σ2=(ξ2+σ−2)=ax2+−2γξ=bx+γ2=c=a(x−b2a)2+(c−b24a)(c−b24a)=γ2−4γ2ξ24(ξ2+σ−2)=γ2(1−ξ2ξ2+σ−2)=γ2(11+ξ2σ2)
dIdγ=12πσexp(−12(c−b24a))2πa−−−√∫∞−∞a2π−−−√exp(−12a(x−b2a)2)dx=12πσexp(−12(c−b24a))2πa−−−√=12πσ2a−−−−−√exp(−12(c−b24a))=12π(1+σ2ξ2)−−−−−−−−−−−√exp(−12γ21+ξ2σ2)
I(γ)=∫γ−∞12π(1+σ2ξ2)−−−−−−−−−−−√exp(−12z21+ξ2σ2)dz=Φ(γ1+ξ2σ2−−−−−−−√)
这意味着
∫∞−∞Φ(ξx)N(x|μ,σ2)dx=I(ξμ)=Φ(ξμ1+ξ2σ2−−−−−−−√).