结构推理 设平稳过程X(t)的自相关函数为RX(τ),证明:
   P{|X(t+τ)-X(t)|≥a}≤2[RX(0)-RX(τ)]/a2,a>0.
【正确答案】命题似与切比雪夫不等式相关联.因X(t)是平稳过程,故有
   E[X(t)]=μX(常数),
   E[X(t)X(t+τ)]=RX(τ),
   E[X2(t)]=E[X2(t+τ)]=RX(0).
   记Y(t)=X(t+τ)-X(t),即有
   E[Y(t)]=E[X(t+τ)]-E[X(t)]=μXX=0,
   D[Y(t)]=E[Y2(t)]=E{[X(t+τ)-X(t)]2}
   =E[X2(t+τ)]-2E[X(t)X(t+τ)]+E[X2(t)]
   =2[RX(0)-RX(τ)].
   对Y(t)=X(t+τ)-X(t)引用切比雪夫不等式,即有
   P{|X(t+τ)-X(t)|≥a}≤2[RX(0)-RX(τ)]/a2。a>0.
【答案解析】